\(\int \sec (c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) [326]

Optimal result

Integrand size = 24, antiderivative size = 139 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {256 i a^4 \sec (c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {64 i a^3 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d} \]

[Out]

256/35*I*a^4*sec(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/2)+64/35*I*a^3*sec(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d+24/35*I*a
^2*sec(d*x+c)*(a+I*a*tan(d*x+c))^(3/2)/d+2/7*I*a*sec(d*x+c)*(a+I*a*tan(d*x+c))^(5/2)/d

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3575, 3574} \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {256 i a^4 \sec (c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {64 i a^3 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d} \]

[In]

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

[Out]

(((256*I)/35)*a^4*Sec[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((64*I)/35)*a^3*Sec[c + d*x]*Sqrt[a + I*a*Ta
n[c + d*x]])/d + (((24*I)/35)*a^2*Sec[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2))/d + (((2*I)/7)*a*Sec[c + d*x]*(a
+ I*a*Tan[c + d*x])^(5/2))/d

Rule 3574

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(
d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rule 3575

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Dist[a*((m + 2*n - 2)/(m + n - 1)), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac {1}{7} (12 a) \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx \\ & = \frac {24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac {1}{35} \left (96 a^2\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx \\ & = \frac {64 i a^3 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac {1}{35} \left (128 a^3\right ) \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx \\ & = \frac {256 i a^4 \sec (c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {64 i a^3 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.78 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {2 a^3 \sec ^2(c+d x) (i \cos (c-2 d x)+\sin (c-2 d x)) (75+102 \cos (2 (c+d x))+19 i \sec (c+d x) \sin (3 (c+d x))+14 i \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{35 d (\cos (d x)+i \sin (d x))^3} \]

[In]

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

[Out]

(2*a^3*Sec[c + d*x]^2*(I*Cos[c - 2*d*x] + Sin[c - 2*d*x])*(75 + 102*Cos[2*(c + d*x)] + (19*I)*Sec[c + d*x]*Sin
[3*(c + d*x)] + (14*I)*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(35*d*(Cos[d*x] + I*Sin[d*x])^3)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (115 ) = 230\).

Time = 7.88 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.69

method	result	size
default	\(\frac {2 \left (-\tan \left (d x +c \right )+i\right )^{3} a^{3} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (128 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-76 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-128 \left (\cos ^{4}\left (d x +c \right )\right )-22 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-204 \left (\cos ^{3}\left (d x +c \right )\right )+5 i \sin \left (d x +c \right )-54 \left (\cos ^{2}\left (d x +c \right )\right )+27 \cos \left (d x +c \right )+5\right )}{35 d \left (8 \left (\cos ^{4}\left (d x +c \right )\right )+4 \left (\cos ^{3}\left (d x +c \right )\right )+8 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-8 \left (\cos ^{2}\left (d x +c \right )\right )+4 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-3 \cos \left (d x +c \right )-4 i \cos \left (d x +c \right ) \sin \left (d x +c \right )+1-i \sin \left (d x +c \right )\right )}\)	\(235\)

[In]

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

[Out]

2/35/d*(-tan(d*x+c)+I)^3*a^3*(a*(1+I*tan(d*x+c)))^(1/2)*(128*I*cos(d*x+c)^3*sin(d*x+c)-76*I*cos(d*x+c)^2*sin(d
*x+c)-128*cos(d*x+c)^4-22*I*cos(d*x+c)*sin(d*x+c)-204*cos(d*x+c)^3+5*I*sin(d*x+c)-54*cos(d*x+c)^2+27*cos(d*x+c
)+5)/(8*cos(d*x+c)^4+4*cos(d*x+c)^3+8*I*sin(d*x+c)*cos(d*x+c)^3-8*cos(d*x+c)^2+4*I*cos(d*x+c)^2*sin(d*x+c)-3*c
os(d*x+c)-4*I*sin(d*x+c)*cos(d*x+c)+1-I*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.78 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {16 \, \sqrt {2} {\left (-35 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 70 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 56 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i \, a^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{35 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

-16/35*sqrt(2)*(-35*I*a^3*e^(6*I*d*x + 6*I*c) - 70*I*a^3*e^(4*I*d*x + 4*I*c) - 56*I*a^3*e^(2*I*d*x + 2*I*c) -
16*I*a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x
+ 2*I*c) + d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

integrate(sec(d*x+c)*(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), x)

Mupad [B] (verification not implemented)

Time = 6.74 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.06 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {a^3\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,16{}\mathrm {i}}{d}-\frac {a^3\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,16{}\mathrm {i}}{d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}+\frac {a^3\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,48{}\mathrm {i}}{5\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {a^3\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,16{}\mathrm {i}}{7\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3} \]

[In]

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

[Out]

(a^3*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*16i)/d - (a
^3*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*16i)/(d*(exp(
c*2i + d*x*2i) + 1)) + (a^3*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i)
+ 1))^(1/2)*48i)/(5*d*(exp(c*2i + d*x*2i) + 1)^2) - (a^3*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i -
1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*16i)/(7*d*(exp(c*2i + d*x*2i) + 1)^3)