3.171 \(\int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx\)

Optimal. Leaf size=123 \[ \frac {c^2 2^{\frac {13}{4}-m} (g \cos (e+f x))^{5/2} (1-\sin (e+f x))^{m-\frac {1}{4}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac {1}{4} (4 m-5),\frac {1}{4} (4 m+5);\frac {1}{4} (4 m+9);\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (4 m+5)} \]

[Out]

2^(13/4-m)*c^2*(g*cos(f*x+e))^(5/2)*hypergeom([5/4+m, -5/4+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)/f/g/(5+4*m)

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Rubi [A]  time = 0.34, 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 = {2853, 2689, 70, 69} \[ \frac {c^2 2^{\frac {13}{4}-m} (g \cos (e+f x))^{5/2} (1-\sin (e+f x))^{m-\frac {1}{4}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac {1}{4} (4 m-5),\frac {1}{4} (4 m+5);\frac {1}{4} (4 m+9);\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (4 m+5)} \]

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])^(1 - m),x]

[Out]

(2^(13/4 - m)*c^2*(g*Cos[e + f*x])^(5/2)*Hypergeometric2F1[(-5 + 4*m)/4, (5 + 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))/(f*g*(5 + 4*m))

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

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Mathematica [F]  time = 176.64, size = 0, normalized size = 0.00 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1-m} \, dx \]

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])^(1 - m),x]

[Out]

Integrate[(g*Cos[e + f*x])^(3/2)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(1 - m), x]

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {g \cos \left (f x + e\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1} g \cos \left (f x + e\right ), x\right ) \]

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))^(1-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 + 1)*g*cos(f*x + e), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1}\,{d x} \]

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-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 + 1), x)

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maple [F]  time = 0.67, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{1-m}\, dx \]

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))^(1-m),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 1}\,{d x} \]

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-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 + 1), x)

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

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 - m),x)

[Out]

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

[Out]

Timed out

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