3.421 \(\int \frac{1}{(e \sec (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=165 \[ -\frac{32 i \sqrt{a+i a \tan (c+d x)}}{35 a d e^2 \sqrt{e \sec (c+d x)}}+\frac{16 i}{35 d e^2 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{12 i \sqrt{a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}+\frac{2 i}{7 d \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}} \]

[Out]

((2*I)/7)/(d*(e*Sec[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) + ((16*I)/35)/(d*e^2*Sqrt[e*Sec[c + d*x]]*Sqrt
[a + I*a*Tan[c + d*x]]) - (((12*I)/35)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*(e*Sec[c + d*x])^(5/2)) - (((32*I)/35)
*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*e^2*Sqrt[e*Sec[c + d*x]])

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Rubi [A]  time = 0.290697, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3502, 3497, 3488} \[ -\frac{32 i \sqrt{a+i a \tan (c+d x)}}{35 a d e^2 \sqrt{e \sec (c+d x)}}+\frac{16 i}{35 d e^2 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{12 i \sqrt{a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}+\frac{2 i}{7 d \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((2*I)/7)/(d*(e*Sec[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) + ((16*I)/35)/(d*e^2*Sqrt[e*Sec[c + d*x]]*Sqrt
[a + I*a*Tan[c + d*x]]) - (((12*I)/35)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*(e*Sec[c + d*x])^(5/2)) - (((32*I)/35)
*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*e^2*Sqrt[e*Sec[c + d*x]])

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rule 3497

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d*
Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(a*f*m), x] + Dist[(a*(m + n))/(m*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

Rule 3488

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

Rubi steps

\begin{align*} \int \frac{1}{(e \sec (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}+\frac{6 \int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx}{7 a}\\ &=\frac{2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}-\frac{12 i \sqrt{a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}+\frac{24 \int \frac{1}{\sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx}{35 e^2}\\ &=\frac{2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}+\frac{16 i}{35 d e^2 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{12 i \sqrt{a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}+\frac{16 \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}} \, dx}{35 a e^2}\\ &=\frac{2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}+\frac{16 i}{35 d e^2 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{12 i \sqrt{a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}-\frac{32 i \sqrt{a+i a \tan (c+d x)}}{35 a d e^2 \sqrt{e \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.339108, size = 79, normalized size = 0.48 \[ -\frac{i (\cos (2 (c+d x))+35 i \tan (c+d x)+3 i \sin (3 (c+d x)) \sec (c+d x)+17)}{35 d e^2 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((-I/35)*(17 + Cos[2*(c + d*x)] + (3*I)*Sec[c + d*x]*Sin[3*(c + d*x)] + (35*I)*Tan[c + d*x]))/(d*e^2*Sqrt[e*Se
c[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.332, size = 115, normalized size = 0.7 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3} \left ( 5\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +2\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+8\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -16\,i \right ) }{35\,ad{e}^{5}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{5}{2}}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35/d/a*cos(d*x+c)^3*(e/cos(d*x+c))^(5/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(5*I*cos(d*x+c)^4+5*
cos(d*x+c)^3*sin(d*x+c)+2*I*cos(d*x+c)^2+8*cos(d*x+c)*sin(d*x+c)-16*I)/e^5

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Maxima [A]  time = 1.9702, size = 240, normalized size = 1.45 \begin{align*} \frac{5 i \, \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) - 7 i \, \cos \left (\frac{5}{7} \, \arctan \left (\sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ), \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )\right )\right ) + 35 i \, \cos \left (\frac{3}{7} \, \arctan \left (\sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ), \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )\right )\right ) - 105 i \, \cos \left (\frac{1}{7} \, \arctan \left (\sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ), \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )\right )\right ) + 5 \, \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 7 \, \sin \left (\frac{5}{7} \, \arctan \left (\sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ), \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac{3}{7} \, \arctan \left (\sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ), \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )\right )\right ) + 105 \, \sin \left (\frac{1}{7} \, \arctan \left (\sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ), \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )\right )\right )}{140 \, \sqrt{a} d e^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/140*(5*I*cos(7/2*d*x + 7/2*c) - 7*I*cos(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 35*I*cos(
3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 105*I*cos(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2
*d*x + 7/2*c))) + 5*sin(7/2*d*x + 7/2*c) + 7*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 35
*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 105*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(
7/2*d*x + 7/2*c))))/(sqrt(a)*d*e^(5/2))

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Fricas [A]  time = 2.08237, size = 304, normalized size = 1.84 \begin{align*} \frac{\sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-7 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 112 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-\frac{7}{2} i \, d x - \frac{7}{2} i \, c\right )}}{140 \, a d e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/140*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-7*I*e^(8*I*d*x + 8*I*c) - 112*I*e^
(6*I*d*x + 6*I*c) - 70*I*e^(4*I*d*x + 4*I*c) + 40*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-7/2*I*d*x - 7/2*I*c)/(a*d*e
^3)

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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(1/(e*sec(d*x+c))**(5/2)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/((e*sec(d*x + c))^(5/2)*sqrt(I*a*tan(d*x + c) + a)), x)