3.121 \(\int \frac{(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=104 \[ -\frac{4 e^2 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 a d \sqrt{\sin (c+d x)}}+\frac{2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac{2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a d} \]

[Out]

(-4*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*a*d*Sqrt[Sin[c + d*x]]) + (2*e*(e*Sin[c + d*
x])^(3/2))/(3*a*d) - (2*e*Cos[c + d*x]*(e*Sin[c + d*x])^(3/2))/(5*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.219942, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3872, 2839, 2564, 30, 2569, 2640, 2639} \[ -\frac{4 e^2 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 a d \sqrt{\sin (c+d x)}}+\frac{2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac{2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*a*d*Sqrt[Sin[c + d*x]]) + (2*e*(e*Sin[c + d*
x])^(3/2))/(3*a*d) - (2*e*Cos[c + d*x]*(e*Sin[c + d*x])^(3/2))/(5*a*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2569

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(b*Sin[e +
f*x])^(n + 1)*(a*Cos[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Sin[e + f*x])^
n*(a*Cos[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m
, 2*n]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx &=-\int \frac{\cos (c+d x) (e \sin (c+d x))^{5/2}}{-a-a \cos (c+d x)} \, dx\\ &=\frac{e^2 \int \cos (c+d x) \sqrt{e \sin (c+d x)} \, dx}{a}-\frac{e^2 \int \cos ^2(c+d x) \sqrt{e \sin (c+d x)} \, dx}{a}\\ &=-\frac{2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a d}+\frac{e \operatorname{Subst}\left (\int \sqrt{x} \, dx,x,e \sin (c+d x)\right )}{a d}-\frac{\left (2 e^2\right ) \int \sqrt{e \sin (c+d x)} \, dx}{5 a}\\ &=\frac{2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac{2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a d}-\frac{\left (2 e^2 \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{5 a \sqrt{\sin (c+d x)}}\\ &=-\frac{4 e^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 a d \sqrt{\sin (c+d x)}}+\frac{2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac{2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a d}\\ \end{align*}

Mathematica [C]  time = 4.71814, size = 232, normalized size = 2.23 \[ \frac{2 \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{5/2} \left (\sqrt{\sin (c+d x)} (10 \sin (c) \cos (d x)-3 \sin (2 c) \cos (2 d x)+10 \cos (c) \sin (d x)-3 \cos (2 c) \sin (2 d x)-12 \tan (c))+\frac{2 \sec (c) e^{-i d x} \sqrt{2-2 e^{2 i (c+d x)}} \left (3 \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},e^{2 i (c+d x)}\right )+e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{2 i (c+d x)}\right )\right )}{\sqrt{-i e^{-i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )}}\right )}{15 a d \sin ^{\frac{5}{2}}(c+d x) (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]^2*Sec[c + d*x]*(e*Sin[c + d*x])^(5/2)*((2*Sqrt[2 - 2*E^((2*I)*(c + d*x))]*(3*Hypergeometri
c2F1[-1/4, 1/2, 3/4, E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))
])*Sec[c])/(E^(I*d*x)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/E^(I*(c + d*x))]) + Sqrt[Sin[c + d*x]]*(10*Cos[d*
x]*Sin[c] - 3*Cos[2*d*x]*Sin[2*c] + 10*Cos[c]*Sin[d*x] - 3*Cos[2*c]*Sin[2*d*x] - 12*Tan[c])))/(15*a*d*(1 + Sec
[c + d*x])*Sin[c + d*x]^(5/2))

________________________________________________________________________________________

Maple [A]  time = 1.322, size = 173, normalized size = 1.7 \begin{align*}{\frac{2\,{e}^{3}}{15\,a\cos \left ( dx+c \right ) d} \left ( 6\,\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},1/2\,\sqrt{2} \right ) -3\,\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},1/2\,\sqrt{2} \right ) +3\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+5\,\cos \left ( dx+c \right ) \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e^3*(6*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*El
lipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-3*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*Ell
ipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+3*cos(d*x+c)^4-5*cos(d*x+c)^3-3*cos(d*x+c)^2+5*cos(d*x+c))/d

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(e^2*cos(d*x + c)^2 - e^2)*sqrt(e*sin(d*x + c))/(a*sec(d*x + c) + a), 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*sin(d*x+c))**(5/2)/(a+a*sec(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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