3.164 \(\int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=114 \[ \frac {g^4 2^{m+\frac {9}{4}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac {5}{4},-m-\frac {1}{4};-\frac {1}{4};\frac {1}{2} (1-\sin (e+f x))\right )}{5 a c^2 f \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}} \]
[Out]
1/5*2^(9/4+m)*g^4*cos(f*x+e)*hypergeom([-5/4, -1/4-m],[-1/4],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(-1/4-m)*(a+a*
sin(f*x+e))^(1+m)/a/c^2/f/(g*cos(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2)
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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} \[ \frac {g^4 2^{m+\frac {9}{4}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac {5}{4},-m-\frac {1}{4};-\frac {1}{4};\frac {1}{2} (1-\sin (e+f x))\right )}{5 a c^2 f \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}} \]
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)*g^4*Cos[e + f*x]*Hypergeometric2F1[-5/4, -1/4 - m, -1/4, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])
^(-1/4 - m)*(a + a*Sin[e + f*x])^(1 + m))/(5*a*c^2*f*(g*Cos[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]])
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 \frac {(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 (g^5 \cos (e+f x)\right ) \int \frac {(a+a \sin (e+f x))^{\frac {5}{2}+m}}{(g \cos (e+f x))^{7/2}} \, dx}{a^2 c^2 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {\left (g^4 \cos (e+f x) (a-a \sin (e+f x))^{5/4} (a+a \sin (e+f x))^{3/4}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{4}+m}}{(a-a x)^{9/4}} \, dx,x,\sin (e+f x)\right )}{c^2 f (g \cos (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {\left (2^{\frac {1}{4}+m} g^4 \cos (e+f x) (a-a \sin (e+f x))^{5/4} (a+a \sin (e+f x))^{1+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{-\frac {1}{4}-m}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{\frac {1}{4}+m}}{(a-a x)^{9/4}} \, dx,x,\sin (e+f x)\right )}{c^2 f (g \cos (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {2^{\frac {9}{4}+m} g^4 \cos (e+f x) \, _2F_1\left (-\frac {5}{4},-\frac {1}{4}-m;-\frac {1}{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))^{1+m}}{5 a c^2 f (g \cos (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}\\ \end {align*}
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Mathematica [C] time = 53.66, size = 2320, normalized size = 20.35 \[ \text {Result too large to show} \]
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])^(5/2),x]
[Out]
((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Cos[(-e + Pi/2 - f*x)/2]*(g*Cos[e + f*x])^(3/2)*(5*AppellF1[3/4, -1/2 - 2*
m, 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 3*AppellF1[-5/4, -1/2 - 2*m, 2*m, -1/4
, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^4)*(Sec[(-e + Pi/2 - f*x)/
4]^2)^(1 + 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(a + a*Sin[e + f*x])^m*Tan[(-e + Pi/2 - f*x)/4])/(60*S
qrt[2]*f*Sqrt[Cos[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)*(Cos[Pi/4 + (e - Pi/2 + f*x)/2] - Sin[Pi/4 + (e - Pi/2
+ f*x)/2])^4*Sqrt[2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2]*(-1/240*((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Sqrt[Cos[e + f
*x]]*(5*AppellF1[3/4, -1/2 - 2*m, 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 3*Appel
lF1[-5/4, -1/2 - 2*m, 2*m, -1/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x
)/4]^4)*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2 + 2*m)*Tan[(-e + Pi/2 - f*x)/4]^2)/(2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2)^
(3/2) - ((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Sqrt[Cos[e + f*x]]*(5*AppellF1[3/4, -1/2 - 2*m, 2*m, 7/4, Tan[(-e
+ Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 3*AppellF1[-5/4, -1/2 - 2*m, 2*m, -1/4, Tan[(-e + Pi/2 - f*
x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^4)*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2 + 2*m))/(480
*Sqrt[2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2]) + (m*(Cos[(-e + Pi/2 - f*x)/4]^2)^(-1 + 2*m)*Sqrt[Cos[e + f*x]]*(5*Ap
pellF1[3/4, -1/2 - 2*m, 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 3*AppellF1[-5/4,
-1/2 - 2*m, 2*m, -1/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^4)*(S
ec[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*Sin[(-e + Pi/2 - f*x)/4]^2)/(120*Sqrt[2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2])
- ((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(5*AppellF1[3/4, -1/2 - 2*m, 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[
(-e + Pi/2 - f*x)/4]^2] + 3*AppellF1[-5/4, -1/2 - 2*m, 2*m, -1/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2
- f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^4)*(Sec[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*Sin[e + f*x]*Tan[(-e + Pi/2 -
f*x)/4])/(240*Sqrt[Cos[e + f*x]]*Sqrt[2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2]) - ((1 + 2*m)*(Cos[(-e + Pi/2 - f*x)/4
]^2)^(2*m)*Sqrt[Cos[e + f*x]]*(5*AppellF1[3/4, -1/2 - 2*m, 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi
/2 - f*x)/4]^2] + 3*AppellF1[-5/4, -1/2 - 2*m, 2*m, -1/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4
]^2]*Cot[(-e + Pi/2 - f*x)/4]^4)*(Sec[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*Tan[(-e + Pi/2 - f*x)/4]^2)/(240*Sqrt[
2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2]) - ((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Sqrt[Cos[e + f*x]]*(Sec[(-e + Pi/2 -
f*x)/4]^2)^(1 + 2*m)*Tan[(-e + Pi/2 - f*x)/4]*(-3*AppellF1[-5/4, -1/2 - 2*m, 2*m, -1/4, Tan[(-e + Pi/2 - f*x)/
4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^3*Csc[(-e + Pi/2 - f*x)/4]^2 + 3*Cot[(-e + Pi/2 -
f*x)/4]^4*(-5*m*AppellF1[-1/4, -1/2 - 2*m, 1 + 2*m, 3/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]
^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4] + (5*(-1/2 - 2*m)*AppellF1[-1/4, 1/2 - 2*m, 2*m, 3/4,
Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/
2) + 5*((-3*m*AppellF1[7/4, -1/2 - 2*m, 1 + 2*m, 11/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2
]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/7 + (3*(-1/2 - 2*m)*AppellF1[7/4, 1/2 - 2*m, 2*m, 11/4,
Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])
/14)))/(120*Sqrt[2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2])))
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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {g \cos \left (f x + e\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} g \cos \left (f x + e\right )}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} - {\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]
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(-sqrt(g*cos(f*x + e))*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m*g*cos(f*x + e)/(3*c^3*cos(f*x
+ e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e)), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]
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)*(a*sin(f*x + e) + a)^m/(-c*sin(f*x + e) + c)^(5/2), x)
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maple [F] time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m}}{\left (c -c \sin \left (f x +e \right )\right )^{\frac {5}{2}}}\, dx \]
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 \[ \int \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]
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)*(a*sin(f*x + e) + a)^m/(-c*sin(f*x + e) + c)^(5/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]
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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