3.220 \(\int \frac{(1+2 \sin (c+d x))^2}{\sin ^{\frac{6}{5}}(c+d x)} \, dx\)

Optimal. Leaf size=73 \[ \frac{5 \sin ^{\frac{4}{5}}(c+d x) \cos (c+d x) \, _2F_1\left (\frac{2}{5},\frac{1}{2};\frac{7}{5};\sin ^2(c+d x)\right )}{d \sqrt{\cos ^2(c+d x)}}-\frac{5 \cos (c+d x)}{d \sqrt [5]{\sin (c+d x)}} \]

[Out]

(-5*Cos[c + d*x])/(d*Sin[c + d*x]^(1/5)) + (5*Cos[c + d*x]*Hypergeometric2F1[2/5, 1/2, 7/5, Sin[c + d*x]^2]*Si
n[c + d*x]^(4/5))/(d*Sqrt[Cos[c + d*x]^2])

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Rubi [A]  time = 0.0738084, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2789, 2643, 3011} \[ \frac{5 \sin ^{\frac{4}{5}}(c+d x) \cos (c+d x) \, _2F_1\left (\frac{2}{5},\frac{1}{2};\frac{7}{5};\sin ^2(c+d x)\right )}{d \sqrt{\cos ^2(c+d x)}}-\frac{5 \cos (c+d x)}{d \sqrt [5]{\sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*Sin[c + d*x])^2/Sin[c + d*x]^(6/5),x]

[Out]

(-5*Cos[c + d*x])/(d*Sin[c + d*x]^(1/5)) + (5*Cos[c + d*x]*Hypergeometric2F1[2/5, 1/2, 7/5, Sin[c + d*x]^2]*Si
n[c + d*x]^(4/5))/(d*Sqrt[Cos[c + d*x]^2])

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3011

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

Rubi steps

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

Mathematica [A]  time = 0.104947, size = 73, normalized size = 1. \[ -\frac{4 \sin ^{\frac{4}{5}}(c+d x) \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{3}{5};\frac{3}{2};\cos ^2(c+d x)\right )}{d \sin ^2(c+d x)^{2/5}}-\frac{5 \cos (c+d x)}{d \sqrt [5]{\sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*Sin[c + d*x])^2/Sin[c + d*x]^(6/5),x]

[Out]

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

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Maple [F]  time = 0.183, size = 0, normalized size = 0. \begin{align*} \int{ \left ( 1+2\,\sin \left ( dx+c \right ) \right ) ^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{-{\frac{6}{5}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+2*sin(d*x+c))^2/sin(d*x+c)^(6/5),x)

[Out]

int((1+2*sin(d*x+c))^2/sin(d*x+c)^(6/5),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, \sin \left (d x + c\right ) + 1\right )}^{2}}{\sin \left (d x + c\right )^{\frac{6}{5}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*sin(d*x+c))^2/sin(d*x+c)^(6/5),x, algorithm="maxima")

[Out]

integrate((2*sin(d*x + c) + 1)^2/sin(d*x + c)^(6/5), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (4 \, \cos \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) - 5\right )} \sin \left (d x + c\right )^{\frac{4}{5}}}{\cos \left (d x + c\right )^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*sin(d*x+c))^2/sin(d*x+c)^(6/5),x, algorithm="fricas")

[Out]

integral((4*cos(d*x + c)^2 - 4*sin(d*x + c) - 5)*sin(d*x + c)^(4/5)/(cos(d*x + c)^2 - 1), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*sin(d*x+c))**2/sin(d*x+c)**(6/5),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, \sin \left (d x + c\right ) + 1\right )}^{2}}{\sin \left (d x + c\right )^{\frac{6}{5}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*sin(d*x+c))^2/sin(d*x+c)^(6/5),x, algorithm="giac")

[Out]

integrate((2*sin(d*x + c) + 1)^2/sin(d*x + c)^(6/5), x)