3.142 \(\int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=157 \[ \frac {a \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{7/2}}+\frac {\frac {1}{2} b \left (b^2-9 a^2\right ) \left (2 \left (a^2+b^2\right )+3 a b \sin (2 (c+d x))\right )-3 \left (3 a^4 b-a^2 b^3+b^5\right ) \cos (2 (c+d x))}{6 d \left (a^2+b^2\right )^3 (a \cos (c+d x)+b \sin (c+d x))^3} \]

[Out]

a*(2*a^2-3*b^2)*arctanh((-b+a*tan(1/2*d*x+1/2*c))/(a^2+b^2)^(1/2))/(a^2+b^2)^(7/2)/d+1/6*(-3*(3*a^4*b-a^2*b^3+
b^5)*cos(2*d*x+2*c)+1/2*b*(-9*a^2+b^2)*(2*a^2+2*b^2+3*a*b*sin(2*d*x+2*c)))/(a^2+b^2)^3/d/(a*cos(d*x+c)+b*sin(d
*x+c))^3

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Rubi [B]  time = 1.17, antiderivative size = 362, normalized size of antiderivative = 2.31, number of steps used = 7, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1660, 12, 618, 206} \[ -\frac {8 b^3 \left (b \left (3 a^2+4 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+a \left (a^2+2 b^2\right )\right )}{3 a^5 d \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 b^2 \left (a \left (30 a^2 b^2+9 a^4+16 b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right )+b \left (18 a^2 b^2+15 a^4+8 b^4\right )\right )}{3 a^5 d \left (a^2+b^2\right )^2 \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b \left (a b \left (6 a^2 b^2+9 a^4+2 b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right )+9 a^4 b^2+12 a^2 b^4+6 a^6+4 b^6\right )}{a^4 d \left (a^2+b^2\right )^3 \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {a \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

-((a*(2*a^2 - 3*b^2)*ArcTanh[(b - a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(7/2)*d)) - (8*b^3*(a*(a^
2 + 2*b^2) + b*(3*a^2 + 4*b^2)*Tan[(c + d*x)/2]))/(3*a^5*(a^2 + b^2)*d*(a + 2*b*Tan[(c + d*x)/2] - a*Tan[(c +
d*x)/2]^2)^3) + (2*b^2*(b*(15*a^4 + 18*a^2*b^2 + 8*b^4) + a*(9*a^4 + 30*a^2*b^2 + 16*b^4)*Tan[(c + d*x)/2]))/(
3*a^5*(a^2 + b^2)^2*d*(a + 2*b*Tan[(c + d*x)/2] - a*Tan[(c + d*x)/2]^2)^2) - (b*(6*a^6 + 9*a^4*b^2 + 12*a^2*b^
4 + 4*b^6 + a*b*(9*a^4 + 6*a^2*b^2 + 2*b^4)*Tan[(c + d*x)/2]))/(a^4*(a^2 + b^2)^3*d*(a + 2*b*Tan[(c + d*x)/2]
- a*Tan[(c + d*x)/2]^2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^3}{\left (a+2 b x-a x^2\right )^4} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d}\\ &=-\frac {8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {4 \left (3 a^6+3 a^4 b^2-12 a^2 b^4-32 b^6\right )}{a^5}+24 b \left (1-\frac {3 b^2}{a^2}-\frac {4 b^4}{a^4}\right ) x+24 \left (a-\frac {b^2}{a}-\frac {2 b^4}{a^3}\right ) x^2-24 b \left (1+\frac {b^2}{a^2}\right ) x^3-12 \left (a+\frac {b^2}{a}\right ) x^4}{\left (a+2 b x-a x^2\right )^3} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{6 \left (a^2+b^2\right ) d}\\ &=-\frac {8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 b^2 \left (b \left (15 a^4+18 a^2 b^2+8 b^4\right )+a \left (9 a^4+30 a^2 b^2+16 b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {\frac {96 \left (a^6-a^4 b^2+7 a^2 b^4+4 b^6\right )}{a^4}-\frac {384 b \left (a^2+b^2\right )^2 x}{a^3}-\frac {96 \left (a^2+b^2\right )^2 x^2}{a^2}}{\left (a+2 b x-a x^2\right )^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{48 \left (a^2+b^2\right )^2 d}\\ &=-\frac {8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 b^2 \left (b \left (15 a^4+18 a^2 b^2+8 b^4\right )+a \left (9 a^4+30 a^2 b^2+16 b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b \left (6 a^6+9 a^4 b^2+12 a^2 b^4+4 b^6+a b \left (9 a^4+6 a^2 b^2+2 b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 \left (a^2+b^2\right )^3 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}-\frac {\operatorname {Subst}\left (\int -\frac {192 a \left (2 a^2-3 b^2\right )}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{192 \left (a^2+b^2\right )^3 d}\\ &=-\frac {8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 b^2 \left (b \left (15 a^4+18 a^2 b^2+8 b^4\right )+a \left (9 a^4+30 a^2 b^2+16 b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b \left (6 a^6+9 a^4 b^2+12 a^2 b^4+4 b^6+a b \left (9 a^4+6 a^2 b^2+2 b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 \left (a^2+b^2\right )^3 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\left (a \left (2 a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^3 d}\\ &=-\frac {8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 b^2 \left (b \left (15 a^4+18 a^2 b^2+8 b^4\right )+a \left (9 a^4+30 a^2 b^2+16 b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b \left (6 a^6+9 a^4 b^2+12 a^2 b^4+4 b^6+a b \left (9 a^4+6 a^2 b^2+2 b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 \left (a^2+b^2\right )^3 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}-\frac {\left (2 a \left (2 a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^3 d}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2} d}-\frac {8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 b^2 \left (b \left (15 a^4+18 a^2 b^2+8 b^4\right )+a \left (9 a^4+30 a^2 b^2+16 b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b \left (6 a^6+9 a^4 b^2+12 a^2 b^4+4 b^6+a b \left (9 a^4+6 a^2 b^2+2 b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 \left (a^2+b^2\right )^3 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}\\ \end {align*}

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Mathematica [C]  time = 1.12, size = 165, normalized size = 1.05 \[ \frac {\frac {6 a \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}+\frac {\frac {1}{2} b \left (b^2-9 a^2\right ) \left (2 \left (a^2+b^2\right )+3 a b \sin (2 (c+d x))\right )-3 \left (3 a^4 b-a^2 b^3+b^5\right ) \cos (2 (c+d x))}{(a-i b)^3 (a+i b)^3 (a \cos (c+d x)+b \sin (c+d x))^3}}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

((6*a*(2*a^2 - 3*b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(7/2) + (-3*(3*a^4*b - a
^2*b^3 + b^5)*Cos[2*(c + d*x)] + (b*(-9*a^2 + b^2)*(2*(a^2 + b^2) + 3*a*b*Sin[2*(c + d*x)]))/2)/((a - I*b)^3*(
a + I*b)^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3))/(6*d)

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fricas [B]  time = 0.76, size = 524, normalized size = 3.34 \[ -\frac {22 \, a^{4} b^{3} + 14 \, a^{2} b^{5} - 8 \, b^{7} + 12 \, {\left (3 \, a^{6} b + 2 \, a^{4} b^{3} + b^{7}\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (9 \, a^{5} b^{2} + 8 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, {\left ({\left (2 \, a^{6} - 9 \, a^{4} b^{2} + 9 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, a^{4} b^{2} - 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right ) + {\left (2 \, a^{3} b^{3} - 3 \, a b^{5} + {\left (6 \, a^{5} b - 11 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right )}{12 \, {\left ({\left (a^{11} + a^{9} b^{2} - 6 \, a^{7} b^{4} - 14 \, a^{5} b^{6} - 11 \, a^{3} b^{8} - 3 \, a b^{10}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{9} b^{2} + 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} + 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) + {\left ({\left (3 \, a^{10} b + 11 \, a^{8} b^{3} + 14 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - a^{2} b^{9} - b^{11}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} b^{3} + 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} + 4 \, a^{2} b^{9} + b^{11}\right )} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/12*(22*a^4*b^3 + 14*a^2*b^5 - 8*b^7 + 12*(3*a^6*b + 2*a^4*b^3 + b^7)*cos(d*x + c)^2 + 6*(9*a^5*b^2 + 8*a^3*
b^4 - a*b^6)*cos(d*x + c)*sin(d*x + c) + 3*((2*a^6 - 9*a^4*b^2 + 9*a^2*b^4)*cos(d*x + c)^3 + 3*(2*a^4*b^2 - 3*
a^2*b^4)*cos(d*x + c) + (2*a^3*b^3 - 3*a*b^5 + (6*a^5*b - 11*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*
sqrt(a^2 + b^2)*log((2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*sqrt(a^2 +
b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2))
)/((a^11 + a^9*b^2 - 6*a^7*b^4 - 14*a^5*b^6 - 11*a^3*b^8 - 3*a*b^10)*d*cos(d*x + c)^3 + 3*(a^9*b^2 + 4*a^7*b^4
+ 6*a^5*b^6 + 4*a^3*b^8 + a*b^10)*d*cos(d*x + c) + ((3*a^10*b + 11*a^8*b^3 + 14*a^6*b^5 + 6*a^4*b^7 - a^2*b^9
- b^11)*d*cos(d*x + c)^2 + (a^8*b^3 + 4*a^6*b^5 + 6*a^4*b^7 + 4*a^2*b^9 + b^11)*d)*sin(d*x + c))

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giac [B]  time = 0.44, size = 524, normalized size = 3.34 \[ -\frac {\frac {3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (27 \, a^{6} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a^{4} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a^{7} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 81 \, a^{5} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 36 \, a^{3} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 108 \, a^{6} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 42 \, a^{4} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a^{7} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, a^{5} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 18 \, a^{3} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 81 \, a^{6} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{4} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a^{7} b + 5 \, a^{5} b^{3} + 2 \, a^{3} b^{5}\right )}}{{\left (a^{9} + 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + a^{3} b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

-1/6*(3*(2*a^3 - 3*a*b^2)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/
2*c) - 2*b + 2*sqrt(a^2 + b^2)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sqrt(a^2 + b^2)) - 2*(27*a^6*b^2*tan(1/2
*d*x + 1/2*c)^5 + 18*a^4*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*a^2*b^6*tan(1/2*d*x + 1/2*c)^5 + 18*a^7*b*tan(1/2*d*x
+ 1/2*c)^4 - 81*a^5*b^3*tan(1/2*d*x + 1/2*c)^4 - 36*a^3*b^5*tan(1/2*d*x + 1/2*c)^4 - 12*a*b^7*tan(1/2*d*x + 1/
2*c)^4 - 108*a^6*b^2*tan(1/2*d*x + 1/2*c)^3 + 42*a^4*b^4*tan(1/2*d*x + 1/2*c)^3 + 8*a^2*b^6*tan(1/2*d*x + 1/2*
c)^3 + 8*b^8*tan(1/2*d*x + 1/2*c)^3 - 36*a^7*b*tan(1/2*d*x + 1/2*c)^2 + 120*a^5*b^3*tan(1/2*d*x + 1/2*c)^2 + 1
8*a^3*b^5*tan(1/2*d*x + 1/2*c)^2 + 12*a*b^7*tan(1/2*d*x + 1/2*c)^2 + 81*a^6*b^2*tan(1/2*d*x + 1/2*c) + 12*a^4*
b^4*tan(1/2*d*x + 1/2*c) + 6*a^2*b^6*tan(1/2*d*x + 1/2*c) + 18*a^7*b + 5*a^5*b^3 + 2*a^3*b^5)/((a^9 + 3*a^7*b^
2 + 3*a^5*b^4 + a^3*b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^3))/d

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maple [B]  time = 0.31, size = 494, normalized size = 3.15 \[ \frac {-\frac {2 \left (-\frac {b^{2} \left (9 a^{4}+6 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}-\frac {b \left (6 a^{6}-27 a^{4} b^{2}-12 a^{2} b^{4}-4 b^{6}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2} \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}+\frac {b^{2} \left (54 a^{6}-21 a^{4} b^{2}-4 a^{2} b^{4}-4 b^{6}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a^{3} \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}+\frac {b \left (6 a^{6}-20 a^{4} b^{2}-3 a^{2} b^{4}-2 b^{6}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}-\frac {b^{2} \left (27 a^{4}+4 a^{2} b^{2}+2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}-\frac {b \left (18 a^{4}+5 a^{2} b^{2}+2 b^{4}\right )}{6 \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}\right )}{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{3}}+\frac {a \left (2 a^{2}-3 b^{2}\right ) \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) \sqrt {a^{2}+b^{2}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

1/d*(-2*(-1/2*b^2*(9*a^4+6*a^2*b^2+2*b^4)/a/(a^6+3*a^4*b^2+3*a^2*b^4+b^6)*tan(1/2*d*x+1/2*c)^5-1/2*b*(6*a^6-27
*a^4*b^2-12*a^2*b^4-4*b^6)/a^2/(a^6+3*a^4*b^2+3*a^2*b^4+b^6)*tan(1/2*d*x+1/2*c)^4+1/3/a^3*b^2*(54*a^6-21*a^4*b
^2-4*a^2*b^4-4*b^6)/(a^6+3*a^4*b^2+3*a^2*b^4+b^6)*tan(1/2*d*x+1/2*c)^3+1/a^2*b*(6*a^6-20*a^4*b^2-3*a^2*b^4-2*b
^6)/(a^6+3*a^4*b^2+3*a^2*b^4+b^6)*tan(1/2*d*x+1/2*c)^2-1/2/a*b^2*(27*a^4+4*a^2*b^2+2*b^4)/(a^6+3*a^4*b^2+3*a^2
*b^4+b^6)*tan(1/2*d*x+1/2*c)-1/6*b*(18*a^4+5*a^2*b^2+2*b^4)/(a^6+3*a^4*b^2+3*a^2*b^4+b^6))/(tan(1/2*d*x+1/2*c)
^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3+a*(2*a^2-3*b^2)/(a^6+3*a^4*b^2+3*a^2*b^4+b^6)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*
a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)))

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maxima [B]  time = 0.45, size = 724, normalized size = 4.61 \[ -\frac {\frac {3 \, {\left (2 \, a^{2} - 3 \, b^{2}\right )} a \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (18 \, a^{7} b + 5 \, a^{5} b^{3} + 2 \, a^{3} b^{5} + \frac {3 \, {\left (27 \, a^{6} b^{2} + 4 \, a^{4} b^{4} + 2 \, a^{2} b^{6}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {6 \, {\left (6 \, a^{7} b - 20 \, a^{5} b^{3} - 3 \, a^{3} b^{5} - 2 \, a b^{7}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, {\left (54 \, a^{6} b^{2} - 21 \, a^{4} b^{4} - 4 \, a^{2} b^{6} - 4 \, b^{8}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, {\left (6 \, a^{7} b - 27 \, a^{5} b^{3} - 12 \, a^{3} b^{5} - 4 \, a b^{7}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {3 \, {\left (9 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 2 \, a^{2} b^{6}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{12} + 3 \, a^{10} b^{2} + 3 \, a^{8} b^{4} + a^{6} b^{6} + \frac {6 \, {\left (a^{11} b + 3 \, a^{9} b^{3} + 3 \, a^{7} b^{5} + a^{5} b^{7}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, {\left (a^{12} - a^{10} b^{2} - 9 \, a^{8} b^{4} - 11 \, a^{6} b^{6} - 4 \, a^{4} b^{8}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, {\left (3 \, a^{11} b + 7 \, a^{9} b^{3} + 3 \, a^{7} b^{5} - 3 \, a^{5} b^{7} - 2 \, a^{3} b^{9}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, {\left (a^{12} - a^{10} b^{2} - 9 \, a^{8} b^{4} - 11 \, a^{6} b^{6} - 4 \, a^{4} b^{8}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, {\left (a^{11} b + 3 \, a^{9} b^{3} + 3 \, a^{7} b^{5} + a^{5} b^{7}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {{\left (a^{12} + 3 \, a^{10} b^{2} + 3 \, a^{8} b^{4} + a^{6} b^{6}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/6*(3*(2*a^2 - 3*b^2)*a*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c)/(c
os(d*x + c) + 1) - sqrt(a^2 + b^2)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sqrt(a^2 + b^2)) + 2*(18*a^7*b + 5*a
^5*b^3 + 2*a^3*b^5 + 3*(27*a^6*b^2 + 4*a^4*b^4 + 2*a^2*b^6)*sin(d*x + c)/(cos(d*x + c) + 1) - 6*(6*a^7*b - 20*
a^5*b^3 - 3*a^3*b^5 - 2*a*b^7)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*(54*a^6*b^2 - 21*a^4*b^4 - 4*a^2*b^6 -
4*b^8)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*(6*a^7*b - 27*a^5*b^3 - 12*a^3*b^5 - 4*a*b^7)*sin(d*x + c)^4/(c
os(d*x + c) + 1)^4 + 3*(9*a^6*b^2 + 6*a^4*b^4 + 2*a^2*b^6)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^12 + 3*a^10
*b^2 + 3*a^8*b^4 + a^6*b^6 + 6*(a^11*b + 3*a^9*b^3 + 3*a^7*b^5 + a^5*b^7)*sin(d*x + c)/(cos(d*x + c) + 1) - 3*
(a^12 - a^10*b^2 - 9*a^8*b^4 - 11*a^6*b^6 - 4*a^4*b^8)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4*(3*a^11*b + 7*a
^9*b^3 + 3*a^7*b^5 - 3*a^5*b^7 - 2*a^3*b^9)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*(a^12 - a^10*b^2 - 9*a^8*b
^4 - 11*a^6*b^6 - 4*a^4*b^8)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 6*(a^11*b + 3*a^9*b^3 + 3*a^7*b^5 + a^5*b^7
)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - (a^12 + 3*a^10*b^2 + 3*a^8*b^4 + a^6*b^6)*sin(d*x + c)^6/(cos(d*x + c)
+ 1)^6))/d

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mupad [B]  time = 2.69, size = 764, normalized size = 4.87 \[ \frac {\ln \left ({\left (a^2+b^2\right )}^{7/2}+a^6\,b+b^7+3\,a^2\,b^5+3\,a^4\,b^3-a^7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,b^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-3\,a^3\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-3\,a^5\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a\,b^2}{2}-a^3\right )}{d\,{\left (a^2+b^2\right )}^{7/2}}-\frac {\frac {18\,a^4\,b+5\,a^2\,b^3+2\,b^5}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-6\,a^6\,b+20\,a^4\,b^3+3\,a^2\,b^5+2\,b^7\right )}{a^2\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-6\,a^6\,b+27\,a^4\,b^3+12\,a^2\,b^5+4\,b^7\right )}{a^2\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (27\,a^4\,b+4\,a^2\,b^3+2\,b^5\right )}{a\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (9\,a^4\,b+6\,a^2\,b^3+2\,b^5\right )}{a\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}-\frac {2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2-2\,b^2\right )\,\left (18\,a^4\,b+5\,a^2\,b^3+2\,b^5\right )}{3\,a^3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (12\,a\,b^2-3\,a^3\right )-a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,a\,b^2-3\,a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (12\,a^2\,b-8\,b^3\right )+a^3+6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {a\,\ln \left ({\left (a^2+b^2\right )}^{7/2}-a^6\,b-b^7-3\,a^2\,b^5-3\,a^4\,b^3+a^7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\,b^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+3\,a^3\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+3\,a^5\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-3\,b^2\right )}{2\,d\,{\left (a^2+b^2\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a*cos(c + d*x) + b*sin(c + d*x))^4,x)

[Out]

(log((a^2 + b^2)^(7/2) + a^6*b + b^7 + 3*a^2*b^5 + 3*a^4*b^3 - a^7*tan(c/2 + (d*x)/2) - a*b^6*tan(c/2 + (d*x)/
2) - 3*a^3*b^4*tan(c/2 + (d*x)/2) - 3*a^5*b^2*tan(c/2 + (d*x)/2))*((3*a*b^2)/2 - a^3))/(d*(a^2 + b^2)^(7/2)) -
((18*a^4*b + 2*b^5 + 5*a^2*b^3)/(3*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^2*(2*b^7 - 6*
a^6*b + 3*a^2*b^5 + 20*a^4*b^3))/(a^2*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) - (tan(c/2 + (d*x)/2)^4*(4*b^7 - 6*
a^6*b + 12*a^2*b^5 + 27*a^4*b^3))/(a^2*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (b*tan(c/2 + (d*x)/2)*(27*a^4*b
+ 2*b^5 + 4*a^2*b^3))/(a*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (b*tan(c/2 + (d*x)/2)^5*(9*a^4*b + 2*b^5 + 6*a
^2*b^3))/(a*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) - (2*b*tan(c/2 + (d*x)/2)^3*(3*a^2 - 2*b^2)*(18*a^4*b + 2*b^5
+ 5*a^2*b^3))/(3*a^3*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))/(d*(tan(c/2 + (d*x)/2)^2*(12*a*b^2 - 3*a^3) - a^3*
tan(c/2 + (d*x)/2)^6 - tan(c/2 + (d*x)/2)^4*(12*a*b^2 - 3*a^3) - tan(c/2 + (d*x)/2)^3*(12*a^2*b - 8*b^3) + a^3
+ 6*a^2*b*tan(c/2 + (d*x)/2) + 6*a^2*b*tan(c/2 + (d*x)/2)^5)) + (a*log((a^2 + b^2)^(7/2) - a^6*b - b^7 - 3*a^
2*b^5 - 3*a^4*b^3 + a^7*tan(c/2 + (d*x)/2) + a*b^6*tan(c/2 + (d*x)/2) + 3*a^3*b^4*tan(c/2 + (d*x)/2) + 3*a^5*b
^2*tan(c/2 + (d*x)/2))*(2*a^2 - 3*b^2))/(2*d*(a^2 + b^2)^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

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