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]

-1/15*2^(9/4+m)*a^3*c^2*(g*cos(f*x+e))^(15/2)*hypergeom([15/4, -1/4-m],[19/4],1/2-1/2*sin(f*x+e))*sec(f*x+e)*(
1+sin(f*x+e))^(-1/4-m)*(a+a*sin(f*x+e))^(-3+m)*(c-c*sin(f*x+e))^(1/2)/f/g^6

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Rubi [A]  time = 0.36, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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 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 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])

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*}

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

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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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (c^{2} g \cos \left (f x + e\right )^{3} + 2 \, c^{2} g \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, c^{2} g \cos \left (f x + e\right )\right )} \sqrt {g \cos \left (f x + e\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

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))^(5/2),x, algorithm="giac")

[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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maple [F]  time = 0.44, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{\frac {5}{2}}\, dx \]

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))^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

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))^(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

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))^(5/2),x)

[Out]

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

[Out]

Timed out

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