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

[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) + (-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)/(6*(a^2 + b^2
)^3*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)

________________________________________________________________________________________

Rubi [B]  time = 1.17309, 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 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
]

Rule 12

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

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 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])

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*}

Mathematica [C]  time = 1.05464, size = 165, normalized size = 1.05 \[ \frac{\frac{6 a \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}+\frac{\frac{1}{2} b \left (b^2-9 a^2\right ) \left (2 \left (a^2+b^2\right )+3 a b \sin (2 (c+d x))\right )-3 \left (-a^2 b^3+3 a^4 b+b^5\right ) \cos (2 (c+d x))}{(a-i b)^3 (a+i b)^3 (a \cos (c+d x)+b \sin (c+d x))^3}}{6 d} \]

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)

________________________________________________________________________________________

Maple [B]  time = 0.234, size = 494, normalized size = 3.2 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) ^{3}} \left ( -1/2\,{\frac{{b}^{2} \left ( 9\,{a}^{4}+6\,{a}^{2}{b}^{2}+2\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{a \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) }}-1/2\,{\frac{b \left ( 6\,{a}^{6}-27\,{a}^{4}{b}^{2}-12\,{a}^{2}{b}^{4}-4\,{b}^{6} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{{a}^{2} \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) }}+1/3\,{\frac{{b}^{2} \left ( 54\,{a}^{6}-21\,{a}^{4}{b}^{2}-4\,{a}^{2}{b}^{4}-4\,{b}^{6} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ){a}^{3}}}+{\frac{b \left ( 6\,{a}^{6}-20\,{a}^{4}{b}^{2}-3\,{a}^{2}{b}^{4}-2\,{b}^{6} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{{a}^{2} \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) }}-1/2\,{\frac{{b}^{2} \left ( 27\,{a}^{4}+4\,{a}^{2}{b}^{2}+2\,{b}^{4} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{a \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) }}-1/6\,{\frac{b \left ( 18\,{a}^{4}+5\,{a}^{2}{b}^{2}+2\,{b}^{4} \right ) }{{a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6}}} \right ) }+{\frac{a \left ( 2\,{a}^{2}-3\,{b}^{2} \right ) }{{a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6}}{\it Artanh} \left ({\frac{1}{2} \left ( 2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ) } \end{align*}

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*tan(1/2*d*x+1/2*c)*b-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)))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 0.595758, size = 1173, normalized size = 7.47 \begin{align*} -\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 )}} \end{align*}

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))

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

________________________________________________________________________________________

Giac [B]  time = 1.27717, size = 707, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

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