3.422 \(\int \frac{1}{(e \sec (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=206 \[ \frac{256 i \sqrt{e \sec (c+d x)}}{315 d e^4 \sqrt{a+i a \tan (c+d x)}}-\frac{128 i \sqrt{a+i a \tan (c+d x)}}{315 a d e^2 (e \sec (c+d x))^{3/2}}+\frac{32 i}{105 d e^2 \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}+\frac{2 i}{9 d \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}} \]

[Out]

((2*I)/9)/(d*(e*Sec[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]) + ((32*I)/105)/(d*e^2*(e*Sec[c + d*x])^(3/2)*S
qrt[a + I*a*Tan[c + d*x]]) + (((256*I)/315)*Sqrt[e*Sec[c + d*x]])/(d*e^4*Sqrt[a + I*a*Tan[c + d*x]]) - (((16*I
)/63)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*(e*Sec[c + d*x])^(7/2)) - (((128*I)/315)*Sqrt[a + I*a*Tan[c + d*x]])/(a
*d*e^2*(e*Sec[c + d*x])^(3/2))

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Rubi [A]  time = 0.387039, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3502, 3497, 3488} \[ \frac{256 i \sqrt{e \sec (c+d x)}}{315 d e^4 \sqrt{a+i a \tan (c+d x)}}-\frac{128 i \sqrt{a+i a \tan (c+d x)}}{315 a d e^2 (e \sec (c+d x))^{3/2}}+\frac{32 i}{105 d e^2 \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}+\frac{2 i}{9 d \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((2*I)/9)/(d*(e*Sec[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]) + ((32*I)/105)/(d*e^2*(e*Sec[c + d*x])^(3/2)*S
qrt[a + I*a*Tan[c + d*x]]) + (((256*I)/315)*Sqrt[e*Sec[c + d*x]])/(d*e^4*Sqrt[a + I*a*Tan[c + d*x]]) - (((16*I
)/63)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*(e*Sec[c + d*x])^(7/2)) - (((128*I)/315)*Sqrt[a + I*a*Tan[c + d*x]])/(a
*d*e^2*(e*Sec[c + d*x])^(3/2))

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))^{7/2} \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)}}+\frac{8 \int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx}{9 a}\\ &=\frac{2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}+\frac{16 \int \frac{1}{(e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}} \, dx}{21 e^2}\\ &=\frac{2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)}}+\frac{32 i}{105 d e^2 (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}+\frac{64 \int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{105 a e^2}\\ &=\frac{2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)}}+\frac{32 i}{105 d e^2 (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}-\frac{128 i \sqrt{a+i a \tan (c+d x)}}{315 a d e^2 (e \sec (c+d x))^{3/2}}+\frac{128 \int \frac{\sqrt{e \sec (c+d x)}}{\sqrt{a+i a \tan (c+d x)}} \, dx}{315 e^4}\\ &=\frac{2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)}}+\frac{32 i}{105 d e^2 (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}+\frac{256 i \sqrt{e \sec (c+d x)}}{315 d e^4 \sqrt{a+i a \tan (c+d x)}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}-\frac{128 i \sqrt{a+i a \tan (c+d x)}}{315 a d e^2 (e \sec (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.367011, size = 87, normalized size = 0.42 \[ \frac{\sqrt{e \sec (c+d x)} (336 \sin (2 (c+d x))+40 \sin (4 (c+d x))-84 i \cos (2 (c+d x))-5 i \cos (4 (c+d x))+945 i)}{1260 d e^4 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

(Sqrt[e*Sec[c + d*x]]*(945*I - (84*I)*Cos[2*(c + d*x)] - (5*I)*Cos[4*(c + d*x)] + 336*Sin[2*(c + d*x)] + 40*Si
n[4*(c + d*x)]))/(1260*d*e^4*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.352, size = 132, normalized size = 0.6 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{4} \left ( 35\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}+35\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+8\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+48\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +64\,i\cos \left ( dx+c \right ) +128\,\sin \left ( dx+c \right ) \right ) }{315\,ad{e}^{7}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{7}{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))^(7/2)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

2/315/d/a*cos(d*x+c)^4*(e/cos(d*x+c))^(7/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(35*I*cos(d*x+c)^5+
35*sin(d*x+c)*cos(d*x+c)^4+8*I*cos(d*x+c)^3+48*cos(d*x+c)^2*sin(d*x+c)+64*I*cos(d*x+c)+128*sin(d*x+c))/e^7

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Maxima [A]  time = 2.02318, size = 305, normalized size = 1.48 \begin{align*} \frac{35 i \, \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) - 45 i \, \cos \left (\frac{7}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 252 i \, \cos \left (\frac{5}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) - 420 i \, \cos \left (\frac{1}{3} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 1890 i \, \cos \left (\frac{1}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 45 \, \sin \left (\frac{7}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 252 \, \sin \left (\frac{5}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 420 \, \sin \left (\frac{1}{3} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 1890 \, \sin \left (\frac{1}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right )}{2520 \, \sqrt{a} d e^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/2520*(35*I*cos(9/2*d*x + 9/2*c) - 45*I*cos(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 252*I*
cos(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 420*I*cos(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos
(9/2*d*x + 9/2*c))) + 1890*I*cos(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 35*sin(9/2*d*x + 9
/2*c) + 45*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 252*sin(5/9*arctan2(sin(9/2*d*x + 9/
2*c), cos(9/2*d*x + 9/2*c))) + 420*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 1890*sin(1/9
*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))/(sqrt(a)*d*e^(7/2))

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Fricas [A]  time = 2.19531, size = 354, normalized size = 1.72 \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 (-45 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 465 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1470 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 2142 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 287 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 35 i\right )} e^{\left (-\frac{9}{2} i \, d x - \frac{9}{2} i \, c\right )}}{2520 \, a d e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/2520*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-45*I*e^(10*I*d*x + 10*I*c) - 465*
I*e^(8*I*d*x + 8*I*c) + 1470*I*e^(6*I*d*x + 6*I*c) + 2142*I*e^(4*I*d*x + 4*I*c) + 287*I*e^(2*I*d*x + 2*I*c) +
35*I)*e^(-9/2*I*d*x - 9/2*I*c)/(a*d*e^4)

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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))**(7/2)/(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{7}{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))^(7/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/((e*sec(d*x + c))^(7/2)*sqrt(I*a*tan(d*x + c) + a)), x)