Integrand size = 28, antiderivative size = 178 \[ \int \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}} \, dx=\frac {154 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {154 i a^4 (e \sec (c+d x))^{3/2}}{15 d e^2}-\frac {154 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}-\frac {22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2} \] Output:
154/5*a^4*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(e*sec( d*x+c))^(1/2)-154/15*I*a^4*(e*sec(d*x+c))^(3/2)/d/e^2-154/5*a^4*(e*sec(d*x +c))^(1/2)*sin(d*x+c)/d/e-4*I*a*(a+I*a*tan(d*x+c))^3/d/(e*sec(d*x+c))^(1/2 )-22/5*I*(e*sec(d*x+c))^(3/2)*(a^4+I*a^4*tan(d*x+c))/d/e^2
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.79 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.69 \[ \int \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {2 i a^4 e^{i (c+d x)} \left (-77-176 e^{2 i (c+d x)}-111 e^{4 i (c+d x)}+77 \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sqrt {e \sec (c+d x)}}{15 d e \left (1+e^{2 i (c+d x)}\right )^2} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^4/Sqrt[e*Sec[c + d*x]],x]
Output:
(((-2*I)/15)*a^4*E^(I*(c + d*x))*(-77 - 176*E^((2*I)*(c + d*x)) - 111*E^(( 4*I)*(c + d*x)) + 77*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2 , 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sqrt[e*Sec[c + d*x]])/(d*e*(1 + E^((2*I )*(c + d*x)))^2)
Time = 0.84 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3977, 3042, 3979, 3042, 3967, 3042, 4255, 3042, 4258, 3042, 3119}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}}dx\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle -\frac {11 a^2 \int (e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)^2dx}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {11 a^2 \int (e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)^2dx}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3979 |
| \(\displaystyle -\frac {11 a^2 \left (\frac {7}{5} a \int (e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {11 a^2 \left (\frac {7}{5} a \int (e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle -\frac {11 a^2 \left (\frac {7}{5} a \left (a \int (e \sec (c+d x))^{3/2}dx+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {11 a^2 \left (\frac {7}{5} a \left (a \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle -\frac {11 a^2 \left (\frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {11 a^2 \left (\frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle -\frac {11 a^2 \left (\frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {11 a^2 \left (\frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {11 a^2 \left (\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}+\frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^4/Sqrt[e*Sec[c + d*x]],x]
Output:
((-4*I)*a*(a + I*a*Tan[c + d*x])^3)/(d*Sqrt[e*Sec[c + d*x]]) - (11*a^2*((7 *a*((((2*I)/3)*a*(e*Sec[c + d*x])^(3/2))/d + a*((-2*e^2*EllipticE[(c + d*x )/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*e*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/d)))/5 + (((2*I)/5)*(e*Sec[c + d*x])^(3/2)*(a^2 + I*a ^2*Tan[c + d*x]))/d))/e^2
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
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] - Simp[b^2*((m + 2*n - 2)/(d^2*m)) Int[(d*Sec[e + f*x]) ^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & & LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]
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] + Simp[a*((m + 2*n - 2)/(m + n - 1)) Int[(d*Se c[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ [2*m, 2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 497 vs. \(2 (155 ) = 310\).
Time = 13.46 (sec) , antiderivative size = 498, normalized size of antiderivative = 2.80
| method | result | size |
| default | \(\frac {2 a^{4} \left (30 i \ln \left (\frac {4 \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}-2 \cos \left (d x +c \right )+2}{\cos \left (d x +c \right )+1}\right )-30 i \ln \left (\frac {2 \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}-\cos \left (d x +c \right )+1}{\cos \left (d x +c \right )+1}\right )+231 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \left (-\cos \left (d x +c \right )-2-\sec \left (d x +c \right )\right )+231 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )+3 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2} \left (40 \cos \left (d x +c \right )^{3}-37 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+20 i \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \left (-6-6 \cos \left (d x +c \right )-\sec \left (d x +c \right )-\sec \left (d x +c \right )^{2}\right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {e \sec \left (d x +c \right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}}\) | \(498\) |
| parts | \(\text {Expression too large to display}\) | \(837\) |
Input:
int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/15*a^4/d*(30*I*ln(2*(2*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2 *(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-cos(d*x+c)+1)/(cos(d*x+c)+1))-30*I*l n((2*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*(-cos(d*x+c)/(cos(d *x+c)+1)^2)^(1/2)-cos(d*x+c)+1)/(cos(d*x+c)+1))+231*I*(-cos(d*x+c)/(cos(d* x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2 )*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)*(-cos(d*x+c)-2-sec(d*x+c))+231*I* (-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/ (cos(d*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*(cos(d*x+c)+2 +sec(d*x+c))+3*tan(d*x+c)*sec(d*x+c)^2*(40*cos(d*x+c)^3-37*cos(d*x+c)^2+co s(d*x+c)+1)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+20*I*(-cos(d*x+c)/(cos(d* x+c)+1)^2)^(1/2)*(-6-6*cos(d*x+c)-sec(d*x+c)-sec(d*x+c)^2))/(cos(d*x+c)+1) /(e*sec(d*x+c))^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)
Time = 0.11 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.92 \[ \int \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (-111 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 176 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} - 77 i \, a^{4} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 231 \, \sqrt {2} {\left (-i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{4}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{15 \, {\left (d e e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e e^{\left (2 i \, d x + 2 i \, c\right )} + d e\right )}} \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(1/2),x, algorithm="fricas")
Output:
-2/15*(sqrt(2)*(-111*I*a^4*e^(5*I*d*x + 5*I*c) - 176*I*a^4*e^(3*I*d*x + 3* I*c) - 77*I*a^4*e^(I*d*x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2* I*d*x + 1/2*I*c) + 231*sqrt(2)*(-I*a^4*e^(4*I*d*x + 4*I*c) - 2*I*a^4*e^(2* I*d*x + 2*I*c) - I*a^4)*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse (-4, 0, e^(I*d*x + I*c))))/(d*e*e^(4*I*d*x + 4*I*c) + 2*d*e*e^(2*I*d*x + 2 *I*c) + d*e)
\[ \int \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}} \, dx=a^{4} \left (\int \frac {1}{\sqrt {e \sec {\left (c + d x \right )}}}\, dx + \int \left (- \frac {6 \tan ^{2}{\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\right )\, dx + \int \frac {\tan ^{4}{\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\, dx + \int \frac {4 i \tan {\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\, dx + \int \left (- \frac {4 i \tan ^{3}{\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\right )\, dx\right ) \] Input:
integrate((a+I*a*tan(d*x+c))**4/(e*sec(d*x+c))**(1/2),x)
Output:
a**4*(Integral(1/sqrt(e*sec(c + d*x)), x) + Integral(-6*tan(c + d*x)**2/sq rt(e*sec(c + d*x)), x) + Integral(tan(c + d*x)**4/sqrt(e*sec(c + d*x)), x) + Integral(4*I*tan(c + d*x)/sqrt(e*sec(c + d*x)), x) + Integral(-4*I*tan( c + d*x)**3/sqrt(e*sec(c + d*x)), x))
\[ \int \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\sqrt {e \sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate((I*a*tan(d*x + c) + a)^4/sqrt(e*sec(d*x + c)), x)
\[ \int \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\sqrt {e \sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(1/2),x, algorithm="giac")
Output:
integrate((I*a*tan(d*x + c) + a)^4/sqrt(e*sec(d*x + c)), x)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(1/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(1/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}} \, dx=\frac {\sqrt {e}\, a^{4} \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x +\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{4}}{\sec \left (d x +c \right )}d x -4 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right )}d x \right ) i -6 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )}d x \right )+4 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )}d x \right ) i \right )}{e} \] Input:
int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(1/2),x)
Output:
(sqrt(e)*a**4*(int(sqrt(sec(c + d*x))/sec(c + d*x),x) + int((sqrt(sec(c + d*x))*tan(c + d*x)**4)/sec(c + d*x),x) - 4*int((sqrt(sec(c + d*x))*tan(c + d*x)**3)/sec(c + d*x),x)*i - 6*int((sqrt(sec(c + d*x))*tan(c + d*x)**2)/s ec(c + d*x),x) + 4*int((sqrt(sec(c + d*x))*tan(c + d*x))/sec(c + d*x),x)*i ))/e