\(\int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx\) [121]

Optimal result

Integrand size = 25, antiderivative size = 104 \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=-\frac {4 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a d \sqrt {\sin (c+d x)}}+\frac {2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a d} \]

[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)

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3957, 2918, 2644, 30, 2649, 2721, 2719} \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=-\frac {4 e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a d \sqrt {\sin (c+d x)}}+\frac {2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a d} \]

[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 2644

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 2649

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 2719

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

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2918

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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*} \text {integral}& = -\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 \text {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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.19 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.86 \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\frac {2 e^3 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (10 \sqrt {\csc ^2(c)} \sin ^2(c+d x)+6 \csc (c) \csc (d x-\arctan (\cot (c))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x-\arctan (\cot (c)))\right ) \sqrt {\sin ^2(d x-\arctan (\cot (c)))}+3 \csc (c) \sec (c) (\sin (c+d x-\arctan (\cot (c)))+3 \sin (c-d x+\arctan (\cot (c))))-3 \sqrt {\csc ^2(c)} \sin (c+d x) (\sin (2 (c+d x))+4 \tan (c))\right )}{15 a d \sqrt {\csc ^2(c)} (1+\sec (c+d x)) \sqrt {e \sin (c+d x)}} \]

[In]

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

[Out]

(2*e^3*Cos[(c + d*x)/2]^2*Sec[c + d*x]*(10*Sqrt[Csc[c]^2]*Sin[c + d*x]^2 + 6*Csc[c]*Csc[d*x - ArcTan[Cot[c]]]*
HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Sin[d*x - ArcTan[Cot[c]]]^2] + 3*Csc[
c]*Sec[c]*(Sin[c + d*x - ArcTan[Cot[c]]] + 3*Sin[c - d*x + ArcTan[Cot[c]]]) - 3*Sqrt[Csc[c]^2]*Sin[c + d*x]*(S
in[2*(c + d*x)] + 4*Tan[c])))/(15*a*d*Sqrt[Csc[c]^2]*(1 + Sec[c + d*x])*Sqrt[e*Sin[c + d*x]])

Maple [A] (verified)

Time = 5.36 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.66

method	result	size
default	\(\frac {2 e^{3} \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+3 \cos \left (d x +c \right )^{4}-5 \cos \left (d x +c \right )^{3}-3 \cos \left (d x +c \right )^{2}+5 \cos \left (d x +c \right )\right )}{15 a \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(173\)

[In]

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

[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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.07 \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=-\frac {2 \, {\left (3 i \, \sqrt {2} \sqrt {-i \, e} e^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} \sqrt {i \, e} e^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + {\left (3 \, e^{2} \cos \left (d x + c\right ) - 5 \, e^{2}\right )} \sqrt {e \sin \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, a d} \]

[In]

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

[Out]

-2/15*(3*I*sqrt(2)*sqrt(-I*e)*e^2*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c
))) - 3*I*sqrt(2)*sqrt(I*e)*e^2*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))
) + (3*e^2*cos(d*x + c) - 5*e^2)*sqrt(e*sin(d*x + c))*sin(d*x + c))/(a*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]

[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)

Giac [F]

\[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]

[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)