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

Optimal. Leaf size=164 \[ \frac{32 i a \sqrt{e \sec (c+d x)}}{35 d e^4 \sqrt{a+i a \tan (c+d x)}}+\frac{12 i a}{35 d e^2 \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{35 d e^2 (e \sec (c+d x))^{3/2}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}} \]

[Out]

(((12*I)/35)*a)/(d*e^2*(e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (((32*I)/35)*a*Sqrt[e*Sec[c + d*x]
])/(d*e^4*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/7)*Sqrt[a + I*a*Tan[c + d*x]])/(d*(e*Sec[c + d*x])^(7/2)) - ((
(16*I)/35)*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^2*(e*Sec[c + d*x])^(3/2))

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Rubi [A]  time = 0.282532, antiderivative size = 164, 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 = {3497, 3502, 3488} \[ \frac{32 i a \sqrt{e \sec (c+d x)}}{35 d e^4 \sqrt{a+i a \tan (c+d x)}}+\frac{12 i a}{35 d e^2 \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{35 d e^2 (e \sec (c+d x))^{3/2}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((12*I)/35)*a)/(d*e^2*(e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (((32*I)/35)*a*Sqrt[e*Sec[c + d*x]
])/(d*e^4*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/7)*Sqrt[a + I*a*Tan[c + d*x]])/(d*(e*Sec[c + d*x])^(7/2)) - ((
(16*I)/35)*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^2*(e*Sec[c + d*x])^(3/2))

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 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 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{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx &=-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}+\frac{(6 a) \int \frac{1}{(e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}} \, dx}{7 e^2}\\ &=\frac{12 i a}{35 d e^2 (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}+\frac{24 \int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{35 e^2}\\ &=\frac{12 i a}{35 d e^2 (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{35 d e^2 (e \sec (c+d x))^{3/2}}+\frac{(16 a) \int \frac{\sqrt{e \sec (c+d x)}}{\sqrt{a+i a \tan (c+d x)}} \, dx}{35 e^4}\\ &=\frac{12 i a}{35 d e^2 (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}+\frac{32 i a \sqrt{e \sec (c+d x)}}{35 d e^4 \sqrt{a+i a \tan (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{35 d e^2 (e \sec (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.218766, size = 80, normalized size = 0.49 \[ \frac{\sqrt{a+i a \tan (c+d x)} (70 \sin (c+d x)+6 \sin (3 (c+d x))+35 i \cos (c+d x)+i \cos (3 (c+d x)))}{70 d e^3 \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((35*I)*Cos[c + d*x] + I*Cos[3*(c + d*x)] + 70*Sin[c + d*x] + 6*Sin[3*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])
/(70*d*e^3*Sqrt[e*Sec[c + d*x]])

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Maple [A]  time = 0.378, size = 102, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+12\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +16\,i\cos \left ( dx+c \right ) +32\,\sin \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{35\,d{e}^{7}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35/d*(I*cos(d*x+c)^3+6*cos(d*x+c)^2*sin(d*x+c)+8*I*cos(d*x+c)+16*sin(d*x+c))*cos(d*x+c)^4*(a*(I*sin(d*x+c)+c
os(d*x+c))/cos(d*x+c))^(1/2)*(e/cos(d*x+c))^(7/2)/e^7

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Maxima [A]  time = 1.94961, size = 240, normalized size = 1.46 \begin{align*} \frac{\sqrt{a}{\left (7 i \, \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) - 5 i \, \cos \left (\frac{7}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) - 35 i \, \cos \left (\frac{3}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 105 i \, \cos \left (\frac{1}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 7 \, \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, \sin \left (\frac{7}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac{3}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 105 \, \sin \left (\frac{1}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right )\right )}}{140 \, d e^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.09596, size = 301, 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 (-5 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 40 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 112 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (-\frac{5}{2} i \, d x - \frac{5}{2} i \, c\right )}}{140 \, d e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(7/2),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))*(-5*I*e^(8*I*d*x + 8*I*c) - 40*I*e^(
6*I*d*x + 6*I*c) + 70*I*e^(4*I*d*x + 4*I*c) + 112*I*e^(2*I*d*x + 2*I*c) + 7*I)*e^(-5/2*I*d*x - 5/2*I*c)/(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((a+I*a*tan(d*x+c))**(1/2)/(e*sec(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{i \, a \tan \left (d x + c\right ) + a}}{\left (e \sec \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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