3.156 \(\int \frac{(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx\)

Optimal. Leaf size=84 \[ -\frac{g 2^{m+\frac{9}{4}} \sqrt{g \cos (e+f x)} (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{4},-m-\frac{1}{4};\frac{5}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{c f} \]

[Out]

-((2^(9/4 + m)*g*Sqrt[g*Cos[e + f*x]]*Hypergeometric2F1[1/4, -1/4 - m, 5/4, (1 - Sin[e + f*x])/2]*(1 + Sin[e +
f*x])^(-1/4 - m)*(a + a*Sin[e + f*x])^m)/(c*f))

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Rubi [A]  time = 0.274795, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2840, 2689, 70, 69} \[ -\frac{g 2^{m+\frac{9}{4}} \sqrt{g \cos (e+f x)} (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{4},-m-\frac{1}{4};\frac{5}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{c f} \]

Antiderivative was successfully verified.

[In]

Int[((g*Cos[e + f*x])^(3/2)*(a + a*Sin[e + f*x])^m)/(c - c*Sin[e + f*x]),x]

[Out]

-((2^(9/4 + m)*g*Sqrt[g*Cos[e + f*x]]*Hypergeometric2F1[1/4, -1/4 - m, 5/4, (1 - Sin[e + f*x])/2]*(1 + Sin[e +
f*x])^(-1/4 - m)*(a + a*Sin[e + f*x])^m)/(c*f))

Rule 2840

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(a^m*c^m)/g^(2*m), Int[(g*Cos[e + f*x])^(2*m + p)*(c + d*Sin[e + f*x])
^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && Integer
Q[m] &&  !(IntegerQ[n] && LtQ[n^2, m^2])

Rule 2689

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int \frac{(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx &=\frac{g^2 \int \frac{(a+a \sin (e+f x))^{1+m}}{\sqrt{g \cos (e+f x)}} \, dx}{a c}\\ &=\frac{\left (a g \sqrt{g \cos (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{4}+m}}{(a-a x)^{3/4}} \, dx,x,\sin (e+f x)\right )}{c f \sqrt [4]{a-a \sin (e+f x)} \sqrt [4]{a+a \sin (e+f x)}}\\ &=\frac{\left (2^{\frac{1}{4}+m} a g \sqrt{g \cos (e+f x)} (a+a \sin (e+f x))^m \left (\frac{a+a \sin (e+f x)}{a}\right )^{-\frac{1}{4}-m}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{x}{2}\right )^{\frac{1}{4}+m}}{(a-a x)^{3/4}} \, dx,x,\sin (e+f x)\right )}{c f \sqrt [4]{a-a \sin (e+f x)}}\\ &=-\frac{2^{\frac{9}{4}+m} g \sqrt{g \cos (e+f x)} \, _2F_1\left (\frac{1}{4},-\frac{1}{4}-m;\frac{5}{4};\frac{1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac{1}{4}-m} (a+a \sin (e+f x))^m}{c f}\\ \end{align*}

Mathematica [A]  time = 0.11968, size = 84, normalized size = 1. \[ -\frac{g 2^{m+\frac{9}{4}} \sqrt{g \cos (e+f x)} (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a (\sin (e+f x)+1))^m \, _2F_1\left (\frac{1}{4},-m-\frac{1}{4};\frac{5}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{c f} \]

Antiderivative was successfully verified.

[In]

Integrate[((g*Cos[e + f*x])^(3/2)*(a + a*Sin[e + f*x])^m)/(c - c*Sin[e + f*x]),x]

[Out]

-((2^(9/4 + m)*g*Sqrt[g*Cos[e + f*x]]*Hypergeometric2F1[1/4, -1/4 - m, 5/4, (1 - Sin[e + f*x])/2]*(1 + Sin[e +
f*x])^(-1/4 - m)*(a*(1 + Sin[e + f*x]))^m)/(c*f))

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Maple [F]  time = 0.253, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}}{c-c\sin \left ( fx+e \right ) } \left ( g\cos \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e)),x)

[Out]

int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{\left (g \cos \left (f x + e\right )\right )^{\frac{3}{2}}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c \sin \left (f x + e\right ) - c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

-integrate((g*cos(f*x + e))^(3/2)*(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral(-sqrt(g*cos(f*x + e))*(a*sin(f*x + e) + a)^m*g*cos(f*x + e)/(c*sin(f*x + e) - c), 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((g*cos(f*x+e))**(3/2)*(a+a*sin(f*x+e))**m/(c-c*sin(f*x+e)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\left (g \cos \left (f x + e\right )\right )^{\frac{3}{2}}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c \sin \left (f x + e\right ) - c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate(-(g*cos(f*x + e))^(3/2)*(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c), x)