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

Optimal. Leaf size=167 \[ -\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} \sqrt [3]{b} d}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b} d}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \sqrt [3]{b} d}-\frac {\log \left (a+b \sin ^3(c+d x)\right )}{3 b d} \]

[Out]

1/3*ln(a^(1/3)+b^(1/3)*sin(d*x+c))/a^(2/3)/b^(1/3)/d-1/6*ln(a^(2/3)-a^(1/3)*b^(1/3)*sin(d*x+c)+b^(2/3)*sin(d*x
+c)^2)/a^(2/3)/b^(1/3)/d-1/3*ln(a+b*sin(d*x+c)^3)/b/d-1/3*arctan(1/3*(a^(1/3)-2*b^(1/3)*sin(d*x+c))/a^(1/3)*3^
(1/2))/a^(2/3)/b^(1/3)/d*3^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3223, 1871, 200, 31, 634, 617, 204, 628, 260} \[ -\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} \sqrt [3]{b} d}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b} d}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \sqrt [3]{b} d}-\frac {\log \left (a+b \sin ^3(c+d x)\right )}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[c + d*x])/(Sqrt[3]*a^(1/3))]/(Sqrt[3]*a^(2/3)*b^(1/3)*d)) + Log[a^(1/3) + b^
(1/3)*Sin[c + d*x]]/(3*a^(2/3)*b^(1/3)*d) - Log[a^(2/3) - a^(1/3)*b^(1/3)*Sin[c + d*x] + b^(2/3)*Sin[c + d*x]^
2]/(6*a^(2/3)*b^(1/3)*d) - Log[a + b*Sin[c + d*x]^3]/(3*b*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, 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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rule 1871

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rule 3223

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x]
, x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0
] || IGtQ[p, 0] || IntegersQ[m, p])

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1-x^2}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{d}-\frac {\operatorname {Subst}\left (\int \frac {x^2}{a+b x^3} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {\log \left (a+b \sin ^3(c+d x)\right )}{3 b d}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,\sin (c+d x)\right )}{3 a^{2/3} d}+\frac {\operatorname {Subst}\left (\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (c+d x)\right )}{3 a^{2/3} d}\\ &=\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b} d}-\frac {\log \left (a+b \sin ^3(c+d x)\right )}{3 b d}+\frac {\operatorname {Subst}\left (\int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (c+d x)\right )}{2 \sqrt [3]{a} d}-\frac {\operatorname {Subst}\left (\int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (c+d x)\right )}{6 a^{2/3} \sqrt [3]{b} d}\\ &=\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b} d}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} \sqrt [3]{b} d}-\frac {\log \left (a+b \sin ^3(c+d x)\right )}{3 b d}+\frac {\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} \sin (c+d x)}{\sqrt [3]{a}}\right )}{a^{2/3} \sqrt [3]{b} d}\\ &=-\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sin (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} a^{2/3} \sqrt [3]{b} d}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} \sqrt [3]{b} d}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (c+d x)+b^{2/3} \sin ^2(c+d x)\right )}{6 a^{2/3} \sqrt [3]{b} d}-\frac {\log \left (a+b \sin ^3(c+d x)\right )}{3 b d}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 139, normalized size = 0.83 \[ \frac {\left ((-1)^{2/3} b^{2/3}-a^{2/3}\right ) \log \left (-(-1)^{2/3} \sqrt [3]{a}-\sqrt [3]{b} \sin (c+d x)\right )+\left (b^{2/3}-a^{2/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )-\left (a^{2/3}+\sqrt [3]{-1} b^{2/3}\right ) \log \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} \sin (c+d x)\right )}{3 a^{2/3} b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-a^(2/3) + (-1)^(2/3)*b^(2/3))*Log[-((-1)^(2/3)*a^(1/3)) - b^(1/3)*Sin[c + d*x]] + (-a^(2/3) + b^(2/3))*Log[
a^(1/3) + b^(1/3)*Sin[c + d*x]] - (a^(2/3) + (-1)^(1/3)*b^(2/3))*Log[a^(1/3) + (-1)^(2/3)*b^(1/3)*Sin[c + d*x]
])/(3*a^(2/3)*b*d)

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fricas [C]  time = 70.62, size = 2298, normalized size = 13.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*sqrt(1/3)*b*d*sqrt((((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d
^3))^(1/3) + 2/(b*d))^2*b^2*d^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a
^2*b^3*d^3))^(1/3) + 2/(b*d))*b*d + 4)/(b^2*d^2))*arctan(-1/8*(2*sqrt(1/3)*sqrt(((1/2)^(1/3)*(I*sqrt(3) + 1)*(
1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))^2*a^2*b^2*d^2 - 4*b^2*cos(d*x + c)^2
- 4*a*b*sin(d*x + c) + 2*(a*b^2*d*sin(d*x + c) - 2*a^2*b*d)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^
2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d)) + 4*a^2 + 4*b^2)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*
d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))*a*b*d^2 - 2*a*d)*sqrt((((1/2)^(1/3)*(I*sqrt
(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))^2*b^2*d^2 - 4*((1/2)^(1/3)
*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))*b*d + 4)/(b^2*d^2)
) + sqrt(1/3)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) +
2/(b*d))^2*a^2*b^2*d^3 - 8*a*b*d*sin(d*x + c) + 4*a^2*d + 4*(a*b^2*d^2*sin(d*x + c) - a^2*b*d^2)*((1/2)^(1/3)*
(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d)))*sqrt((((1/2)^(1/3)
*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))^2*b^2*d^2 - 4*((1/
2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))*b*d + 4)/(
b^2*d^2)))/b) - 6*sqrt(1/3)*b*d*sqrt((((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/
(a^2*b^3*d^3))^(1/3) + 2/(b*d))^2*b^2*d^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2
- b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))*b*d + 4)/(b^2*d^2))*arctan(-1/8*(2*sqrt(1/3)*sqrt(((1/2)^(1/3)*(I*sqrt
(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))^2*a^2*b^2*d^2 - 4*b^2*cos(
d*x + c)^2 - 4*a*b*sin(d*x + c) + 2*(a*b^2*d*sin(d*x + c) - 2*a^2*b*d)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^
3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d)) + 4*a^2 + 4*b^2)*(((1/2)^(1/3)*(I*sqrt(3) + 1
)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))*a*b*d^2 - 2*a*d)*sqrt((((1/2)^(1/
3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))^2*b^2*d^2 - 4*((
1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))*b*d + 4)
/(b^2*d^2)) - sqrt(1/3)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3)
)^(1/3) + 2/(b*d))^2*a^2*b^2*d^3 - 8*a*b*d*sin(d*x + c) + 4*a^2*d + 4*(a*b^2*d^2*sin(d*x + c) - a^2*b*d^2)*((1
/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d)))*sqrt((((
1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))^2*b^2*d^
2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))*
b*d + 4)/(b^2*d^2)))/b) - ((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3
))^(1/3) + 2/(b*d))*b*d*log(1/4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b
^3*d^3))^(1/3) + 2/(b*d))^2*a^2*b^2*d^2 - b^2*cos(d*x + c)^2 + 2*a*b*sin(d*x + c) - (a*b^2*d*sin(d*x + c) + a^
2*b*d)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d))
+ a^2 + b^2) + (((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3)
+ 2/(b*d))*b*d - 6)*log(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))
^(1/3) + 2/(b*d))^2*a^2*b^2*d^2 - 4*b^2*cos(d*x + c)^2 - 4*a*b*sin(d*x + c) + 2*(a*b^2*d*sin(d*x + c) - 2*a^2*
b*d)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/(b^3*d^3) + 1/(a^2*b*d^3) - (a^2 - b^2)/(a^2*b^3*d^3))^(1/3) + 2/(b*d)) +
4*a^2 + 4*b^2))/(b*d)

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giac [A]  time = 0.18, size = 156, normalized size = 0.93 \[ -\frac {\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (d x + c\right ) \right |}\right )}{a} + \frac {2 \, \log \left ({\left | b \sin \left (d x + c\right )^{3} + a \right |}\right )}{b} - \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, \sin \left (d x + c\right )\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (\sin \left (d x + c\right )^{2} + \left (-\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x + c\right ) + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*(-a/b)^(1/3)*log(abs(-(-a/b)^(1/3) + sin(d*x + c)))/a + 2*log(abs(b*sin(d*x + c)^3 + a))/b - 2*sqrt(3)
*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*((-a/b)^(1/3) + 2*sin(d*x + c))/(-a/b)^(1/3))/(a*b) - (-a*b^2)^(1/3)*log(si
n(d*x + c)^2 + (-a/b)^(1/3)*sin(d*x + c) + (-a/b)^(2/3))/(a*b))/d

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maple [A]  time = 0.84, size = 141, normalized size = 0.84 \[ \frac {\ln \left (\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 d b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\sin ^{2}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 d b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (a +b \left (\sin ^{3}\left (d x +c \right )\right )\right )}{3 b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/d/b/(a/b)^(2/3)*ln(sin(d*x+c)+(a/b)^(1/3))-1/6/d/b/(a/b)^(2/3)*ln(sin(d*x+c)^2-(a/b)^(1/3)*sin(d*x+c)+(a/b
)^(2/3))+1/3/d/b/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*sin(d*x+c)-1))-1/3*ln(a+b*sin(d*x+c)^3)
/b/d

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maxima [A]  time = 0.45, size = 159, normalized size = 0.95 \[ \frac {\frac {2 \, \sqrt {3} {\left (b {\left (3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}} - \frac {2 \, a}{b}\right )} + 2 \, a\right )} \arctan \left (-\frac {\sqrt {3} {\left (\left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, \sin \left (d x + c\right )\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b} - \frac {3 \, {\left (2 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} + 1\right )} \log \left (\sin \left (d x + c\right )^{2} - \left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (d x + c\right ) + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {6 \, {\left (\left (\frac {a}{b}\right )^{\frac {2}{3}} - 1\right )} \log \left (\left (\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (d x + c\right )\right )}{b \left (\frac {a}{b}\right )^{\frac {2}{3}}}}{18 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(2*sqrt(3)*(b*(3*(a/b)^(1/3) - 2*a/b) + 2*a)*arctan(-1/3*sqrt(3)*((a/b)^(1/3) - 2*sin(d*x + c))/(a/b)^(1/
3))/(a*b) - 3*(2*(a/b)^(2/3) + 1)*log(sin(d*x + c)^2 - (a/b)^(1/3)*sin(d*x + c) + (a/b)^(2/3))/(b*(a/b)^(2/3))
- 6*((a/b)^(2/3) - 1)*log((a/b)^(1/3) + sin(d*x + c))/(b*(a/b)^(2/3)))/d

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mupad [B]  time = 15.02, size = 153, normalized size = 0.92 \[ \frac {\sum _{k=1}^3\ln \left (\left (\mathrm {root}\left (27\,a^2\,b^3\,d^3+27\,a^2\,b^2\,d^2+9\,a^2\,b\,d-b^2+a^2,d,k\right )\,b\,3+1\right )\,\left (a+b\,\sin \left (c+d\,x\right )+\mathrm {root}\left (27\,a^2\,b^3\,d^3+27\,a^2\,b^2\,d^2+9\,a^2\,b\,d-b^2+a^2,d,k\right )\,a\,b\,3\right )\right )\,\mathrm {root}\left (27\,a^2\,b^3\,d^3+27\,a^2\,b^2\,d^2+9\,a^2\,b\,d-b^2+a^2,d,k\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log((3*root(27*a^2*b^3*d^3 + 27*a^2*b^2*d^2 + 9*a^2*b*d - b^2 + a^2, d, k)*b + 1)*(a + b*sin(c + d*x) +
3*root(27*a^2*b^3*d^3 + 27*a^2*b^2*d^2 + 9*a^2*b*d - b^2 + a^2, d, k)*a*b))*root(27*a^2*b^3*d^3 + 27*a^2*b^2*
d^2 + 9*a^2*b*d - b^2 + a^2, d, k), k, 1, 3)/d

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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+b*sin(d*x+c)**3),x)

[Out]

Timed out

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