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

Optimal. Leaf size=106 \[ -\frac {a 2^{m+\frac {9}{4}} \cos (e+f x) (g \cos (e+f x))^{3/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (\frac {3}{4},-m-\frac {1}{4};\frac {7}{4};\frac {1}{2} (1-\sin (e+f x))\right )}{3 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

-1/3*2^(9/4+m)*a*cos(f*x+e)*(g*cos(f*x+e))^(3/2)*hypergeom([3/4, -1/4-m],[7/4],1/2-1/2*sin(f*x+e))*(1+sin(f*x+
e))^(-1/4-m)*(a+a*sin(f*x+e))^(-1+m)/f/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 106, 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} \[ -\frac {a 2^{m+\frac {9}{4}} \cos (e+f x) (g \cos (e+f x))^{3/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (\frac {3}{4},-m-\frac {1}{4};\frac {7}{4};\frac {1}{2} (1-\sin (e+f x))\right )}{3 f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2^(9/4 + m)*a*Cos[e + f*x]*(g*Cos[e + f*x])^(3/2)*Hypergeometric2F1[3/4, -1/4 - m, 7/4, (1 - Sin[e + f*x])/2
]*(1 + Sin[e + f*x])^(-1/4 - m)*(a + a*Sin[e + f*x])^(-1 + m))/(3*f*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}{\sqrt {c-c \sin (e+f x)}} \, dx &=\frac {(g \cos (e+f x)) \int \sqrt {g \cos (e+f x)} (a+a \sin (e+f x))^{\frac {1}{2}+m} \, dx}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {\left (a^2 \cos (e+f x) (g \cos (e+f x))^{3/2}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{4}+m}}{\sqrt [4]{a-a x}} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{3/4} (a+a \sin (e+f x))^{5/4} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {\left (2^{\frac {1}{4}+m} a^2 \cos (e+f x) (g \cos (e+f x))^{3/2} (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}}{\sqrt [4]{a-a x}} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{3/4} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {2^{\frac {9}{4}+m} a \cos (e+f x) (g \cos (e+f x))^{3/2} \, _2F_1\left (\frac {3}{4},-\frac {1}{4}-m;\frac {7}{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}}{3 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 13.59, size = 832, normalized size = 7.85 \[ -\frac {88 \sqrt {2} \cos ^6\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) \cos (e+f x) (g \cos (e+f x))^{3/2} \csc \left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) \sec ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (\sin (e+f x) a+a)^m \left (\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )-1\right ) \left (3 F_1\left (\frac {7}{4};-2 m-\frac {1}{2},2 m+3;\frac {11}{4};\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right ) \tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )-7 F_1\left (\frac {3}{4};-2 m-\frac {1}{2},2 m+3;\frac {7}{4};\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )\right )}{3 f \left (616 F_1\left (\frac {3}{4};-2 m-\frac {1}{2},2 m+3;\frac {7}{4};\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right ) \cos ^4\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )-4 \sin ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) \left (88 (2 m+3) F_1\left (\frac {7}{4};-2 m-\frac {1}{2},2 m+4;\frac {11}{4};\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right ) \cos ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )+44 (4 m+1) F_1\left (\frac {7}{4};\frac {1}{2}-2 m,2 m+3;\frac {11}{4};\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right ) \cos ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )-28 \left ((4 m+6) F_1\left (\frac {11}{4};-2 m-\frac {1}{2},2 m+4;\frac {15}{4};\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )+(4 m+1) F_1\left (\frac {11}{4};\frac {1}{2}-2 m,2 m+3;\frac {15}{4};\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )\right ) \sin ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )+77 F_1\left (\frac {7}{4};-2 m-\frac {1}{2},2 m+3;\frac {11}{4};\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right ) \left (\cos \left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )+1\right )\right )\right ) \sqrt {c-c \sin (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-88*Sqrt[2]*Cos[(-e + Pi/2 - f*x)/4]^6*Cos[e + f*x]*(g*Cos[e + f*x])^(3/2)*Csc[(-e + Pi/2 - f*x)/2]*Sec[(-e +
Pi/2 - f*x)/2]^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a + a*Sin[e + f*x])^m*(-1 + Tan[(-e + Pi/2 - f*x)/4]^
2)*(-7*AppellF1[3/4, -1/2 - 2*m, 3 + 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 3*Ap
pellF1[7/4, -1/2 - 2*m, 3 + 2*m, 11/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2
- f*x)/4]^2))/(3*f*(616*AppellF1[3/4, -1/2 - 2*m, 3 + 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 -
f*x)/4]^2]*Cos[(-e + Pi/2 - f*x)/4]^4 - 4*Sin[(-e + Pi/2 - f*x)/4]^2*(88*(3 + 2*m)*AppellF1[7/4, -1/2 - 2*m,
4 + 2*m, 11/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cos[(-e + Pi/2 - f*x)/4]^2 + 44*(1 + 4
*m)*AppellF1[7/4, 1/2 - 2*m, 3 + 2*m, 11/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cos[(-e +
Pi/2 - f*x)/4]^2 + 77*AppellF1[7/4, -1/2 - 2*m, 3 + 2*m, 11/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 -
f*x)/4]^2]*(1 + Cos[(-e + Pi/2 - f*x)/2]) - 28*((6 + 4*m)*AppellF1[11/4, -1/2 - 2*m, 4 + 2*m, 15/4, Tan[(-e +
Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + (1 + 4*m)*AppellF1[11/4, 1/2 - 2*m, 3 + 2*m, 15/4, Tan[(-e +
Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Sin[(-e + Pi/2 - f*x)/4]^2))*Sqrt[c - c*Sin[e + f*x]])

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {g \cos \left (f x + e\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} g \cos \left (f x + e\right )}{c \sin \left (f x + e\right ) - c}, x\right ) \]

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))^(1/2),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)/(c*sin(f*x + e)
- c), x)

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

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))^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m}}{\sqrt {c -c \sin \left (f x +e \right )}}\, dx \]

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))^(1/2),x)

[Out]

int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(1/2),x)

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

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))^(1/2),x, algorithm="maxima")

[Out]

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

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

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))^(1/2),x)

[Out]

int(((g*cos(e + f*x))^(3/2)*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(1/2), 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))**(1/2),x)

[Out]

Timed out

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