3.572 \(\int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=149 \[ \frac {4 \left (8 a^2+b^2\right ) \tan ^{-1}\left (\frac {2 a \tan (c+d x)+b}{\sqrt {4 a^2-b^2}}\right )}{d \left (4 a^2-b^2\right )^{5/2}}+\frac {12 a b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right )^2 (2 a+b \sin (2 c+2 d x))}+\frac {2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))^2} \]

[Out]

4*(8*a^2+b^2)*arctan((b+2*a*tan(d*x+c))/(4*a^2-b^2)^(1/2))/(4*a^2-b^2)^(5/2)/d+2*b*cos(2*d*x+2*c)/(4*a^2-b^2)/
d/(2*a+b*sin(2*d*x+2*c))^2+12*a*b*cos(2*d*x+2*c)/(4*a^2-b^2)^2/d/(2*a+b*sin(2*d*x+2*c))

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Rubi [A]  time = 0.18, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2666, 2664, 2754, 12, 2660, 618, 204} \[ \frac {4 \left (8 a^2+b^2\right ) \tan ^{-1}\left (\frac {2 a \tan (c+d x)+b}{\sqrt {4 a^2-b^2}}\right )}{d \left (4 a^2-b^2\right )^{5/2}}+\frac {12 a b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right )^2 (2 a+b \sin (2 c+2 d x))}+\frac {2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(8*a^2 + b^2)*ArcTan[(b + 2*a*Tan[c + d*x])/Sqrt[4*a^2 - b^2]])/((4*a^2 - b^2)^(5/2)*d) + (2*b*Cos[2*c + 2*
d*x])/((4*a^2 - b^2)*d*(2*a + b*Sin[2*c + 2*d*x])^2) + (12*a*b*Cos[2*c + 2*d*x])/((4*a^2 - b^2)^2*d*(2*a + b*S
in[2*c + 2*d*x]))

Rule 12

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^3} \, dx &=\int \frac {1}{\left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^3} \, dx\\ &=\frac {2 b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^2}-\frac {2 \int \frac {-2 a+\frac {1}{2} b \sin (2 c+2 d x)}{\left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^2} \, dx}{4 a^2-b^2}\\ &=\frac {2 b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^2}+\frac {12 a b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right )^2 d (2 a+b \sin (2 c+2 d x))}+\frac {8 \int \frac {8 a^2+b^2}{4 \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )} \, dx}{\left (4 a^2-b^2\right )^2}\\ &=\frac {2 b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^2}+\frac {12 a b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right )^2 d (2 a+b \sin (2 c+2 d x))}+\frac {\left (2 \left (8 a^2+b^2\right )\right ) \int \frac {1}{a+\frac {1}{2} b \sin (2 c+2 d x)} \, dx}{\left (4 a^2-b^2\right )^2}\\ &=\frac {2 b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^2}+\frac {12 a b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right )^2 d (2 a+b \sin (2 c+2 d x))}+\frac {\left (2 \left (8 a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (2 c+2 d x)\right )\right )}{\left (4 a^2-b^2\right )^2 d}\\ &=\frac {2 b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^2}+\frac {12 a b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right )^2 d (2 a+b \sin (2 c+2 d x))}-\frac {\left (4 \left (8 a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 a^2+b^2-x^2} \, dx,x,b+2 a \tan \left (\frac {1}{2} (2 c+2 d x)\right )\right )}{\left (4 a^2-b^2\right )^2 d}\\ &=\frac {4 \left (8 a^2+b^2\right ) \tan ^{-1}\left (\frac {b+2 a \tan (c+d x)}{\sqrt {4 a^2-b^2}}\right )}{\left (4 a^2-b^2\right )^{5/2} d}+\frac {2 b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^2}+\frac {12 a b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right )^2 d (2 a+b \sin (2 c+2 d x))}\\ \end {align*}

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Mathematica [A]  time = 0.93, size = 120, normalized size = 0.81 \[ \frac {2 \left (\frac {2 \left (8 a^2+b^2\right ) \tan ^{-1}\left (\frac {2 a \tan (c+d x)+b}{\sqrt {4 a^2-b^2}}\right )}{\left (4 a^2-b^2\right )^{5/2}}+\frac {b \cos (2 (c+d x)) \left (16 a^2+6 a b \sin (2 (c+d x))-b^2\right )}{\left (b^2-4 a^2\right )^2 (2 a+b \sin (2 (c+d x)))^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((2*(8*a^2 + b^2)*ArcTan[(b + 2*a*Tan[c + d*x])/Sqrt[4*a^2 - b^2]])/(4*a^2 - b^2)^(5/2) + (b*Cos[2*(c + d*x
)]*(16*a^2 - b^2 + 6*a*b*Sin[2*(c + d*x)]))/((-4*a^2 + b^2)^2*(2*a + b*Sin[2*(c + d*x)])^2)))/d

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fricas [B]  time = 1.06, size = 969, normalized size = 6.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/2*(64*a^4*b - 20*a^2*b^3 + b^5 - 2*(64*a^4*b - 20*a^2*b^3 + b^5)*cos(d*x + c)^2 - 2*((8*a^2*b^2 + b^4)*cos(
d*x + c)^4 - 8*a^4 - a^2*b^2 - (8*a^2*b^2 + b^4)*cos(d*x + c)^2 - 2*(8*a^3*b + a*b^3)*cos(d*x + c)*sin(d*x + c
))*sqrt(-4*a^2 + b^2)*log(-(2*(8*a^2 - b^2)*cos(d*x + c)^4 - 4*a*b*cos(d*x + c)*sin(d*x + c) - 2*(8*a^2 - b^2)
*cos(d*x + c)^2 + 2*a^2 - b^2 + (2*b*cos(d*x + c)^2 + 4*(2*a*cos(d*x + c)^3 - a*cos(d*x + c))*sin(d*x + c) - b
)*sqrt(-4*a^2 + b^2))/(b^2*cos(d*x + c)^4 - b^2*cos(d*x + c)^2 - 2*a*b*cos(d*x + c)*sin(d*x + c) - a^2)) - 12*
(2*(4*a^3*b^2 - a*b^4)*cos(d*x + c)^3 - (4*a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((64*a^6*b^2 - 48*a^4*
b^4 + 12*a^2*b^6 - b^8)*d*cos(d*x + c)^4 - (64*a^6*b^2 - 48*a^4*b^4 + 12*a^2*b^6 - b^8)*d*cos(d*x + c)^2 - 2*(
64*a^7*b - 48*a^5*b^3 + 12*a^3*b^5 - a*b^7)*d*cos(d*x + c)*sin(d*x + c) - (64*a^8 - 48*a^6*b^2 + 12*a^4*b^4 -
a^2*b^6)*d), 1/2*(64*a^4*b - 20*a^2*b^3 + b^5 - 2*(64*a^4*b - 20*a^2*b^3 + b^5)*cos(d*x + c)^2 - 4*((8*a^2*b^2
+ b^4)*cos(d*x + c)^4 - 8*a^4 - a^2*b^2 - (8*a^2*b^2 + b^4)*cos(d*x + c)^2 - 2*(8*a^3*b + a*b^3)*cos(d*x + c)
*sin(d*x + c))*sqrt(4*a^2 - b^2)*arctan(-(4*a*cos(d*x + c)*sin(d*x + c) + b)*sqrt(4*a^2 - b^2)/(2*(4*a^2 - b^2
)*cos(d*x + c)^2 - 4*a^2 + b^2)) - 12*(2*(4*a^3*b^2 - a*b^4)*cos(d*x + c)^3 - (4*a^3*b^2 - a*b^4)*cos(d*x + c)
)*sin(d*x + c))/((64*a^6*b^2 - 48*a^4*b^4 + 12*a^2*b^6 - b^8)*d*cos(d*x + c)^4 - (64*a^6*b^2 - 48*a^4*b^4 + 12
*a^2*b^6 - b^8)*d*cos(d*x + c)^2 - 2*(64*a^7*b - 48*a^5*b^3 + 12*a^3*b^5 - a*b^7)*d*cos(d*x + c)*sin(d*x + c)
- (64*a^8 - 48*a^6*b^2 + 12*a^4*b^4 - a^2*b^6)*d)]

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giac [A]  time = 0.19, size = 252, normalized size = 1.69 \[ \frac {\frac {8 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {2 \, a \tan \left (d x + c\right ) + b}{\sqrt {4 \, a^{2} - b^{2}}}\right )\right )} {\left (8 \, a^{2} + b^{2}\right )}}{{\left (16 \, a^{4} - 8 \, a^{2} b^{2} + b^{4}\right )} \sqrt {4 \, a^{2} - b^{2}}} + \frac {20 \, a^{3} b^{2} \tan \left (d x + c\right )^{3} - 2 \, a b^{4} \tan \left (d x + c\right )^{3} + 32 \, a^{4} b \tan \left (d x + c\right )^{2} + 14 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} - b^{5} \tan \left (d x + c\right )^{2} + 44 \, a^{3} b^{2} \tan \left (d x + c\right ) - 2 \, a b^{4} \tan \left (d x + c\right ) + 32 \, a^{4} b - 2 \, a^{2} b^{3}}{{\left (16 \, a^{6} - 8 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(8*(pi*floor((d*x + c)/pi + 1/2)*sgn(a) + arctan((2*a*tan(d*x + c) + b)/sqrt(4*a^2 - b^2)))*(8*a^2 + b^2)/
((16*a^4 - 8*a^2*b^2 + b^4)*sqrt(4*a^2 - b^2)) + (20*a^3*b^2*tan(d*x + c)^3 - 2*a*b^4*tan(d*x + c)^3 + 32*a^4*
b*tan(d*x + c)^2 + 14*a^2*b^3*tan(d*x + c)^2 - b^5*tan(d*x + c)^2 + 44*a^3*b^2*tan(d*x + c) - 2*a*b^4*tan(d*x
+ c) + 32*a^4*b - 2*a^2*b^3)/((16*a^6 - 8*a^4*b^2 + a^2*b^4)*(a*tan(d*x + c)^2 + b*tan(d*x + c) + a)^2))/d

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maple [B]  time = 0.49, size = 640, normalized size = 4.30 \[ \frac {10 b^{2} a \left (\tan ^{3}\left (d x +c \right )\right )}{d \left (\left (\tan ^{2}\left (d x +c \right )\right ) a +b \tan \left (d x +c \right )+a \right )^{2} \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right )}-\frac {b^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{d \left (\left (\tan ^{2}\left (d x +c \right )\right ) a +b \tan \left (d x +c \right )+a \right )^{2} \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right ) a}+\frac {16 b \,a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{d \left (\left (\tan ^{2}\left (d x +c \right )\right ) a +b \tan \left (d x +c \right )+a \right )^{2} \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right )}+\frac {7 b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{d \left (\left (\tan ^{2}\left (d x +c \right )\right ) a +b \tan \left (d x +c \right )+a \right )^{2} \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right )}-\frac {b^{5} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d \left (\left (\tan ^{2}\left (d x +c \right )\right ) a +b \tan \left (d x +c \right )+a \right )^{2} \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {22 b^{2} a \tan \left (d x +c \right )}{d \left (\left (\tan ^{2}\left (d x +c \right )\right ) a +b \tan \left (d x +c \right )+a \right )^{2} \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right )}-\frac {b^{4} \tan \left (d x +c \right )}{d \left (\left (\tan ^{2}\left (d x +c \right )\right ) a +b \tan \left (d x +c \right )+a \right )^{2} a \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right )}+\frac {16 b \,a^{2}}{d \left (\left (\tan ^{2}\left (d x +c \right )\right ) a +b \tan \left (d x +c \right )+a \right )^{2} \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right )}-\frac {b^{3}}{d \left (\left (\tan ^{2}\left (d x +c \right )\right ) a +b \tan \left (d x +c \right )+a \right )^{2} \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right )}+\frac {32 \arctan \left (\frac {b +2 a \tan \left (d x +c \right )}{\sqrt {4 a^{2}-b^{2}}}\right ) a^{2}}{d \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right ) \sqrt {4 a^{2}-b^{2}}}+\frac {4 \arctan \left (\frac {b +2 a \tan \left (d x +c \right )}{\sqrt {4 a^{2}-b^{2}}}\right ) b^{2}}{d \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right ) \sqrt {4 a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(d*x+c)*sin(d*x+c))^3,x)

[Out]

10/d/(tan(d*x+c)^2*a+b*tan(d*x+c)+a)^2*b^2/(16*a^4-8*a^2*b^2+b^4)*a*tan(d*x+c)^3-1/d/(tan(d*x+c)^2*a+b*tan(d*x
+c)+a)^2*b^4/(16*a^4-8*a^2*b^2+b^4)/a*tan(d*x+c)^3+16/d/(tan(d*x+c)^2*a+b*tan(d*x+c)+a)^2*b/(16*a^4-8*a^2*b^2+
b^4)*a^2*tan(d*x+c)^2+7/d/(tan(d*x+c)^2*a+b*tan(d*x+c)+a)^2*b^3/(16*a^4-8*a^2*b^2+b^4)*tan(d*x+c)^2-1/2/d/(tan
(d*x+c)^2*a+b*tan(d*x+c)+a)^2*b^5/(16*a^4-8*a^2*b^2+b^4)/a^2*tan(d*x+c)^2+22/d/(tan(d*x+c)^2*a+b*tan(d*x+c)+a)
^2*b^2*a/(16*a^4-8*a^2*b^2+b^4)*tan(d*x+c)-1/d/(tan(d*x+c)^2*a+b*tan(d*x+c)+a)^2*b^4/a/(16*a^4-8*a^2*b^2+b^4)*
tan(d*x+c)+16/d/(tan(d*x+c)^2*a+b*tan(d*x+c)+a)^2*b/(16*a^4-8*a^2*b^2+b^4)*a^2-1/d/(tan(d*x+c)^2*a+b*tan(d*x+c
)+a)^2*b^3/(16*a^4-8*a^2*b^2+b^4)+32/d/(16*a^4-8*a^2*b^2+b^4)/(4*a^2-b^2)^(1/2)*arctan((b+2*a*tan(d*x+c))/(4*a
^2-b^2)^(1/2))*a^2+4/d/(16*a^4-8*a^2*b^2+b^4)/(4*a^2-b^2)^(1/2)*arctan((b+2*a*tan(d*x+c))/(4*a^2-b^2)^(1/2))*b
^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b^2-4*a^2>0)', see `assume?` f
or more details)Is b^2-4*a^2 positive or negative?

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mupad [B]  time = 3.87, size = 396, normalized size = 2.66 \[ \frac {\frac {16\,a^2\,b-b^3}{16\,a^4-8\,a^2\,b^2+b^4}+\frac {b\,\mathrm {tan}\left (c+d\,x\right )\,\left (22\,a^2\,b-b^3\right )}{a\,\left (16\,a^4-8\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (16\,a^2\,b-b^3\right )\,\left (2\,a^2+b^2\right )}{2\,a^2\,\left (16\,a^4-8\,a^2\,b^2+b^4\right )}+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (10\,a^2\,b-b^3\right )}{a\,\left (16\,a^4-8\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a^2+b^2\right )+a^2+a^2\,{\mathrm {tan}\left (c+d\,x\right )}^4+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )}+\frac {4\,\mathrm {atan}\left (\frac {\left (\frac {4\,a\,\mathrm {tan}\left (c+d\,x\right )\,\left (8\,a^2+b^2\right )}{{\left (2\,a+b\right )}^{5/2}\,{\left (2\,a-b\right )}^{5/2}}+\frac {2\,\left (8\,a^2+b^2\right )\,\left (16\,a^4\,b-8\,a^2\,b^3+b^5\right )}{{\left (2\,a+b\right )}^{5/2}\,{\left (2\,a-b\right )}^{5/2}\,\left (16\,a^4-8\,a^2\,b^2+b^4\right )}\right )\,\left (16\,a^4-8\,a^2\,b^2+b^4\right )}{16\,a^2+2\,b^2}\right )\,\left (8\,a^2+b^2\right )}{d\,{\left (2\,a+b\right )}^{5/2}\,{\left (2\,a-b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(c + d*x)*sin(c + d*x))^3,x)

[Out]

((16*a^2*b - b^3)/(16*a^4 + b^4 - 8*a^2*b^2) + (b*tan(c + d*x)*(22*a^2*b - b^3))/(a*(16*a^4 + b^4 - 8*a^2*b^2)
) + (tan(c + d*x)^2*(16*a^2*b - b^3)*(2*a^2 + b^2))/(2*a^2*(16*a^4 + b^4 - 8*a^2*b^2)) + (b*tan(c + d*x)^3*(10
*a^2*b - b^3))/(a*(16*a^4 + b^4 - 8*a^2*b^2)))/(d*(tan(c + d*x)^2*(2*a^2 + b^2) + a^2 + a^2*tan(c + d*x)^4 + 2
*a*b*tan(c + d*x) + 2*a*b*tan(c + d*x)^3)) + (4*atan((((4*a*tan(c + d*x)*(8*a^2 + b^2))/((2*a + b)^(5/2)*(2*a
- b)^(5/2)) + (2*(8*a^2 + b^2)*(16*a^4*b + b^5 - 8*a^2*b^3))/((2*a + b)^(5/2)*(2*a - b)^(5/2)*(16*a^4 + b^4 -
8*a^2*b^2)))*(16*a^4 + b^4 - 8*a^2*b^2))/(16*a^2 + 2*b^2))*(8*a^2 + b^2))/(d*(2*a + b)^(5/2)*(2*a - b)^(5/2))

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

[Out]

Timed out

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