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

Optimal. Leaf size=59 \[ \frac{2 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d}-\frac{4 i (a+i a \tan (c+d x))^{9/2}}{9 a^2 d} \]

[Out]

(((-4*I)/9)*(a + I*a*Tan[c + d*x])^(9/2))/(a^2*d) + (((2*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^3*d)

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Rubi [A]  time = 0.0705784, 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} \[ \frac{2 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d}-\frac{4 i (a+i a \tan (c+d x))^{9/2}}{9 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-4*I)/9)*(a + I*a*Tan[c + d*x])^(9/2))/(a^2*d) + (((2*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(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 \sec ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x) (a+x)^{7/2} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (2 a (a+x)^{7/2}-(a+x)^{9/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{4 i (a+i a \tan (c+d x))^{9/2}}{9 a^2 d}+\frac{2 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.531602, size = 85, normalized size = 1.44 \[ -\frac{2 a^2 (9 \tan (c+d x)+13 i) \sec ^4(c+d x) \sqrt{a+i a \tan (c+d x)} (\cos (4 c+6 d x)+i \sin (4 c+6 d x))}{99 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Sec[c + d*x]^4*(Cos[4*c + 6*d*x] + I*Sin[4*c + 6*d*x])*(13*I + 9*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*
x]])/(99*d*(Cos[d*x] + I*Sin[d*x])^2)

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Maple [B]  time = 0.327, size = 117, normalized size = 2. \begin{align*} -{\frac{2\,{a}^{2} \left ( 32\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}-32\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-20\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -23\,i\cos \left ( dx+c \right ) +9\,\sin \left ( dx+c \right ) \right ) }{99\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}\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]

-2/99/d*a^2*(32*I*cos(d*x+c)^5-32*sin(d*x+c)*cos(d*x+c)^4+4*I*cos(d*x+c)^3-20*cos(d*x+c)^2*sin(d*x+c)-23*I*cos
(d*x+c)+9*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^5

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Maxima [A]  time = 1.07692, size = 54, normalized size = 0.92 \begin{align*} \frac{2 i \,{\left (9 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{11}{2}} - 22 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{9}{2}} a\right )}}{99 \, 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="maxima")

[Out]

2/99*I*(9*(I*a*tan(d*x + c) + a)^(11/2) - 22*(I*a*tan(d*x + c) + a)^(9/2)*a)/(a^3*d)

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Fricas [B]  time = 2.61219, size = 360, normalized size = 6.1 \begin{align*} \frac{\sqrt{2}{\left (-128 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} - 704 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}}{99 \,{\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 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))^(5/2),x, algorithm="fricas")

[Out]

1/99*sqrt(2)*(-128*I*a^2*e^(10*I*d*x + 10*I*c) - 704*I*a^2*e^(8*I*d*x + 8*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*e^(I*d*x + I*c)/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*
I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + 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{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{4}\,{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((I*a*tan(d*x + c) + a)^(5/2)*sec(d*x + c)^4, x)