\(\int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx\) [155]

Optimal result

Integrand size = 25, antiderivative size = 88 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=-\frac {2^{\frac {9}{4}+m} a (g \cos (e+f x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-\frac {1}{4}-m,\frac {9}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{4}-m} (a+a \sin (e+f x))^{-1+m}}{5 f g} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2768, 72, 71} \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=-\frac {a 2^{m+\frac {9}{4}} (g \cos (e+f x))^{5/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-m-\frac {1}{4},\frac {9}{4},\frac {1}{2} (1-\sin (e+f x))\right )}{5 f g} \]

[In]

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

[Out]

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2 (g \cos (e+f x))^{5/2}\right ) \text {Subst}\left (\int \sqrt [4]{a-a x} (a+a x)^{\frac {1}{4}+m} \, dx,x,\sin (e+f x)\right )}{f g (a-a \sin (e+f x))^{5/4} (a+a \sin (e+f x))^{5/4}} \\ & = \frac {\left (2^{\frac {1}{4}+m} a^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{-1+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{-\frac {1}{4}-m}\right ) \text {Subst}\left (\int \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 g (a-a \sin (e+f x))^{5/4}} \\ & = -\frac {2^{\frac {9}{4}+m} a (g \cos (e+f x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-\frac {1}{4}-m,\frac {9}{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 f g} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.97 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=-\frac {2^{\frac {9}{4}+m} (g \cos (e+f x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-\frac {1}{4}-m,\frac {9}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {5}{4}-m} (a (1+\sin (e+f x)))^m}{5 f g} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]

[In]

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

[Out]

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