\(\int \sec ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) [320]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {8 i (a+i a \tan (c+d x))^{13/2}}{13 a^3 d}+\frac {8 i (a+i a \tan (c+d x))^{15/2}}{15 a^4 d}-\frac {2 i (a+i a \tan (c+d x))^{17/2}}{17 a^5 d} \]

[Out]

-8/13*I*(a+I*a*tan(d*x+c))^(13/2)/a^3/d+8/15*I*(a+I*a*tan(d*x+c))^(15/2)/a^4/d-2/17*I*(a+I*a*tan(d*x+c))^(17/2
)/a^5/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 45} \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{17/2}}{17 a^5 d}+\frac {8 i (a+i a \tan (c+d x))^{15/2}}{15 a^4 d}-\frac {8 i (a+i a \tan (c+d x))^{13/2}}{13 a^3 d} \]

[In]

Int[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((-8*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^3*d) + (((8*I)/15)*(a + I*a*Tan[c + d*x])^(15/2))/(a^4*d) - (((
2*I)/17)*(a + I*a*Tan[c + d*x])^(17/2))/(a^5*d)

Rule 45

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])

Rule 3568

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int (a-x)^2 (a+x)^{11/2} \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {i \text {Subst}\left (\int \left (4 a^2 (a+x)^{11/2}-4 a (a+x)^{13/2}+(a+x)^{15/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {8 i (a+i a \tan (c+d x))^{13/2}}{13 a^3 d}+\frac {8 i (a+i a \tan (c+d x))^{15/2}}{15 a^4 d}-\frac {2 i (a+i a \tan (c+d x))^{17/2}}{17 a^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {2 a^3 (-i+\tan (c+d x))^6 \sqrt {a+i a \tan (c+d x)} \left (331 i+494 \tan (c+d x)-195 i \tan ^2(c+d x)\right )}{3315 d} \]

[In]

Integrate[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(2*a^3*(-I + Tan[c + d*x])^6*Sqrt[a + I*a*Tan[c + d*x]]*(331*I + 494*Tan[c + d*x] - (195*I)*Tan[c + d*x]^2))/(
3315*d)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72

\[\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {17}{2}}}{17}+\frac {4 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {15}{2}}}{15}-\frac {4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}\right )}{d \,a^{5}}\]

[In]

int(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

2*I/d/a^5*(-1/17*(a+I*a*tan(d*x+c))^(17/2)+4/15*a*(a+I*a*tan(d*x+c))^(15/2)-4/13*a^2*(a+I*a*tan(d*x+c))^(13/2)
)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (64) = 128\).

Time = 0.27 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.86 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {512 \, \sqrt {2} {\left (8 i \, a^{3} e^{\left (17 i \, d x + 17 i \, c\right )} + 68 i \, a^{3} e^{\left (15 i \, d x + 15 i \, c\right )} + 255 i \, a^{3} e^{\left (13 i \, d x + 13 i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3315 \, {\left (d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-512/3315*sqrt(2)*(8*I*a^3*e^(17*I*d*x + 17*I*c) + 68*I*a^3*e^(15*I*d*x + 15*I*c) + 255*I*a^3*e^(13*I*d*x + 13
*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(16*I*d*x + 16*I*c) + 8*d*e^(14*I*d*x + 14*I*c) + 28*d*e^(12*I*d
*x + 12*I*c) + 56*d*e^(10*I*d*x + 10*I*c) + 70*d*e^(8*I*d*x + 8*I*c) + 56*d*e^(6*I*d*x + 6*I*c) + 28*d*e^(4*I*
d*x + 4*I*c) + 8*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F(-1)]

Timed out. \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**6*(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.66 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {2 i \, {\left (195 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {17}{2}} - 884 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {15}{2}} a + 1020 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a^{2}\right )}}{3315 \, a^{5} d} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-2/3315*I*(195*(I*a*tan(d*x + c) + a)^(17/2) - 884*(I*a*tan(d*x + c) + a)^(15/2)*a + 1020*(I*a*tan(d*x + c) +
a)^(13/2)*a^2)/(a^5*d)

Giac [F]

\[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \sec \left (d x + c\right )^{6} \,d x } \]

[In]

integrate(sec(d*x+c)^6*(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)^6, x)

Mupad [B] (verification not implemented)

Time = 17.40 (sec) , antiderivative size = 562, normalized size of antiderivative = 6.39 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {a^3\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,4096{}\mathrm {i}}{3315\,d}-\frac {a^3\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,2048{}\mathrm {i}}{3315\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {a^3\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{1105\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {a^3\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,56320{}\mathrm {i}}{663\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {a^3\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,205312{}\mathrm {i}}{663\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}+\frac {a^3\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,540672{}\mathrm {i}}{1105\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {a^3\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1341952{}\mathrm {i}}{3315\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}+\frac {a^3\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,44032{}\mathrm {i}}{255\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7}-\frac {a^3\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{17\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^8} \]

[In]

int((a + a*tan(c + d*x)*1i)^(7/2)/cos(c + d*x)^6,x)

[Out]

(a^3*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*56320i)/(663*d*(exp(c*2i + d*x*2
i) + 1)^3) - (a^3*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*2048i)/(3315*d*(exp
(c*2i + d*x*2i) + 1)) - (a^3*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/(1
105*d*(exp(c*2i + d*x*2i) + 1)^2) - (a^3*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1
/2)*4096i)/(3315*d) - (a^3*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*205312i)/(
663*d*(exp(c*2i + d*x*2i) + 1)^4) + (a^3*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1
/2)*540672i)/(1105*d*(exp(c*2i + d*x*2i) + 1)^5) - (a^3*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d
*x*2i) + 1))^(1/2)*1341952i)/(3315*d*(exp(c*2i + d*x*2i) + 1)^6) + (a^3*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1
i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*44032i)/(255*d*(exp(c*2i + d*x*2i) + 1)^7) - (a^3*(a - (a*(exp(c*2i + d*x*2
i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/(17*d*(exp(c*2i + d*x*2i) + 1)^8)