3.365 \(\int \frac{\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=55 \[ \frac{2 i \sqrt{a+i a \tan (c+d x)}}{a^3 d}+\frac{4 i}{a^2 d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0701862, antiderivative size = 55, 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} \[ \frac{2 i \sqrt{a+i a \tan (c+d x)}}{a^3 d}+\frac{4 i}{a^2 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(4*I)/(a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + ((2*I)*Sqrt[a + I*a*Tan[c + d*x]])/(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))^{5/2}} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{a-x}{(a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (\frac{2 a}{(a+x)^{3/2}}-\frac{1}{\sqrt{a+x}}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=\frac{4 i}{a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{2 i \sqrt{a+i a \tan (c+d x)}}{a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.254093, size = 36, normalized size = 0.65 \[ \frac{-2 \tan (c+d x)+6 i}{a^2 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

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

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Maple [A]  time = 0.266, size = 65, normalized size = 1.2 \begin{align*} 2\,{\frac{2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +i+2\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d{a}^{3}}\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))^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.11321, size = 59, normalized size = 1.07 \begin{align*} \frac{2 i \,{\left (\frac{\sqrt{i \, a \tan \left (d x + c\right ) + a}}{a^{2}} + \frac{2}{\sqrt{i \, a \tan \left (d x + c\right ) + a} a}\right )}}{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))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.14453, size = 135, normalized size = 2.45 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-i \, d x - i \, c\right )}}{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))^(5/2),x, algorithm="fricas")

[Out]

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

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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(sec(d*x+c)**4/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/(I*a*tan(d*x + c) + a)^(5/2), x)