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

Optimal. Leaf size=93 \[ \frac {g^3 2^{m+\frac {9}{4}} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac {3}{4},-m-\frac {1}{4};\frac {1}{4};\frac {1}{2} (1-\sin (e+f x))\right )}{3 a c^2 f (g \cos (e+f x))^{3/2}} \]

[Out]

1/3*2^(9/4+m)*g^3*hypergeom([-3/4, -1/4-m],[1/4],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(-1/4-m)*(a+a*sin(f*x+e))^
(1+m)/a/c^2/f/(g*cos(f*x+e))^(3/2)

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Rubi [A]  time = 0.28, antiderivative size = 93, normalized size of antiderivative = 1.00, 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} \[ \frac {g^3 2^{m+\frac {9}{4}} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac {3}{4},-m-\frac {1}{4};\frac {1}{4};\frac {1}{2} (1-\sin (e+f x))\right )}{3 a c^2 f (g \cos (e+f x))^{3/2}} \]

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])^2,x]

[Out]

(2^(9/4 + m)*g^3*Hypergeometric2F1[-3/4, -1/4 - m, 1/4, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/4 - m)*(a
+ a*Sin[e + f*x])^(1 + m))/(3*a*c^2*f*(g*Cos[e + f*x])^(3/2))

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]))

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 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 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])

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))^2} \, dx &=\frac {g^4 \int \frac {(a+a \sin (e+f x))^{2+m}}{(g \cos (e+f x))^{5/2}} \, dx}{a^2 c^2}\\ &=\frac {\left (g^3 (a-a \sin (e+f x))^{3/4} (a+a \sin (e+f x))^{3/4}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{4}+m}}{(a-a x)^{7/4}} \, dx,x,\sin (e+f x)\right )}{c^2 f (g \cos (e+f x))^{3/2}}\\ &=\frac {\left (2^{\frac {1}{4}+m} g^3 (a-a \sin (e+f x))^{3/4} (a+a \sin (e+f x))^{1+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)^{7/4}} \, dx,x,\sin (e+f x)\right )}{c^2 f (g \cos (e+f x))^{3/2}}\\ &=\frac {2^{\frac {9}{4}+m} g^3 \, _2F_1\left (-\frac {3}{4},-\frac {1}{4}-m;\frac {1}{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))^{1+m}}{3 a c^2 f (g \cos (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 96, normalized size = 1.03 \[ -\frac {g 2^{m+\frac {9}{4}} \sqrt {g \cos (e+f x)} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a (\sin (e+f x)+1))^m \, _2F_1\left (-\frac {3}{4},-m-\frac {1}{4};\frac {1}{4};\frac {1}{2} (1-\sin (e+f x))\right )}{3 c^2 f (\sin (e+f x)-1)} \]

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])^2,x]

[Out]

-1/3*(2^(9/4 + m)*g*Sqrt[g*Cos[e + f*x]]*Hypergeometric2F1[-3/4, -1/4 - m, 1/4, (1 - Sin[e + f*x])/2]*(1 + Sin
[e + f*x])^(-1/4 - m)*(a*(1 + Sin[e + f*x]))^m)/(c^2*f*(-1 + Sin[e + f*x]))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {g \cos \left (f x + e\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} g \cos \left (f x + e\right )}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{2}}, x\right ) \]

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))^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}}\,{d x} \]

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))^2,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)^2, x)

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maple [F]  time = 1.36, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m}}{\left (c -c \sin \left (f x +e \right )\right )^{2}}\, dx \]

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))^2,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}}\,{d x} \]

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))^2,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)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**2,x)

[Out]

Timed out

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