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

Optimal. Leaf size=91 \[ -\frac {a^2 c 2^{m+\frac {9}{4}} (g \cos (e+f x))^{9/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (\frac {9}{4},-m-\frac {1}{4};\frac {13}{4};\frac {1}{2} (1-\sin (e+f x))\right )}{9 f g^3} \]

[Out]

-1/9*2^(9/4+m)*a^2*c*(g*cos(f*x+e))^(9/2)*hypergeom([9/4, -1/4-m],[13/4],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(-
1/4-m)*(a+a*sin(f*x+e))^(-2+m)/f/g^3

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Rubi [A]  time = 0.21, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2840, 2689, 70, 69} \[ -\frac {a^2 c 2^{m+\frac {9}{4}} (g \cos (e+f x))^{9/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (\frac {9}{4},-m-\frac {1}{4};\frac {13}{4};\frac {1}{2} (1-\sin (e+f x))\right )}{9 f g^3} \]

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)*a^2*c*(g*Cos[e + f*x])^(9/2)*Hypergeometric2F1[9/4, -1/4 - m, 13/4, (1 - Sin[e + f*x])/2]*(1 + S
in[e + f*x])^(-1/4 - m)*(a + a*Sin[e + f*x])^(-2 + m))/(9*f*g^3)

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

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Mathematica [F]  time = 169.60, size = 0, normalized size = 0.00 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx \]

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]),x]

[Out]

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

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (c g \cos \left (f x + e\right ) \sin \left (f x + e\right ) - c g \cos \left (f x + e\right )\right )} \sqrt {g \cos \left (f x + e\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

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

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

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)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.81, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )\, dx \]

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

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)),x, algorithm="maxima")

[Out]

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

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

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)),x)

[Out]

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

[Out]

Timed out

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