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

Optimal. Leaf size=97 \[ -\frac {8\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{7 d (\sin (c+d x)+1)^{5/6}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d} \]

[Out]

-8/7*2^(5/6)*a*cos(d*x+c)*hypergeom([-5/6, 1/2],[3/2],1/2-1/2*sin(d*x+c))*(a+a*sin(d*x+c))^(1/3)/d/(1+sin(d*x+
c))^(5/6)-3/7*cos(d*x+c)*(a+a*sin(d*x+c))^(4/3)/d

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Rubi [A]  time = 0.07, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2751, 2652, 2651} \[ -\frac {8\ 2^{5/6} a \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{7 d (\sin (c+d x)+1)^{5/6}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*2^(5/6)*a*Cos[c + d*x]*Hypergeometric2F1[-5/6, 1/2, 3/2, (1 - Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^(1/3))
/(7*d*(1 + Sin[c + d*x])^(5/6)) - (3*Cos[c + d*x]*(a + a*Sin[c + d*x])^(4/3))/(7*d)

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)]

Rubi steps

\begin {align*} \int \sin (c+d x) (a+a \sin (c+d x))^{4/3} \, dx &=-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{7 d}+\frac {4}{7} \int (a+a \sin (c+d x))^{4/3} \, dx\\ &=-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{7 d}+\frac {\left (4 a \sqrt [3]{a+a \sin (c+d x)}\right ) \int (1+\sin (c+d x))^{4/3} \, dx}{7 \sqrt [3]{1+\sin (c+d x)}}\\ &=-\frac {8\ 2^{5/6} a \cos (c+d x) \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [3]{a+a \sin (c+d x)}}{7 d (1+\sin (c+d x))^{5/6}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{7 d}\\ \end {align*}

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Mathematica [C]  time = 2.00, size = 351, normalized size = 3.62 \[ \frac {(a (\sin (c+d x)+1))^{4/3} \left (-\frac {3}{2} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (\sin (2 (c+d x))+4 \cos (c+d x)-10)+\frac {3 (-1)^{3/4} e^{-\frac {3}{2} i (c+d x)} \left (e^{i (c+d x)}+i\right ) \left (-2 \left (1+i e^{-i (c+d x)}\right )^{2/3} \left (1+e^{2 i (c+d x)}\right ) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\sin ^2\left (\frac {1}{4} (2 c+2 d x+\pi )\right )\right )+5 i \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-i e^{-i (c+d x)}\right ) \sqrt {2-2 \sin (c+d x)}+20 e^{i (c+d x)} \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {2}{3};-i e^{-i (c+d x)}\right ) \sqrt {\cos ^2\left (\frac {1}{4} (2 c+2 d x+\pi )\right )}\right )}{2 \sqrt {2} \left (1+i e^{-i (c+d x)}\right )^{2/3} \sqrt {i e^{-i (c+d x)} \left (e^{i (c+d x)}-i\right )^2}}\right )}{7 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(4/3)*((3*(-1)^(3/4)*(I + E^(I*(c + d*x)))*(20*E^(I*(c + d*x))*Sqrt[Cos[(2*c + Pi + 2*
d*x)/4]^2]*Hypergeometric2F1[-1/3, 1/3, 2/3, (-I)/E^(I*(c + d*x))] - 2*(1 + I/E^(I*(c + d*x)))^(2/3)*(1 + E^((
2*I)*(c + d*x)))*Hypergeometric2F1[1/2, 5/6, 11/6, Sin[(2*c + Pi + 2*d*x)/4]^2] + (5*I)*Hypergeometric2F1[1/3,
2/3, 5/3, (-I)/E^(I*(c + d*x))]*Sqrt[2 - 2*Sin[c + d*x]]))/(2*Sqrt[2]*E^(((3*I)/2)*(c + d*x))*(1 + I/E^(I*(c
+ d*x)))^(2/3)*Sqrt[(I*(-I + E^(I*(c + d*x)))^2)/E^(I*(c + d*x))]) - (3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*
(-10 + 4*Cos[c + d*x] + Sin[2*(c + d*x)]))/2))/(7*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a \cos \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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