### 3.12 $$\int \frac{1}{(a+a \sin (c+d x))^{2/3}} \, dx$$

Optimal. Leaf size=66 $-\frac{\sqrt [6]{\sin (c+d x)+1} \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{\sqrt [6]{2} d (a \sin (c+d x)+a)^{2/3}}$

[Out]

-((Cos[c + d*x]*Hypergeometric2F1[1/2, 7/6, 3/2, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(1/6))/(2^(1/6)*d*(a
+ a*Sin[c + d*x])^(2/3)))

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Rubi [A]  time = 0.0332749, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {2652, 2651} $-\frac{\sqrt [6]{\sin (c+d x)+1} \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{\sqrt [6]{2} d (a \sin (c+d x)+a)^{2/3}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + a*Sin[c + d*x])^(-2/3),x]

[Out]

-((Cos[c + d*x]*Hypergeometric2F1[1/2, 7/6, 3/2, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(1/6))/(2^(1/6)*d*(a
+ a*Sin[c + d*x])^(2/3)))

Rule 2652

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Sin[c + d*x])^FracPart
[n])/(1 + (b*Sin[c + d*x])/a)^FracPart[n], Int[(1 + (b*Sin[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sin (c+d x))^{2/3}} \, dx &=\frac{(1+\sin (c+d x))^{2/3} \int \frac{1}{(1+\sin (c+d x))^{2/3}} \, dx}{(a+a \sin (c+d x))^{2/3}}\\ &=-\frac{\cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right ) \sqrt [6]{1+\sin (c+d x)}}{\sqrt [6]{2} d (a+a \sin (c+d x))^{2/3}}\\ \end{align*}

Mathematica [C]  time = 6.09823, size = 604, normalized size = 9.15 $\frac{2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (\frac{3 \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}-3\right )}{d (a (\sin (c+d x)+1))^{2/3}}-\frac{2 \sqrt{2} \sqrt [6]{\sin (c+d x)+1} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (\frac{3 \sin \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right ) \cos ^2\left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right ) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2\left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )\right )}{5 \sqrt{\sin ^2\left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )} \sqrt [6]{\cos \left (2 \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )\right )+1}}-\frac{i \left (-\frac{3 i \left (e^{-i \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )}+e^{i \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )}\right )^{2/3} \, _2F_1\left (-\frac{1}{3},\frac{1}{3};\frac{2}{3};-e^{2 i \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )}\right )}{2^{2/3} \left (1+e^{2 i \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )}\right )^{2/3}}-\frac{3 i e^{i \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )} \sqrt [3]{1+e^{2 i \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )}} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-e^{2 i \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )}\right )}{2\ 2^{2/3} \sqrt [3]{e^{-i \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )}+e^{i \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )}}}\right ) \sqrt [3]{\cos \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )}}{2 \sqrt [6]{\cos \left (2 \left (\frac{1}{2} (-c-d x)+\frac{\pi }{4}\right )\right )+1}}\right )}{d (a (\sin (c+d x)+1))^{2/3}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + a*Sin[c + d*x])^(-2/3),x]

[Out]

(2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(-3 + (3*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/
(d*(a*(1 + Sin[c + d*x]))^(2/3)) - (2*Sqrt[2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(1 + Sin[c + d*x])^(1/6)*(
((-I/2)*Cos[Pi/4 + (-c - d*x)/2]^(1/3)*(((-3*I)*(E^((-I)*(Pi/4 + (-c - d*x)/2)) + E^(I*(Pi/4 + (-c - d*x)/2)))
^(2/3)*Hypergeometric2F1[-1/3, 1/3, 2/3, -E^((2*I)*(Pi/4 + (-c - d*x)/2))])/(2^(2/3)*(1 + E^((2*I)*(Pi/4 + (-c
- d*x)/2)))^(2/3)) - (((3*I)/2)*E^(I*(Pi/4 + (-c - d*x)/2))*(1 + E^((2*I)*(Pi/4 + (-c - d*x)/2)))^(1/3)*Hyper
geometric2F1[1/3, 2/3, 5/3, -E^((2*I)*(Pi/4 + (-c - d*x)/2))])/(2^(2/3)*(E^((-I)*(Pi/4 + (-c - d*x)/2)) + E^(I
*(Pi/4 + (-c - d*x)/2)))^(1/3))))/(1 + Cos[2*(Pi/4 + (-c - d*x)/2)])^(1/6) + (3*Cos[Pi/4 + (-c - d*x)/2]^2*Hyp
ergeometric2F1[1/2, 5/6, 11/6, Cos[Pi/4 + (-c - d*x)/2]^2]*Sin[Pi/4 + (-c - d*x)/2])/(5*(1 + Cos[2*(Pi/4 + (-c
- d*x)/2)])^(1/6)*Sqrt[Sin[Pi/4 + (-c - d*x)/2]^2])))/(d*(a*(1 + Sin[c + d*x]))^(2/3))

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(d*x+c))^(2/3),x)

[Out]

int(1/(a+a*sin(d*x+c))^(2/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(-2/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{2}{3}}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

integral((a*sin(d*x + c) + a)^(-2/3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sin{\left (c + d x \right )} + a\right )^{\frac{2}{3}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))**(2/3),x)

[Out]

Integral((a*sin(c + d*x) + a)**(-2/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^(2/3),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^(-2/3), x)