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

Optimal. Leaf size=245 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}} \]

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

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Rubi [A]  time = 0.25, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3213, 2660, 618, 204} \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x]^3)^(-1),x]

[Out]

(2*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*a^(2/3)*Sqrt[a^(2/3) - b^(2/3)]*d)
+ (2*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^(2/3)*S
qrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*d) - (2*ArcTan[((-1)^(1/3)*(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x)/2]))/
Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3*a^(2/3)*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*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 3213

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*
x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f, n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rubi steps

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

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Mathematica [C]  time = 0.16, size = 126, normalized size = 0.51 \[ -\frac {2 i \text {RootSum}\left [i \text {$\#$1}^6 b-3 i \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 i \text {$\#$1}^2 b-i b\& ,\frac {2 \text {$\#$1} \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i \text {$\#$1} \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )}{\text {$\#$1}^4 b-2 \text {$\#$1}^2 b-4 i \text {$\#$1} a+b}\& \right ]}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x]^3)^(-1),x]

[Out]

(((-2*I)/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)]*#1 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1)/(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(1/(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 \[ \int \frac {1}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.57, size = 83, normalized size = 0.34 \[ \frac {\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/d*sum((_R^4+2*_R^2+1)/(_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 \[ \int \frac {1}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 16.71, size = 609, normalized size = 2.49 \[ \frac {\sum _{k=1}^6\ln \left (-\frac {8192\,a\,b^3\,\left (-729\,a^5+243\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-324\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )+972\,a^3\,b^2+a^3\,b\,\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )\,243-162\,a^3\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^2+648\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2\,\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )+216\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^2-72\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^3+36\,a\,b\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^3-9\,a\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^4+24\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^4-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^5\right )}{{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^5}\right )\,\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^6\,d^6-243\,a^4\,d^4-27\,a^2\,d^2-1,d,k\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(c + d*x)^3),x)

[Out]

symsum(log(-(8192*a*b^3*(972*a^3*b^2 - 729*a^5 - 9*a*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2)
, d, k)^4 - 162*a^3*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^2 - 4*tan(c/2 + (d*x)/2)*
root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^5 + 243*a^4*b*tan(c/2 + (d*x)/2) - 324*tan(c/
2 + (d*x)/2)*a^4*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k) + 24*b*tan(c/2 + (d*x)/2)*ro
ot(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^4 - 72*a^2*tan(c/2 + (d*x)/2)*root(d^6 + 27*a^2
*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^3 + 36*a*b*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2
- b^2), d, k)^3 + 243*b*a^3*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k) + 648*tan(c/2 + (
d*x)/2)*a^2*b^2*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k) + 216*a^2*b*tan(c/2 + (d*x)/2
)*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^2))/root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 7
29*a^4*(a^2 - b^2), d, k)^5)*root(729*a^4*b^2*d^6 - 729*a^6*d^6 - 243*a^4*d^4 - 27*a^2*d^2 - 1, d, k), k, 1, 6
)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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