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

Optimal. Leaf size=123 \[ \frac{2^{\frac{1}{4}-m} (g \cos (e+f x))^{5/2} (1-\sin (e+f x))^{m-\frac{1}{4}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac{1}{4} (4 m+5),\frac{1}{4} (4 m+7);\frac{1}{4} (4 m+9);\frac{1}{2} (\sin (e+f x)+1)\right )}{c f g (4 m+5)} \]

[Out]

(2^(1/4 - m)*(g*Cos[e + f*x])^(5/2)*Hypergeometric2F1[(5 + 4*m)/4, (7 + 4*m)/4, (9 + 4*m)/4, (1 + Sin[e + f*x]
)/2]*(1 - Sin[e + f*x])^(-1/4 + m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(c*f*g*(5 + 4*m))

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Rubi [A]  time = 0.36663, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2853, 2689, 70, 69} \[ \frac{2^{\frac{1}{4}-m} (g \cos (e+f x))^{5/2} (1-\sin (e+f x))^{m-\frac{1}{4}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac{1}{4} (4 m+5),\frac{1}{4} (4 m+7);\frac{1}{4} (4 m+9);\frac{1}{2} (\sin (e+f x)+1)\right )}{c f g (4 m+5)} \]

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 - m),x]

[Out]

(2^(1/4 - m)*(g*Cos[e + f*x])^(5/2)*Hypergeometric2F1[(5 + 4*m)/4, (7 + 4*m)/4, (9 + 4*m)/4, (1 + Sin[e + f*x]
)/2]*(1 - Sin[e + f*x])^(-1/4 + m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(c*f*g*(5 + 4*m))

Rule 2853

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^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e
+ f*x])^FracPart[m])/(g^(2*IntPart[m])*(g*Cos[e + f*x])^(2*FracPart[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, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 -
b^2, 0] && (FractionQ[m] ||  !FractionQ[n])

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

Mathematica [A]  time = 6.16755, size = 202, normalized size = 1.64 \[ \frac{g 2^{-m-1} \csc ^2\left (\frac{1}{8} (-2 e-2 f x+\pi )\right ) \sqrt{g \cos (e+f x)} \cos ^{-2 m}\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \left (1-\tan ^2\left (\frac{1}{8} (2 e+2 f x-\pi )\right )\right )^{-2 m-\frac{1}{2}} (a (\sin (e+f x)+1))^m (c-c \sin (e+f x))^{-m} \, _2F_1\left (-2 m-\frac{3}{2},-m-\frac{3}{4};\frac{1}{4}-m;\tan ^2\left (\frac{1}{8} (-2 e-2 f x+\pi )\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{2 (m+2)}}{c^2 f (4 m+3) (\sin (e+f x)-1)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(-1 - m)*g*Sqrt[g*Cos[e + f*x]]*Csc[(-2*e + Pi - 2*f*x)/8]^2*Hypergeometric2F1[-3/2 - 2*m, -3/4 - m, 1/4 -
m, Tan[(-2*e + Pi - 2*f*x)/8]^2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(2*(2 + m))*(a*(1 + Sin[e + f*x]))^m*(1
- Tan[(2*e - Pi + 2*f*x)/8]^2)^(-1/2 - 2*m))/(c^2*f*(3 + 4*m)*Cos[(2*e + Pi + 2*f*x)/4]^(2*m)*(-1 + Sin[e + f
*x])^2*(c - c*Sin[e + f*x])^m)

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Maple [F]  time = 0.338, size = 0, normalized size = 0. \begin{align*} \int \left ( g\cos \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-2-m}\, 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))^(-2-m),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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 )}^{-m - 2}\,{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))^(-2-m),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)^(-m - 2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{g \cos \left (f x + e\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} g \cos \left (f x + e\right ), 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))^(-2-m),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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))^(-2-m),x, algorithm="giac")

[Out]

Exception raised: AttributeError