3.2.67 \(\int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx\) [167]

Optimal. Leaf size=123 \[ \frac {2^{-\frac {3}{4}-m} (g \cos (e+f x))^{5/2} \, _2F_1\left (\frac {1}{4} (5+4 m),\frac {1}{4} (11+4 m);\frac {1}{4} (9+4 m);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{-\frac {1}{4}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c^2 f g (5+4 m)} \]

[Out]

2^(-3/4-m)*(g*cos(f*x+e))^(5/2)*hypergeom([5/4+m, 11/4+m],[9/4+m],1/2+1/2*sin(f*x+e))*(1-sin(f*x+e))^(-1/4+m)*
(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1-m)/c^2/f/g/(5+4*m)

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Rubi [A]
time = 0.26, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2932, 2768, 72, 71} \begin {gather*} \frac {2^{-m-\frac {3}{4}} (g \cos (e+f x))^{5/2} (1-\sin (e+f x))^{m-\frac {1}{4}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac {1}{4} (4 m+5),\frac {1}{4} (4 m+11);\frac {1}{4} (4 m+9);\frac {1}{2} (\sin (e+f x)+1)\right )}{c^2 f g (4 m+5)} \end {gather*}

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 - m),x]

[Out]

(2^(-3/4 - m)*(g*Cos[e + f*x])^(5/2)*Hypergeometric2F1[(5 + 4*m)/4, (11 + 4*m)/4, (9 + 4*m)/4, (1 + Sin[e + f*
x])/2]*(1 - Sin[e + f*x])^(-1/4 + m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(c^2*f*g*(5 + 4*m))

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 2768

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 2932

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))^{-3-m} \, dx &=\left ((g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int (g \cos (e+f x))^{\frac {3}{2}+2 m} (c-c \sin (e+f x))^{-3-2 m} \, dx\\ &=\frac {\left (c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac {1}{2} \left (-\frac {5}{2}-2 m\right )+m} (c+c \sin (e+f x))^{\frac {1}{2} \left (-\frac {5}{2}-2 m\right )}\right ) \text {Subst}\left (\int (c-c x)^{-3-2 m+\frac {1}{2} \left (\frac {1}{2}+2 m\right )} (c+c x)^{\frac {1}{2} \left (\frac {1}{2}+2 m\right )} \, dx,x,\sin (e+f x)\right )}{f g}\\ &=\frac {\left (2^{-\frac {11}{4}-m} (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac {1}{4}+\frac {1}{2} \left (-\frac {5}{2}-2 m\right )} \left (\frac {c-c \sin (e+f x)}{c}\right )^{-\frac {1}{4}+m} (c+c \sin (e+f x))^{\frac {1}{2} \left (-\frac {5}{2}-2 m\right )}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-3-2 m+\frac {1}{2} \left (\frac {1}{2}+2 m\right )} (c+c x)^{\frac {1}{2} \left (\frac {1}{2}+2 m\right )} \, dx,x,\sin (e+f x)\right )}{c f g}\\ &=\frac {2^{-\frac {3}{4}-m} (g \cos (e+f x))^{5/2} \, _2F_1\left (\frac {1}{4} (5+4 m),\frac {1}{4} (11+4 m);\frac {1}{4} (9+4 m);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{-\frac {1}{4}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c^2 f g (5+4 m)}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(382\) vs. \(2(123)=246\).
time = 22.31, size = 382, normalized size = 3.11 \begin {gather*} \frac {2^{-4-m} (g \cos (e+f x))^{3/2} \left (\left (-3+8 m+16 m^2\right ) \cot ^4\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right ) \, _2F_1\left (-\frac {3}{2}-2 m,-\frac {7}{4}-m;-\frac {3}{4}-m;\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right )+(7+4 m) \left (2 (-1+4 m) \cot ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right ) \, _2F_1\left (-\frac {3}{2}-2 m,-\frac {3}{4}-m;\frac {1}{4}-m;\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right )+(3+4 m) \, _2F_1\left (-\frac {3}{2}-2 m,\frac {1}{4}-m;\frac {5}{4}-m;\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right )\right )\right ) \sec ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right ) \sec (e+f x) \sin ^{-2 m}\left (\frac {1}{2} \left (-e+\frac {\pi }{2}-f x\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{-2 (-3-m)} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \left (1-\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right )^{-\frac {1}{2}-2 m}}{f (-1+4 m) (3+4 m) (7+4 m)} \end {gather*}

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])^(-3 - m),x]

[Out]

(2^(-4 - m)*(g*Cos[e + f*x])^(3/2)*((-3 + 8*m + 16*m^2)*Cot[(-e + Pi/2 - f*x)/4]^4*Hypergeometric2F1[-3/2 - 2*
m, -7/4 - m, -3/4 - m, Tan[(-e + Pi/2 - f*x)/4]^2] + (7 + 4*m)*(2*(-1 + 4*m)*Cot[(-e + Pi/2 - f*x)/4]^2*Hyperg
eometric2F1[-3/2 - 2*m, -3/4 - m, 1/4 - m, Tan[(-e + Pi/2 - f*x)/4]^2] + (3 + 4*m)*Hypergeometric2F1[-3/2 - 2*
m, 1/4 - m, 5/4 - m, Tan[(-e + Pi/2 - f*x)/4]^2]))*Sec[(-e + Pi/2 - f*x)/4]^2*Sec[e + f*x]*(a + a*Sin[e + f*x]
)^m*(c - c*Sin[e + f*x])^(-3 - m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(-1/2 - 2*m))/(f*(-1 + 4*m)*(3 + 4*m)*(7 +
4*m)*Sin[(-e + Pi/2 - f*x)/2]^(2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(2*(-3 - m)))

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Maple [F]
time = 0.30, size = 0, normalized size = 0.00 \[\int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-3-m}\, dx\]

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))^(-3-m),x)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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-m),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)^(-m - 3), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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-m),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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-m),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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-m),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)^(-m - 3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 )}^{m+3}} \,d x \end {gather*}

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))^(m + 3),x)

[Out]

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

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