3.133 \(\int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=119 \[ \frac {\left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}}-\frac {b \left (\left (4 a^2+b^2\right ) \cos (c+d x)+3 a b \sin (c+d x)\right )}{2 d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.59, antiderivative size = 225, normalized size of antiderivative = 1.89, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1660, 12, 618, 206} \[ \frac {2 b^2 \left (\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+a b\right )}{a^3 d \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b \left (a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+3 a^2 b^2+4 a^4+2 b^4\right )}{a^3 d \left (a^2+b^2\right )^2 \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {\left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

-(((2*a^2 - b^2)*ArcTanh[(b - a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(5/2)*d)) + (2*b^2*(a*b + (a^
2 + 2*b^2)*Tan[(c + d*x)/2]))/(a^3*(a^2 + b^2)*d*(a + 2*b*Tan[(c + d*x)/2] - a*Tan[(c + d*x)/2]^2)^2) - (b*(4*
a^4 + 3*a^2*b^2 + 2*b^4 + a*b*(5*a^2 + 2*b^2)*Tan[(c + d*x)/2]))/(a^3*(a^2 + b^2)^2*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 ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{\left (a+2 b x-a x^2\right )^3} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d}\\ &=\frac {2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \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 )^2}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {8 \left (a^4+2 b^4\right )}{a^3}+16 b \left (1+\frac {b^2}{a^2}\right ) x+8 \left (a+\frac {b^2}{a}\right ) x^2}{\left (a+2 b x-a x^2\right )^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 \left (a^2+b^2\right ) d}\\ &=\frac {2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \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 )^2}-\frac {b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \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 )}+\frac {\operatorname {Subst}\left (\int \frac {16 \left (2 a^2-b^2\right )}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{16 \left (a^2+b^2\right )^2 d}\\ &=\frac {2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \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 )^2}-\frac {b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \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 )}+\frac {\left (2 a^2-b^2\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 )^2 d}\\ &=\frac {2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \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 )^2}-\frac {b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \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 )}-\frac {\left (2 \left (2 a^2-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 )^2 d}\\ &=-\frac {\left (2 a^2-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 )^{5/2} d}+\frac {2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \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 )^2}-\frac {b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \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 )}\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 119, normalized size = 1.00 \[ \frac {\frac {2 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {b \left (\left (4 a^2+b^2\right ) \cos (c+d x)+3 a b \sin (c+d x)\right )}{\left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

((2*(2*a^2 - b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (b*((4*a^2 + b^2)*Co
s[c + d*x] + 3*a*b*Sin[c + d*x]))/((a^2 + b^2)^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2))/(2*d)

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fricas [B]  time = 0.67, size = 352, normalized size = 2.96 \[ -\frac {{\left (2 \, a^{2} b^{2} - b^{4} + {\left (2 \, a^{4} - 3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \, {\left (4 \, a^{4} b + 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} + 2 \, a^{6} b^{2} - 2 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*((2*a^2*b^2 - b^4 + (2*a^4 - 3*a^2*b^2 + b^4)*cos(d*x + c)^2 + 2*(2*a^3*b - a*b^3)*cos(d*x + c)*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)) + 2*(4*a^4*b + 5*a^2*b^3 + b^5)*cos(d*x + c) + 6*(a^3*b^2 + a*b^4)*sin(d*x + c))/((a^8 + 2*a^6*b^2 - 2*a
^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*cos(d*x + c)*sin(d*x + c) + (a^6*
b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d)

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giac [B]  time = 2.19, size = 293, normalized size = 2.46 \[ -\frac {\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{4} b - a^{2} b^{3}\right )}}{{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*((2*a^2 - 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^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(5*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 2
*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*a^4*b*tan(1/2*d*x + 1/2*c)^2 - 7*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 - 2*b^5*tan(
1/2*d*x + 1/2*c)^2 - 11*a^3*b^2*tan(1/2*d*x + 1/2*c) - 2*a*b^4*tan(1/2*d*x + 1/2*c) - 4*a^4*b - a^2*b^3)/((a^6
+ 2*a^4*b^2 + a^2*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^2))/d

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maple [B]  time = 0.25, size = 280, normalized size = 2.35 \[ \frac {-\frac {2 \left (-\frac {b^{2} \left (5 a^{2}+2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (4 a^{4}-7 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{2}+b^{2}\right )}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}\right )}{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

1/d*(-2*(-1/2*b^2*(5*a^2+2*b^2)/a/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^3-1/2*b*(4*a^4-7*a^2*b^2-2*b^4)/(a^4+
2*a^2*b^2+b^4)/a^2*tan(1/2*d*x+1/2*c)^2+1/2*b^2*(11*a^2+2*b^2)/(a^4+2*a^2*b^2+b^4)/a*tan(1/2*d*x+1/2*c)+1/2*b*
(4*a^2+b^2)/(a^4+2*a^2*b^2+b^4))/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^2+(2*a^2-b^2)/(a^4+2*a^2*b^
2+b^4)/(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 = 0.44, size = 412, normalized size = 3.46 \[ -\frac {\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (4 \, a^{4} b + a^{2} b^{3} + \frac {{\left (11 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {{\left (4 \, a^{4} b - 7 \, a^{2} b^{3} - 2 \, b^{5}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {{\left (5 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4} + \frac {4 \, {\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, {\left (a^{8} - 3 \, a^{4} b^{4} - 2 \, a^{2} b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, {\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {{\left (a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.72, size = 443, normalized size = 3.72 \[ \frac {\ln \left ({\left (a^2+b^2\right )}^{5/2}-a^4\,b-b^5-2\,a^2\,b^3+a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-\frac {b^2}{2}\right )}{d\,{\left (a^2+b^2\right )}^{5/2}}-\frac {\ln \left ({\left (a^2+b^2\right )}^{5/2}+a^4\,b+b^5+2\,a^2\,b^3-a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-b^2\right )}{2\,d\,{\left (a^2+b^2\right )}^{5/2}}-\frac {\frac {4\,a^2\,b+b^3}{a^4+2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2-2\,b^2\right )\,\left (4\,a^2\,b+b^3\right )}{a^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (11\,a^2\,b+2\,b^3\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,a^2\,b+2\,b^3\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-4\,b^2\right )+a^2-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(a*cos(c + d*x) + b*sin(c + d*x))^3,x)

[Out]

(log((a^2 + b^2)^(5/2) - a^4*b - b^5 - 2*a^2*b^3 + a^5*tan(c/2 + (d*x)/2) + a*b^4*tan(c/2 + (d*x)/2) + 2*a^3*b
^2*tan(c/2 + (d*x)/2))*(a^2 - b^2/2))/(d*(a^2 + b^2)^(5/2)) - (log((a^2 + b^2)^(5/2) + a^4*b + b^5 + 2*a^2*b^3
- a^5*tan(c/2 + (d*x)/2) - a*b^4*tan(c/2 + (d*x)/2) - 2*a^3*b^2*tan(c/2 + (d*x)/2))*(2*a^2 - b^2))/(2*d*(a^2
+ b^2)^(5/2)) - ((4*a^2*b + b^3)/(a^4 + b^4 + 2*a^2*b^2) - (tan(c/2 + (d*x)/2)^2*(a^2 - 2*b^2)*(4*a^2*b + b^3)
)/(a^2*(a^4 + b^4 + 2*a^2*b^2)) + (b*tan(c/2 + (d*x)/2)*(11*a^2*b + 2*b^3))/(a*(a^4 + b^4 + 2*a^2*b^2)) - (b*t
an(c/2 + (d*x)/2)^3*(5*a^2*b + 2*b^3))/(a*(a^4 + b^4 + 2*a^2*b^2)))/(d*(a^2*tan(c/2 + (d*x)/2)^4 - tan(c/2 + (
d*x)/2)^2*(2*a^2 - 4*b^2) + a^2 - 4*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b*tan(c/2 + (d*x)/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)**2/(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Timed out

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