3.3.19 \(\int \sin ^{-1-\frac {a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx\) [219]

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

[Out]

-(a^2+b^2)*cos(d*x+c)/d/(sin(d*x+c)^(a^2/(a^2+b^2)))+2*a*(a^2+b^2)*cos(d*x+c)*hypergeom([1/2, 1/2*b^2/(a^2+b^2
)],[3/2-1/2*a^2/(a^2+b^2)],sin(d*x+c)^2)*sin(d*x+c)^(b^2/(a^2+b^2))/b/d/(cos(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 142, normalized size of antiderivative = 1.00, 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 = {2868, 2722, 3090} \begin {gather*} \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} \end {gather*}

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 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 \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*}

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Mathematica [A]
time = 0.21, size = 188, normalized size = 1.32 \begin {gather*} -\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 (2 a b \, _2F_1\left (\frac {1}{2},\frac {1}{2} \left (1+\frac {a^2}{a^2+b^2}\right );\frac {3}{2};\cos ^2(c+d x)\right ) \sin (c+d x)+\left (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 )+a^2 \, _2F_1\left (\frac {1}{2},1+\frac {a^2}{2 \left (a^2+b^2\right )};\frac {3}{2};\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}\right )}{d} \end {gather*}

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 = 0.94, size = 0, normalized size = 0.00 \[\int \left (\sin ^{-1-\frac {a^{2}}{a^{2}+b^{2}}}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{2}\, dx\]

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 = 0.00, 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(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 = 0.00, 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(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sin ^{- \frac {a^{2}}{a^{2} + b^{2}} - 1}{\left (c + d x \right )}\, dx \end {gather*}

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]

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

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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(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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2}{{\sin \left (c+d\,x\right )}^{\frac {a^2}{a^2+b^2}+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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