3.219 \(\int \sin ^{-1-\frac{a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=142 \[ \frac{2 a \left (a^2+b^2\right ) \cos (c+d x) \sin ^{\frac{b^2}{a^2+b^2}}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{b^2}{2 \left (a^2+b^2\right )};\frac{1}{2} \left (3-\frac{a^2}{a^2+b^2}\right );\sin ^2(c+d x)\right )}{b d \sqrt{\cos ^2(c+d x)}}-\frac{\left (a^2+b^2\right ) \cos (c+d x) \sin ^{-\frac{a^2}{a^2+b^2}}(c+d x)}{d} \]

[Out]

-(((a^2 + b^2)*Cos[c + d*x])/(d*Sin[c + d*x]^(a^2/(a^2 + b^2)))) + (2*a*(a^2 + b^2)*Cos[c + d*x]*Hypergeometri
c2F1[1/2, b^2/(2*(a^2 + b^2)), (3 - a^2/(a^2 + b^2))/2, Sin[c + d*x]^2]*Sin[c + d*x]^(b^2/(a^2 + b^2)))/(b*d*S
qrt[Cos[c + d*x]^2])

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Rubi [A]  time = 0.12323, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2789, 2643, 3011} \[ \frac{2 a \left (a^2+b^2\right ) \cos (c+d x) \sin ^{\frac{b^2}{a^2+b^2}}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{b^2}{2 \left (a^2+b^2\right )};\frac{1}{2} \left (3-\frac{a^2}{a^2+b^2}\right );\sin ^2(c+d x)\right )}{b d \sqrt{\cos ^2(c+d x)}}-\frac{\left (a^2+b^2\right ) \cos (c+d x) \sin ^{-\frac{a^2}{a^2+b^2}}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^(-1 - a^2/(a^2 + b^2))*(a + b*Sin[c + d*x])^2,x]

[Out]

-(((a^2 + b^2)*Cos[c + d*x])/(d*Sin[c + d*x]^(a^2/(a^2 + b^2)))) + (2*a*(a^2 + b^2)*Cos[c + d*x]*Hypergeometri
c2F1[1/2, b^2/(2*(a^2 + b^2)), (3 - a^2/(a^2 + b^2))/2, Sin[c + d*x]^2]*Sin[c + d*x]^(b^2/(a^2 + b^2)))/(b*d*S
qrt[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 \sin ^{-1-\frac{a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \sin ^{-\frac{a^2}{a^2+b^2}}(c+d x) \, dx+\int \sin ^{-1-\frac{a^2}{a^2+b^2}}(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac{\left (a^2+b^2\right ) \cos (c+d x) \sin ^{-\frac{a^2}{a^2+b^2}}(c+d x)}{d}+\frac{2 a \left (a^2+b^2\right ) \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{b^2}{2 \left (a^2+b^2\right )};\frac{1}{2} \left (3-\frac{a^2}{a^2+b^2}\right );\sin ^2(c+d x)\right ) \sin ^{\frac{b^2}{a^2+b^2}}(c+d x)}{b d \sqrt{\cos ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.310594, size = 188, normalized size = 1.32 \[ -\frac{\cos (c+d x) \sin ^{-\frac{a^2}{a^2+b^2}}(c+d x) \sin ^2(c+d x)^{-\frac{b^2}{2 \left (a^2+b^2\right )}} \left (\sqrt{\sin ^2(c+d x)} \left (a^2 \, _2F_1\left (\frac{1}{2},\frac{a^2}{2 \left (a^2+b^2\right )}+1;\frac{3}{2};\cos ^2(c+d x)\right )+b^2 \, _2F_1\left (\frac{1}{2},\frac{a^2}{2 \left (a^2+b^2\right )};\frac{3}{2};\cos ^2(c+d x)\right )\right )+2 a b \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} \left (\frac{a^2}{a^2+b^2}+1\right );\frac{3}{2};\cos ^2(c+d x)\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^(-1 - a^2/(a^2 + b^2))*(a + b*Sin[c + d*x])^2,x]

[Out]

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

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Maple [F]  time = 5.474, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( dx+c \right ) \right ) ^{-1-{\frac{{a}^{2}}{{a}^{2}+{b}^{2}}}} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(d*x + c) + a)^2*sin(d*x + c)^(-a^2/(a^2 + b^2) - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(d*x + c) + a)^2*sin(d*x + c)^(-a^2/(a^2 + b^2) - 1), x)