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]

2/3*e*(e*sin(d*x+c))^(3/2)/a/d-2/5*e*cos(d*x+c)*(e*sin(d*x+c))^(3/2)/a/d+4/5*e^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)
^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/a/d/sin(d*x
+c)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, 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 30

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

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 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 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]

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 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 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]

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*}

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Mathematica [C]  time = 4.90, size = 232, normalized size = 2.23 \[ \frac {2 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{5/2} \left (\sqrt {\sin (c+d x)} (10 \sin (c) \cos (d x)-3 \sin (2 c) \cos (2 d x)+10 \cos (c) \sin (d x)-3 \cos (2 c) \sin (2 d x)-12 \tan (c))+\frac {2 \sec (c) e^{-i d x} \sqrt {2-2 e^{2 i (c+d x)}} \left (3 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};e^{2 i (c+d x)}\right )+e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};e^{2 i (c+d x)}\right )\right )}{\sqrt {-i e^{-i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )}}\right )}{15 a d \sin ^{\frac {5}{2}}(c+d x) (\sec (c+d x)+1)} \]

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))

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (e^{2} \cos \left (d x + c\right )^{2} - e^{2}\right )} \sqrt {e \sin \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \]

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)

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maple [A]  time = 3.49, size = 173, normalized size = 1.66 \[ \frac {2 e^{3} \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+3 \left (\cos ^{4}\left (d x +c \right )\right )-5 \left (\cos ^{3}\left (d x +c \right )\right )-3 \left (\cos ^{2}\left (d x +c \right )\right )+5 \cos \left (d x +c \right )\right )}{15 a \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d} \]

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*sin(d*x+c)+2)^(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*sin(d*x+c)+2)^(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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((cos(c + d*x)*(e*sin(c + d*x))^(5/2))/(a*(cos(c + d*x) + 1)), 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((e*sin(d*x+c))**(5/2)/(a+a*sec(d*x+c)),x)

[Out]

Timed out

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