3.7.61 \(\int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx\) [661]

Optimal. Leaf size=96 \[ \frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{3 d (e \cos (c+d x))^{5/2}} \]

[Out]

2/5*I*a/d/(e*cos(d*x+c))^(5/2)+2/3*a*cos(d*x+c)^(5/2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
F(sin(1/2*d*x+1/2*c),2^(1/2))/d/(e*cos(d*x+c))^(5/2)+2/3*a*cos(d*x+c)*sin(d*x+c)/d/(e*cos(d*x+c))^(5/2)

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3596, 3567, 3853, 3856, 2720} \begin {gather*} \frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d (e \cos (c+d x))^{5/2}}+\frac {2 a \sin (c+d x) \cos (c+d x)}{3 d (e \cos (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])/(e*Cos[c + d*x])^(5/2),x]

[Out]

(((2*I)/5)*a)/(d*(e*Cos[c + d*x])^(5/2)) + (2*a*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2])/(3*d*(e*Cos[c +
d*x])^(5/2)) + (2*a*Cos[c + d*x]*Sin[c + d*x])/(3*d*(e*Cos[c + d*x])^(5/2))

Rule 2720

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

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rule 3853

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

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx &=\frac {\int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {a \int (e \sec (c+d x))^{5/2} \, dx}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{3 d (e \cos (c+d x))^{5/2}}+\frac {\left (a e^2\right ) \int \sqrt {e \sec (c+d x)} \, dx}{3 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{3 d (e \cos (c+d x))^{5/2}}+\frac {\left (a \cos ^{\frac {5}{2}}(c+d x)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 (e \cos (c+d x))^{5/2}}\\ &=\frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{3 d (e \cos (c+d x))^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.38, size = 57, normalized size = 0.59 \begin {gather*} \frac {a \left (6 i+10 \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \sin (2 (c+d x))\right )}{15 d (e \cos (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])/(e*Cos[c + d*x])^(5/2),x]

[Out]

(a*(6*I + 10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 5*Sin[2*(c + d*x)]))/(15*d*(e*Cos[c + d*x])^(5/2))

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (107 ) = 214\).
time = 2.14, size = 283, normalized size = 2.95

method	result	size
default	\(-\frac {2 \left (20 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-20 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{15 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\)	\(283\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))/(e*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/15/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)
/e^2*(20*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*s
in(1/2*d*x+1/2*c)^4+20*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-20*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(c
os(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-10*sin(1/2*d*x+1/2*c)^2*cos(1/2*d
*x+1/2*c)+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)-3*I*sin(1/2*d*x+1/2*c))*a/d

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-5/2)*integrate((I*a*tan(d*x + c) + a)/cos(d*x + c)^(5/2), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.08, size = 178, normalized size = 1.85 \begin {gather*} -\frac {2 \, {\left (2 \, \sqrt {\frac {1}{2}} {\left (5 i \, a e^{\left (5 i \, d x + 5 i \, c\right )} - 12 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 5 i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 5 \, {\left (i \, \sqrt {2} a e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, \sqrt {2} a e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, \sqrt {2} a e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {2} a\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{15 \, {\left (d e^{\frac {5}{2}} + d e^{\left (6 i \, d x + 6 i \, c + \frac {5}{2}\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c + \frac {5}{2}\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(2*sqrt(1/2)*(5*I*a*e^(5*I*d*x + 5*I*c) - 12*I*a*e^(3*I*d*x + 3*I*c) - 5*I*a*e^(I*d*x + I*c))*sqrt(e^(2*
I*d*x + 2*I*c) + 1)*e^(-1/2*I*d*x - 1/2*I*c) + 5*(I*sqrt(2)*a*e^(6*I*d*x + 6*I*c) + 3*I*sqrt(2)*a*e^(4*I*d*x +
4*I*c) + 3*I*sqrt(2)*a*e^(2*I*d*x + 2*I*c) + I*sqrt(2)*a)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))/(d*e^(
5/2) + d*e^(6*I*d*x + 6*I*c + 5/2) + 3*d*e^(4*I*d*x + 4*I*c + 5/2) + 3*d*e^(2*I*d*x + 2*I*c + 5/2))

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))/(e*cos(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((I*a*tan(d*x + c) + a)*e^(-5/2)/cos(d*x + c)^(5/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________