3.159 \(\int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=114 \[ -\frac{a^3 c^2 2^{m+\frac{9}{4}} \sec (e+f x) \sqrt{c-c \sin (e+f x)} (g \cos (e+f x))^{15/2} (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a \sin (e+f x)+a)^{m-3} \, _2F_1\left (\frac{15}{4},-m-\frac{1}{4};\frac{19}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{15 f g^6} \]

[Out]

-(2^(9/4 + m)*a^3*c^2*(g*Cos[e + f*x])^(15/2)*Hypergeometric2F1[15/4, -1/4 - m, 19/4, (1 - Sin[e + f*x])/2]*Se
c[e + f*x]*(1 + Sin[e + f*x])^(-1/4 - m)*(a + a*Sin[e + f*x])^(-3 + m)*Sqrt[c - c*Sin[e + f*x]])/(15*f*g^6)

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Rubi [A]  time = 0.356974, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2853, 2689, 70, 69} \[ -\frac{a^3 c^2 2^{m+\frac{9}{4}} \sec (e+f x) \sqrt{c-c \sin (e+f x)} (g \cos (e+f x))^{15/2} (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a \sin (e+f x)+a)^{m-3} \, _2F_1\left (\frac{15}{4},-m-\frac{1}{4};\frac{19}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{15 f g^6} \]

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])^(5/2),x]

[Out]

-(2^(9/4 + m)*a^3*c^2*(g*Cos[e + f*x])^(15/2)*Hypergeometric2F1[15/4, -1/4 - m, 19/4, (1 - Sin[e + f*x])/2]*Se
c[e + f*x]*(1 + Sin[e + f*x])^(-1/4 - m)*(a + a*Sin[e + f*x])^(-3 + m)*Sqrt[c - c*Sin[e + f*x]])/(15*f*g^6)

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))^{5/2} \, dx &=\frac{\left (a^2 c^2 \sec (e+f x) \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}\right ) \int (g \cos (e+f x))^{13/2} (a+a \sin (e+f x))^{-\frac{5}{2}+m} \, dx}{g^5}\\ &=\frac{\left (a^4 c^2 (g \cos (e+f x))^{15/2} \sec (e+f x) \sqrt{c-c \sin (e+f x)}\right ) \operatorname{Subst}\left (\int (a-a x)^{11/4} (a+a x)^{\frac{1}{4}+m} \, dx,x,\sin (e+f x)\right )}{f g^6 (a-a \sin (e+f x))^{15/4} (a+a \sin (e+f x))^{13/4}}\\ &=\frac{\left (2^{\frac{1}{4}+m} a^4 c^2 (g \cos (e+f x))^{15/2} \sec (e+f x) (a+a \sin (e+f x))^{-3+m} \left (\frac{a+a \sin (e+f x)}{a}\right )^{-\frac{1}{4}-m} \sqrt{c-c \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{x}{2}\right )^{\frac{1}{4}+m} (a-a x)^{11/4} \, dx,x,\sin (e+f x)\right )}{f g^6 (a-a \sin (e+f x))^{15/4}}\\ &=-\frac{2^{\frac{9}{4}+m} a^3 c^2 (g \cos (e+f x))^{15/2} \, _2F_1\left (\frac{15}{4},-\frac{1}{4}-m;\frac{19}{4};\frac{1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{-\frac{1}{4}-m} (a+a \sin (e+f x))^{-3+m} \sqrt{c-c \sin (e+f x)}}{15 f g^6}\\ \end{align*}

Mathematica [F]  time = 180.037, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[(g*Cos[e + f*x])^(3/2)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

$Aborted

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Maple [F]  time = 0.243, 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 ) ^{{\frac{5}{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))^(5/2),x)

[Out]

int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),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 (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((g*cos(f*x + e))^(3/2)*(-c*sin(f*x + e) + c)^(5/2)*(a*sin(f*x + e) + a)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (c^{2} g \cos \left (f x + e\right )^{3} + 2 \, c^{2} g \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, c^{2} g \cos \left (f x + e\right )\right )} \sqrt{g \cos \left (f x + e\right )} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}, 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))^(5/2),x, algorithm="fricas")

[Out]

integral(-(c^2*g*cos(f*x + e)^3 + 2*c^2*g*cos(f*x + e)*sin(f*x + e) - 2*c^2*g*cos(f*x + e))*sqrt(g*cos(f*x + e
))*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m, 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))**(5/2),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))^(5/2),x, algorithm="giac")

[Out]

Exception raised: AttributeError