3.321 \(\int \sec ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=59 \[ \frac{2 i (a+i a \tan (c+d x))^{13/2}}{13 a^3 d}-\frac{4 i (a+i a \tan (c+d x))^{11/2}}{11 a^2 d} \]

[Out]

(((-4*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^2*d) + (((2*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^3*d)

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Rubi [A]  time = 0.0679677, 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} \[ \frac{2 i (a+i a \tan (c+d x))^{13/2}}{13 a^3 d}-\frac{4 i (a+i a \tan (c+d x))^{11/2}}{11 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((-4*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^2*d) + (((2*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(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))^{7/2} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x) (a+x)^{9/2} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (2 a (a+x)^{9/2}-(a+x)^{11/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{4 i (a+i a \tan (c+d x))^{11/2}}{11 a^2 d}+\frac{2 i (a+i a \tan (c+d x))^{13/2}}{13 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.590619, size = 85, normalized size = 1.44 \[ -\frac{2 a^3 (11 \tan (c+d x)+15 i) \sec ^5(c+d x) \sqrt{a+i a \tan (c+d x)} (\cos (5 c+8 d x)+i \sin (5 c+8 d x))}{143 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(-2*a^3*Sec[c + d*x]^5*(Cos[5*c + 8*d*x] + I*Sin[5*c + 8*d*x])*(15*I + 11*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d
*x]])/(143*d*(Cos[d*x] + I*Sin[d*x])^3)

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Maple [B]  time = 0.365, size = 127, normalized size = 2.2 \begin{align*}{\frac{2\,{a}^{3} \left ( -64\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+64\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) -8\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+40\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +68\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-40\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -11\,i \right ) }{143\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}\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))^(7/2),x)

[Out]

2/143/d*a^3*(-64*I*cos(d*x+c)^6+64*cos(d*x+c)^5*sin(d*x+c)-8*I*cos(d*x+c)^4+40*cos(d*x+c)^3*sin(d*x+c)+68*I*co
s(d*x+c)^2-40*cos(d*x+c)*sin(d*x+c)-11*I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^6

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Maxima [A]  time = 1.09262, size = 54, normalized size = 0.92 \begin{align*} \frac{2 i \,{\left (11 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{13}{2}} - 26 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{11}{2}} a\right )}}{143 \, 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))^(7/2),x, algorithm="maxima")

[Out]

2/143*I*(11*(I*a*tan(d*x + c) + a)^(13/2) - 26*(I*a*tan(d*x + c) + a)^(11/2)*a)/(a^3*d)

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Fricas [B]  time = 1.98042, size = 405, normalized size = 6.86 \begin{align*} \frac{\sqrt{2}{\left (-256 i \, a^{3} e^{\left (12 i \, d x + 12 i \, c\right )} - 1664 i \, a^{3} e^{\left (10 i \, d x + 10 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 )}}{143 \,{\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, 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))^(7/2),x, algorithm="fricas")

[Out]

1/143*sqrt(2)*(-256*I*a^3*e^(12*I*d*x + 12*I*c) - 1664*I*a^3*e^(10*I*d*x + 10*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c
) + 1))*e^(I*d*x + I*c)/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) + 20*d
*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*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))**(7/2),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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(7/2)*sec(d*x + c)^4, x)