3.349 \(\int \frac{\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=57 \[ \frac{2 i (a+i a \tan (c+d x))^{3/2}}{3 a^3 d}-\frac{4 i \sqrt{a+i a \tan (c+d x)}}{a^2 d} \]

[Out]

((-4*I)*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d) + (((2*I)/3)*(a + I*a*Tan[c + d*x])^(3/2))/(a^3*d)

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Rubi [A]  time = 0.0716966, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3487, 43} \[ \frac{2 i (a+i a \tan (c+d x))^{3/2}}{3 a^3 d}-\frac{4 i \sqrt{a+i a \tan (c+d x)}}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((-4*I)*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d) + (((2*I)/3)*(a + I*a*Tan[c + d*x])^(3/2))/(a^3*d)

Rule 3487

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]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{a-x}{\sqrt{a+x}} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (\frac{2 a}{\sqrt{a+x}}-\sqrt{a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{4 i \sqrt{a+i a \tan (c+d x)}}{a^2 d}+\frac{2 i (a+i a \tan (c+d x))^{3/2}}{3 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.191338, size = 80, normalized size = 1.4 \[ \frac{2 i (\tan (c+d x)+5 i) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))}{3 a d (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((2*I)/3)*Sec[c + d*x]^2*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*(5*I + Tan[c + d*x]))/(a*d*(-I + Tan[c + d*x
])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.262, size = 61, normalized size = 1.1 \begin{align*} -{\frac{10\,i\cos \left ( dx+c \right ) +2\,\sin \left ( dx+c \right ) }{3\,{a}^{2}d\cos \left ( dx+c \right ) }\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(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

-2/3/d/a^2*(5*I*cos(d*x+c)+sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)

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Maxima [A]  time = 1.05252, size = 51, normalized size = 0.89 \begin{align*} \frac{2 i \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} - 6 \, \sqrt{i \, a \tan \left (d x + c\right ) + a} a\right )}}{3 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.07209, size = 180, normalized size = 3.16 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 12 i\right )} e^{\left (i \, d x + i \, c\right )}}{3 \,{\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/3*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-8*I*e^(2*I*d*x + 2*I*c) - 12*I)*e^(I*d*x + I*c)/(a^2*d*e^(2*I*
d*x + 2*I*c) + a^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\left (c + d x \right )}}{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**4/(a*(I*tan(c + d*x) + 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/(I*a*tan(d*x + c) + a)^(3/2), x)