\(\int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx\) [627]

Optimal result

Integrand size = 25, antiderivative size = 154 \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 e \operatorname {AppellF1}\left (-\frac {1}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{-1+p} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}}}{b d \sqrt {a+b \sin (c+d x)}} \]

[Out]

-2*e*AppellF1(-1/2,1/2-1/2*p,1/2-1/2*p,1/2,(a+b*sin(d*x+c))/(a-b),(a+b*sin(d*x+c))/(a+b))*(e*cos(d*x+c))^(-1+p
)*(1+(-a-b*sin(d*x+c))/(a-b))^(1/2-1/2*p)*(1+(-a-b*sin(d*x+c))/(a+b))^(1/2-1/2*p)/b/d/(a+b*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2783, 143} \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 e (e \cos (c+d x))^{p-1} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (-\frac {1}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{b d \sqrt {a+b \sin (c+d x)}} \]

[In]

Int[(e*Cos[c + d*x])^p/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-2*e*AppellF1[-1/2, (1 - p)/2, (1 - p)/2, 1/2, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(a + b)]*(e
*Cos[c + d*x])^(-1 + p)*(1 - (a + b*Sin[c + d*x])/(a - b))^((1 - p)/2)*(1 - (a + b*Sin[c + d*x])/(a + b))^((1
- p)/2))/(b*d*Sqrt[a + b*Sin[c + d*x]])

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 2783

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[g*((g*
Cos[e + f*x])^(p - 1)/(f*(1 - (a + b*Sin[e + f*x])/(a - b))^((p - 1)/2)*(1 - (a + b*Sin[e + f*x])/(a + b))^((p
- 1)/2))), Subst[Int[(-b/(a - b) - b*(x/(a - b)))^((p - 1)/2)*(b/(a + b) - b*(x/(a + b)))^((p - 1)/2)*(a + b*
x)^m, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] &&  !IGtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (e (e \cos (c+d x))^{-1+p} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}}\right ) \text {Subst}\left (\int \frac {\left (-\frac {b}{a-b}-\frac {b x}{a-b}\right )^{\frac {1}{2} (-1+p)} \left (\frac {b}{a+b}-\frac {b x}{a+b}\right )^{\frac {1}{2} (-1+p)}}{(a+b x)^{3/2}} \, dx,x,\sin (c+d x)\right )}{d} \\ & = -\frac {2 e \operatorname {AppellF1}\left (-\frac {1}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{-1+p} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}}}{b d \sqrt {a+b \sin (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.30 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.20 \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 e \operatorname {AppellF1}\left (-\frac {1}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {1}{2},\frac {a+b \sin (c+d x)}{a-\sqrt {b^2}},\frac {a+b \sin (c+d x)}{a+\sqrt {b^2}}\right ) (e \cos (c+d x))^{-1+p} \left (\frac {\sqrt {b^2}-b \sin (c+d x)}{a+\sqrt {b^2}}\right )^{\frac {1-p}{2}} \left (\frac {\sqrt {b^2}+b \sin (c+d x)}{-a+\sqrt {b^2}}\right )^{\frac {1-p}{2}}}{b d \sqrt {a+b \sin (c+d x)}} \]

[In]

Integrate[(e*Cos[c + d*x])^p/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-2*e*AppellF1[-1/2, (1 - p)/2, (1 - p)/2, 1/2, (a + b*Sin[c + d*x])/(a - Sqrt[b^2]), (a + b*Sin[c + d*x])/(a
+ Sqrt[b^2])]*(e*Cos[c + d*x])^(-1 + p)*((Sqrt[b^2] - b*Sin[c + d*x])/(a + Sqrt[b^2]))^((1 - p)/2)*((Sqrt[b^2]
+ b*Sin[c + d*x])/(-a + Sqrt[b^2]))^((1 - p)/2))/(b*d*Sqrt[a + b*Sin[c + d*x]])

Maple [F]

\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}d x\]

[In]

int((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^(3/2),x)

[Out]

int((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^(3/2),x)

Fricas [F]

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

[In]

integrate((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*(e*cos(d*x + c))^p/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2), x
)

Sympy [F]

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

[In]

integrate((e*cos(d*x+c))**p/(a+b*sin(d*x+c))**(3/2),x)

[Out]

Integral((e*cos(c + d*x))**p/(a + b*sin(c + d*x))**(3/2), x)

Maxima [F]

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

[In]

integrate((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^p/(b*sin(d*x + c) + a)^(3/2), x)

Giac [F]

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

[In]

integrate((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^p/(b*sin(d*x + c) + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int((e*cos(c + d*x))^p/(a + b*sin(c + d*x))^(3/2),x)

[Out]

int((e*cos(c + d*x))^p/(a + b*sin(c + d*x))^(3/2), x)