3.111 \(\int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=129 \[ \frac {13 \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{5\ 2^{5/6} a d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}-\frac {3 \cos (c+d x)}{2 a d \sqrt [3]{a \sin (c+d x)+a}}-\frac {3 \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}} \]

[Out]

-3/5*cos(d*x+c)/d/(a+a*sin(d*x+c))^(4/3)-3/2*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/3)+13/10*2^(1/6)*cos(d*x+c)*hy
pergeom([1/2, 5/6],[3/2],1/2-1/2*sin(d*x+c))/a/d/(1+sin(d*x+c))^(1/6)/(a+a*sin(d*x+c))^(1/3)

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Rubi [A]  time = 0.14, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2758, 2751, 2652, 2651} \[ \frac {13 \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{5\ 2^{5/6} a d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}-\frac {3 \cos (c+d x)}{2 a d \sqrt [3]{a \sin (c+d x)+a}}-\frac {3 \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^2/(a + a*Sin[c + d*x])^(4/3),x]

[Out]

(-3*Cos[c + d*x])/(5*d*(a + a*Sin[c + d*x])^(4/3)) - (3*Cos[c + d*x])/(2*a*d*(a + a*Sin[c + d*x])^(1/3)) + (13
*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Sin[c + d*x])/2])/(5*2^(5/6)*a*d*(1 + Sin[c + d*x])^(1/6)*
(a + a*Sin[c + d*x])^(1/3))

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]

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 2751

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

Rule 2758

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

Rubi steps

\begin {align*} \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx &=-\frac {3 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}+\frac {3 \int \frac {-\frac {4 a}{3}+\frac {5}{3} a \sin (c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx}{5 a^2}\\ &=-\frac {3 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac {3 \cos (c+d x)}{2 a d \sqrt [3]{a+a \sin (c+d x)}}-\frac {13 \int \frac {1}{\sqrt [3]{a+a \sin (c+d x)}} \, dx}{10 a}\\ &=-\frac {3 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac {3 \cos (c+d x)}{2 a d \sqrt [3]{a+a \sin (c+d x)}}-\frac {\left (13 \sqrt [3]{1+\sin (c+d x)}\right ) \int \frac {1}{\sqrt [3]{1+\sin (c+d x)}} \, dx}{10 a \sqrt [3]{a+a \sin (c+d x)}}\\ &=-\frac {3 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac {3 \cos (c+d x)}{2 a d \sqrt [3]{a+a \sin (c+d x)}}+\frac {13 \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{5\ 2^{5/6} a d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 108, normalized size = 0.84 \[ -\frac {3 \cos (c+d x) \left (13 \sqrt {2} (\sin (c+d x)+1) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\sin ^2\left (\frac {1}{4} (2 c+2 d x+\pi )\right )\right )+\sqrt {1-\sin (c+d x)} (5 \sin (c+d x)+7)\right )}{10 d \sqrt {1-\sin (c+d x)} (a (\sin (c+d x)+1))^{4/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^2/(a + a*Sin[c + d*x])^(4/3),x]

[Out]

(-3*Cos[c + d*x]*(13*Sqrt[2]*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[(2*c + Pi + 2*d*x)/4]^2]*(1 + Sin[c + d*x])
+ Sqrt[1 - Sin[c + d*x]]*(7 + 5*Sin[c + d*x])))/(10*d*Sqrt[1 - Sin[c + d*x]]*(a*(1 + Sin[c + d*x]))^(4/3))

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (\cos \left (d x + c\right )^{2} - 1\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2/(a+a*sin(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

integral((cos(d*x + c)^2 - 1)*(a*sin(d*x + c) + a)^(2/3)/(a^2*cos(d*x + c)^2 - 2*a^2*sin(d*x + c) - 2*a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2/(a+a*sin(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^2/(a*sin(d*x + c) + a)^(4/3), x)

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maple [F]  time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}\left (d x +c \right )}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^2/(a+a*sin(d*x+c))^(4/3),x)

[Out]

int(sin(d*x+c)^2/(a+a*sin(d*x+c))^(4/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2/(a+a*sin(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^2/(a*sin(d*x + c) + a)^(4/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (c+d\,x\right )}^2}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^2/(a + a*sin(c + d*x))^(4/3),x)

[Out]

int(sin(c + d*x)^2/(a + a*sin(c + d*x))^(4/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**2/(a+a*sin(d*x+c))**(4/3),x)

[Out]

Integral(sin(c + d*x)**2/(a*(sin(c + d*x) + 1))**(4/3), x)

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