3.4.66 \(\int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [366]

Optimal. Leaf size=29 \[ \frac {2 i}{3 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 32} \begin {gather*} \frac {2 i}{3 a d (a+i a \tan (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3568

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

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=-\frac {i \text {Subst}\left (\int \frac {1}{(a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{a d}\\ &=\frac {2 i}{3 a d (a+i a \tan (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 39, normalized size = 1.34 \begin {gather*} \frac {2}{3 a^2 d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.23, size = 24, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {2 i}{3 a d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\)	\(24\)
default	\(\frac {2 i}{3 a d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 21, normalized size = 0.72 \begin {gather*} \frac {2 i}{3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (21) = 42\).
time = 0.38, size = 61, normalized size = 2.10 \begin {gather*} \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{6 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(I*e^(4*I*d*x + 4*I*c) + 2*I*e^(2*I*d*x + 2*I*c) + I)*e^(-3*I*d*
x - 3*I*c)/(a^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)**2/(I*a*(tan(c + d*x) - I))**(5/2), x)

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

[Out]

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

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Mupad [B]
time = 3.58, size = 23, normalized size = 0.79 \begin {gather*} \frac {2{}\mathrm {i}}{3\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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