3.343 \(\int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=35 \[ \frac {2 i a \sec ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

2/3*I*a*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.05, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {3493} \[ \frac {2 i a \sec ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((2*I)/3)*a*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2))

Rule 3493

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

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {2 i a \sec ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 40, normalized size = 1.14 \[ \frac {2 (\tan (c+d x)+i) \sec (c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(2*Sec[c + d*x]*(I + Tan[c + d*x]))/(3*d*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [A]  time = 0.71, size = 40, normalized size = 1.14 \[ \frac {4 i \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3 \, {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

4/3*I*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(a*d*e^(2*I*d*x + 2*I*c) + a*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{3}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^3/sqrt(I*a*tan(d*x + c) + a), x)

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maple [B]  time = 1.18, size = 73, normalized size = 2.09 \[ \frac {2 \left (2 i \left (\cos ^{2}\left (d x +c \right )\right )+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )-i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{3 d \cos \left (d x +c \right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

2/3/d*(2*I*cos(d*x+c)^2+2*cos(d*x+c)*sin(d*x+c)-I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)/a

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maxima [B]  time = 0.69, size = 206, normalized size = 5.89 \[ -\frac {2 \, {\left (-i \, \sqrt {a} - \frac {2 \, \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {i \, \sqrt {a} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1}}{3 \, {\left (a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} d \sqrt {-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-2/3*(-I*sqrt(a) - 2*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 2*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 +
I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*sqrt(sin(d*x + c)/(cos(d*x + c) + 1) + 1)*sqrt(sin(d*x + c)/(c
os(d*x + c) + 1) - 1)/((a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*d
*sqrt(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1))

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mupad [B]  time = 1.02, size = 98, normalized size = 2.80 \[ \frac {2\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,1{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}\right )}{3\,a\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(1/2)),x)

[Out]

(2*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos(c + d*x)*1i + sin(c +
d*x) + cos(3*c + 3*d*x)*1i + sin(3*c + 3*d*x)))/(3*a*d*(cos(2*c + 2*d*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**3/sqrt(I*a*(tan(c + d*x) - I)), x)

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