3.628 \(\int \frac{(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=156 \[ -\frac{2 e (e \cos (c+d x))^{p-1} \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{\frac{1-p}{2}} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{\frac{1-p}{2}} F_1\left (-\frac{3}{2};\frac{1-p}{2},\frac{1-p}{2};-\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )}{3 b d (a+b \sin (c+d x))^{3/2}} \]

[Out]

(-2*e*AppellF1[-3/2, (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))/(3*b*d*(a + b*Sin[c + d*x])^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.117712, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2704, 138} \[ -\frac{2 e (e \cos (c+d x))^{p-1} \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{\frac{1-p}{2}} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{\frac{1-p}{2}} F_1\left (-\frac{3}{2};\frac{1-p}{2},\frac{1-p}{2};-\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )}{3 b d (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^p/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-2*e*AppellF1[-3/2, (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))/(3*b*d*(a + b*Sin[c + d*x])^(3/2))

Rule 2704

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]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), 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])

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{5/2}} \, dx &=\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 ) \operatorname{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)^{5/2}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{2 e F_1\left (-\frac{3}{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}}}{3 b d (a+b \sin (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 3.07107, size = 187, normalized size = 1.2 \[ -\frac{2 e (e \cos (c+d x))^{p-1} \left (\frac{\sqrt{b^2}-b \sin (c+d x)}{a+\sqrt{b^2}}\right )^{\frac{1-p}{2}} \left (\frac{\sqrt{b^2}+b \sin (c+d x)}{\sqrt{b^2}-a}\right )^{\frac{1-p}{2}} F_1\left (-\frac{3}{2};\frac{1-p}{2},\frac{1-p}{2};-\frac{1}{2};\frac{a+b \sin (c+d x)}{a-\sqrt{b^2}},\frac{a+b \sin (c+d x)}{a+\sqrt{b^2}}\right )}{3 b d (a+b \sin (c+d x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Cos[c + d*x])^p/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-2*e*AppellF1[-3/2, (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))/(3*b*d*(a + b*Sin[c + d*x])^(3/2))

________________________________________________________________________________________

Maple [F]  time = 0.124, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\cos \left ( dx+c \right ) \right ) ^{p} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^(5/2),x)

[Out]

int((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^(5/2),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^p/(b*sin(d*x + c) + a)^(5/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{b \sin \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{p}}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} +{\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**p/(a+b*sin(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^p/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^p/(b*sin(d*x + c) + a)^(5/2), x)