3.335 \(\int \frac{\sec ^4(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=59 \[ \frac{2 i (a+i a \tan (c+d x))^{5/2}}{5 a^3 d}-\frac{4 i (a+i a \tan (c+d x))^{3/2}}{3 a^2 d} \]

[Out]

(((-4*I)/3)*(a + I*a*Tan[c + d*x])^(3/2))/(a^2*d) + (((2*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(a^3*d)

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Rubi [A]  time = 0.0643235, antiderivative size = 59, 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} \[ \frac{2 i (a+i a \tan (c+d x))^{5/2}}{5 a^3 d}-\frac{4 i (a+i a \tan (c+d x))^{3/2}}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((-4*I)/3)*(a + I*a*Tan[c + d*x])^(3/2))/(a^2*d) + (((2*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(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)}{\sqrt{a+i a \tan (c+d x)}} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x) \sqrt{a+x} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (2 a \sqrt{a+x}-(a+x)^{3/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{4 i (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}+\frac{2 i (a+i a \tan (c+d x))^{5/2}}{5 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.175313, size = 65, normalized size = 1.1 \[ -\frac{2 (3 \tan (c+d x)+7 i) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))}{15 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(-2*Sec[c + d*x]^2*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*(7*I + 3*Tan[c + d*x]))/(15*d*Sqrt[a + I*a*Tan[c +
d*x]])

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Maple [A]  time = 0.319, size = 73, normalized size = 1.2 \begin{align*} -{\frac{8\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-8\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +6\,i}{15\,ad \left ( \cos \left ( dx+c \right ) \right ) ^{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(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

-2/15/d/a*(4*I*cos(d*x+c)^2-4*cos(d*x+c)*sin(d*x+c)+3*I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*
x+c)^2

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Maxima [A]  time = 1.09604, size = 107, normalized size = 1.81 \begin{align*} -\frac{2 i \,{\left (15 \, \sqrt{i \, a \tan \left (d x + c\right ) + a} - \frac{3 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} - 10 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a + 15 \, \sqrt{i \, a \tan \left (d x + c\right ) + a} a^{2}}{a^{2}}\right )}}{15 \, a 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))^(1/2),x, algorithm="maxima")

[Out]

-2/15*I*(15*sqrt(I*a*tan(d*x + c) + a) - (3*(I*a*tan(d*x + c) + a)^(5/2) - 10*(I*a*tan(d*x + c) + a)^(3/2)*a +
15*sqrt(I*a*tan(d*x + c) + a)*a^2)/a^2)/(a*d)

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Fricas [A]  time = 1.91261, size = 242, normalized size = 4.1 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-16 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 40 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (i \, d x + i \, c\right )}}{15 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a 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))^(1/2),x, algorithm="fricas")

[Out]

1/15*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-16*I*e^(4*I*d*x + 4*I*c) - 40*I*e^(2*I*d*x + 2*I*c))*e^(I*d*x
+ I*c)/(a*d*e^(4*I*d*x + 4*I*c) + 2*a*d*e^(2*I*d*x + 2*I*c) + a*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 )}}{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )}}\, 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))**(1/2),x)

[Out]

Integral(sec(c + d*x)**4/sqrt(a*(I*tan(c + d*x) + 1)), 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}}{\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(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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