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

Optimal. Leaf size=267 \[ \frac {2 (-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \sqrt [3]{-1} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3220, 2660, 618, 204} \[ \frac {2 (-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \sqrt [3]{-1} \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rubi steps

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

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Mathematica [C]  time = 0.18, size = 172, normalized size = 0.64 \[ -\frac {\text {RootSum}\left [i \text {$\#$1}^6 b-3 i \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 i \text {$\#$1}^2 b-i b\& ,\frac {-i \text {$\#$1}^2 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )+i \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )+2 \text {$\#$1}^2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )}{\text {$\#$1}^4 b-2 \text {$\#$1}^2 b-4 i \text {$\#$1} a+b}\& \right ]}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^6 & , (-2*ArcTan[Sin[c + d*x]/(Cos[c + d
*x] - #1)] + I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - I*Log[1 -
2*Cos[c + d*x]*#1 + #1^2]*#1^2)/(b - (4*I)*a*#1 - 2*b*#1^2 + b*#1^4) & ]/d

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.53, size = 78, normalized size = 0.29 \[ \frac {2 \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d*sum((_R^3+_R)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3
*b+3*_Z^2*a+a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 16.12, size = 652, normalized size = 2.44 \[ \frac {\sum _{k=1}^6\ln \left (-8192\,a^3\,b+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^2\,a^3\,b^3\,294912+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^3\,a^4\,b^3\,1548288+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^4\,a^5\,b^3\,1990656-{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^5\,a^4\,b^5\,7962624+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^5\,a^6\,b^3\,5971968+65536\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,196608+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^2\,a^4\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,294912-{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^3\,a^3\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1769472+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^3\,a^5\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,221184+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^4\,a^4\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,2654208-{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^5\,a^5\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1990656\right )\,\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log(294912*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^2*a^3*b^3 - 8192*a^3*b +
1548288*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^3*a^4*b^3 + 1990656*root(729*a^4*
b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^4*a^5*b^3 - 7962624*root(729*a^4*b^2*d^6 - 729*a^2*b^4*
d^6 + 243*a^2*b^2*d^4 + 1, d, k)^5*a^4*b^5 + 5971968*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4
+ 1, d, k)^5*a^6*b^3 + 65536*a^2*b^2*tan(c/2 + (d*x)/2) + 196608*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*
a^2*b^2*d^4 + 1, d, k)*a^3*b^2*tan(c/2 + (d*x)/2) + 294912*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^
2*d^4 + 1, d, k)^2*a^4*b^2*tan(c/2 + (d*x)/2) - 1769472*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d
^4 + 1, d, k)^3*a^3*b^4*tan(c/2 + (d*x)/2) + 221184*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 +
1, d, k)^3*a^5*b^2*tan(c/2 + (d*x)/2) + 2654208*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1,
d, k)^4*a^4*b^4*tan(c/2 + (d*x)/2) - 1990656*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d,
k)^5*a^5*b^4*tan(c/2 + (d*x)/2))*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k), k, 1, 6
)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin {\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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