3.3.20 \(\int \frac {(1+2 \sin (c+d x))^2}{\sin ^{\frac {6}{5}}(c+d x)} \, dx\) [220]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2868, 2722, 3090} \begin {gather*} \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)}} \end {gather*}

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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2868

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 3090

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*}

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Mathematica [A]
time = 0.07, size = 73, normalized size = 1.00 \begin {gather*} -\frac {5 \cos (c+d x)}{d \sqrt [5]{\sin (c+d x)}}-\frac {4 \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {3}{5};\frac {3}{2};\cos ^2(c+d x)\right ) \sin ^{\frac {4}{5}}(c+d x)}{d \sin ^2(c+d x)^{2/5}} \end {gather*}

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 = 0.42, size = 0, normalized size = 0.00 \[\int \frac {\left (1+2 \sin \left (d x +c \right )\right )^{2}}{\sin \left (d x +c \right )^{\frac {6}{5}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 7.47, size = 127, normalized size = 1.74 \begin {gather*} -\frac {4\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^{4/5}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{5};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{2/5}}-\frac {\cos \left (c+d\,x\right )\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{1/10}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{10};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\sin \left (c+d\,x\right )}^{1/5}}-\frac {4\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^{9/5}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{10},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{9/10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (4*cos(c + d*x)*sin(c + d*x)^(4/5)*hypergeom([1/2, 3/5], 3/2, cos(c + d*x)^2))/(d*(sin(c + d*x)^2)^(2/5)) -
(cos(c + d*x)*(sin(c + d*x)^2)^(1/10)*hypergeom([1/2, 11/10], 3/2, cos(c + d*x)^2))/(d*sin(c + d*x)^(1/5)) - (
4*cos(c + d*x)*sin(c + d*x)^(9/5)*hypergeom([1/10, 1/2], 3/2, cos(c + d*x)^2))/(d*(sin(c + d*x)^2)^(9/10))

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