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

Optimal. Leaf size=93 \[ -\frac{a^4 c^3 2^{m+\frac{9}{4}} (g \cos (e+f x))^{17/2} (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a \sin (e+f x)+a)^{m-4} \, _2F_1\left (\frac{17}{4},-m-\frac{1}{4};\frac{21}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{17 f g^7} \]

[Out]

-(2^(9/4 + m)*a^4*c^3*(g*Cos[e + f*x])^(17/2)*Hypergeometric2F1[17/4, -1/4 - m, 21/4, (1 - Sin[e + f*x])/2]*(1
+ Sin[e + f*x])^(-1/4 - m)*(a + a*Sin[e + f*x])^(-4 + m))/(17*f*g^7)

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Rubi [A]  time = 0.284945, antiderivative size = 93, 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} \[ -\frac{a^4 c^3 2^{m+\frac{9}{4}} (g \cos (e+f x))^{17/2} (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a \sin (e+f x)+a)^{m-4} \, _2F_1\left (\frac{17}{4},-m-\frac{1}{4};\frac{21}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{17 f g^7} \]

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])^3,x]

[Out]

-(2^(9/4 + m)*a^4*c^3*(g*Cos[e + f*x])^(17/2)*Hypergeometric2F1[17/4, -1/4 - m, 21/4, (1 - Sin[e + f*x])/2]*(1
+ Sin[e + f*x])^(-1/4 - m)*(a + a*Sin[e + f*x])^(-4 + m))/(17*f*g^7)

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 (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^3 \, dx &=\frac{\left (a^3 c^3\right ) \int (g \cos (e+f x))^{15/2} (a+a \sin (e+f x))^{-3+m} \, dx}{g^6}\\ &=\frac{\left (a^5 c^3 (g \cos (e+f x))^{17/2}\right ) \operatorname{Subst}\left (\int (a-a x)^{13/4} (a+a x)^{\frac{1}{4}+m} \, dx,x,\sin (e+f x)\right )}{f g^7 (a-a \sin (e+f x))^{17/4} (a+a \sin (e+f x))^{17/4}}\\ &=\frac{\left (2^{\frac{1}{4}+m} a^5 c^3 (g \cos (e+f x))^{17/2} (a+a \sin (e+f x))^{-4+m} \left (\frac{a+a \sin (e+f x)}{a}\right )^{-\frac{1}{4}-m}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{x}{2}\right )^{\frac{1}{4}+m} (a-a x)^{13/4} \, dx,x,\sin (e+f x)\right )}{f g^7 (a-a \sin (e+f x))^{17/4}}\\ &=-\frac{2^{\frac{9}{4}+m} a^4 c^3 (g \cos (e+f x))^{17/2} \, _2F_1\left (\frac{17}{4},-\frac{1}{4}-m;\frac{21}{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))^{-4+m}}{17 f g^7}\\ \end{align*}

Mathematica [F]  time = 180.007, 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])^3,x]

[Out]

$Aborted

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Maple [F]  time = 0.603, 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 ) ^{3}\, 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))^3,x)

[Out]

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

[Out]

-integrate((g*cos(f*x + e))^(3/2)*(c*sin(f*x + e) - c)^3*(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 (3 \, c^{3} g \cos \left (f x + e\right )^{3} - 4 \, c^{3} g \cos \left (f x + e\right ) -{\left (c^{3} g \cos \left (f x + e\right )^{3} - 4 \, c^{3} g \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt{g \cos \left (f x + e\right )}{\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))^3,x, algorithm="fricas")

[Out]

integral(-(3*c^3*g*cos(f*x + e)^3 - 4*c^3*g*cos(f*x + e) - (c^3*g*cos(f*x + e)^3 - 4*c^3*g*cos(f*x + e))*sin(f
*x + e))*sqrt(g*cos(f*x + e))*(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))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \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))^3,x, algorithm="giac")

[Out]

sage2