Integrand size = 28, antiderivative size = 183 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^4 \, dx=\frac {78 i a^4 \sqrt {e \sec (c+d x)}}{7 d}+\frac {78 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{7 d}+\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3}{7 d}+\frac {26 i \sqrt {e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )^2}{35 d}+\frac {78 i \sqrt {e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}{35 d} \] Output:
78/7*I*a^4*(e*sec(d*x+c))^(1/2)/d+78/7*a^4*cos(d*x+c)^(1/2)*InverseJacobiA M(1/2*d*x+1/2*c,2^(1/2))*(e*sec(d*x+c))^(1/2)/d+2/7*I*a*(e*sec(d*x+c))^(1/ 2)*(a+I*a*tan(d*x+c))^3/d+26/35*I*(e*sec(d*x+c))^(1/2)*(a^2+I*a^2*tan(d*x+ c))^2/d+78/35*I*(e*sec(d*x+c))^(1/2)*(a^4+I*a^4*tan(d*x+c))/d
Time = 2.68 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.55 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 \sec ^4(c+d x) \sqrt {e \sec (c+d x)} \left (728 i+1008 i \cos (2 (c+d x))+280 i \cos (4 (c+d x))+1560 \cos ^{\frac {9}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-150 \sin (2 (c+d x))-85 \sin (4 (c+d x))\right )}{140 d} \] Input:
Integrate[Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^4,x]
Output:
(a^4*Sec[c + d*x]^4*Sqrt[e*Sec[c + d*x]]*(728*I + (1008*I)*Cos[2*(c + d*x) ] + (280*I)*Cos[4*(c + d*x)] + 1560*Cos[c + d*x]^(9/2)*EllipticF[(c + d*x) /2, 2] - 150*Sin[2*(c + d*x)] - 85*Sin[4*(c + d*x)]))/(140*d)
Time = 0.87 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3979, 3042, 3979, 3042, 3979, 3042, 3967, 3042, 4258, 3042, 3120}
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.
| \(\displaystyle \int (a+i a \tan (c+d x))^4 \sqrt {e \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (c+d x))^4 \sqrt {e \sec (c+d x)}dx\) |
| \(\Big \downarrow \) 3979 |
| \(\displaystyle \frac {13}{7} a \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^3dx+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13}{7} a \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^3dx+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3979 |
| \(\displaystyle \frac {13}{7} a \left (\frac {9}{5} a \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^2dx+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13}{7} a \left (\frac {9}{5} a \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^2dx+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3979 |
| \(\displaystyle \frac {13}{7} a \left (\frac {9}{5} a \left (\frac {5}{3} a \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13}{7} a \left (\frac {9}{5} a \left (\frac {5}{3} a \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle \frac {13}{7} a \left (\frac {9}{5} a \left (\frac {5}{3} a \left (a \int \sqrt {e \sec (c+d x)}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13}{7} a \left (\frac {9}{5} a \left (\frac {5}{3} a \left (a \int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {13}{7} a \left (\frac {9}{5} a \left (\frac {5}{3} a \left (a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13}{7} a \left (\frac {9}{5} a \left (\frac {5}{3} a \left (a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {13}{7} a \left (\frac {9}{5} a \left (\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {5}{3} a \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{d}+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )\right )+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d}\) |
Input:
Int[Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^4,x]
Output:
(((2*I)/7)*a*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3)/d + (13*a*(((( 2*I)/5)*a*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^2)/d + (9*a*((5*a*(( (2*I)*a*Sqrt[e*Sec[c + d*x]])/d + (2*a*Sqrt[Cos[c + d*x]]*EllipticF[(c + d *x)/2, 2]*Sqrt[e*Sec[c + d*x]])/d))/3 + (((2*I)/3)*Sqrt[e*Sec[c + d*x]]*(a ^2 + I*a^2*Tan[c + d*x]))/d))/5))/7
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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[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*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]
Time = 14.79 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.65
| method | result | size |
| default | \(\frac {a^{4} \left (-\frac {34 \tan \left (d x +c \right )}{7}+\frac {2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{7}+16 i-\frac {8 i \sec \left (d x +c \right )^{2}}{5}+\frac {78 i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )}{7}\right ) \sqrt {e \sec \left (d x +c \right )}}{d}\) | \(119\) |
| parts | \(-\frac {2 i a^{4} \left (\cos \left (d x +c \right )+1\right ) \sqrt {e \sec \left (d x +c \right )}\, \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}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{d}+\frac {a^{4} \left (-\frac {6 \tan \left (d x +c \right )}{7}+\frac {2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{7}-\frac {8 i \left (\cos \left (d x +c \right )+1\right ) \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}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{7}\right ) \sqrt {e \sec \left (d x +c \right )}}{d}+\frac {2 i a^{4} \sqrt {e \sec \left (d x +c \right )}\, \left (4 \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \left (5+5 \cos \left (d x +c \right )-\sec \left (d x +c \right )-\sec \left (d x +c \right )^{2}\right )-5 \cos \left (d x +c \right ) \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 )+5 \cos \left (d x +c \right ) \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 )\right )}{5 d \left (\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}}}}+\frac {8 i a^{4} \sqrt {e \sec \left (d x +c \right )}}{d}-\frac {6 a^{4} \left (\frac {2 \tan \left (d x +c \right )}{3}+\frac {4 i \left (\cos \left (d x +c \right )+1\right ) \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}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{3}\right ) \sqrt {e \sec \left (d x +c \right )}}{d}\) | \(556\) |
Input:
int((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
a^4/d*(-34/7*tan(d*x+c)+2/7*tan(d*x+c)*sec(d*x+c)^2+16*I-8/5*I*sec(d*x+c)^ 2+78/7*I*(cos(d*x+c)+1)*(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))*(e*sec(d*x+c))^(1/2)
Time = 0.09 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.03 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^4 \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (-365 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 793 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 663 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 195 i \, a^{4}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 195 \, \sqrt {2} {\left (i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{4}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{35 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
Output:
-2/35*(sqrt(2)*(-365*I*a^4*e^(6*I*d*x + 6*I*c) - 793*I*a^4*e^(4*I*d*x + 4* I*c) - 663*I*a^4*e^(2*I*d*x + 2*I*c) - 195*I*a^4)*sqrt(e/(e^(2*I*d*x + 2*I *c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 195*sqrt(2)*(I*a^4*e^(6*I*d*x + 6*I*c) + 3*I*a^4*e^(4*I*d*x + 4*I*c) + 3*I*a^4*e^(2*I*d*x + 2*I*c) + I*a^4)*sqrt (e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))/(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)
\[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^4 \, dx=a^{4} \left (\int \sqrt {e \sec {\left (c + d x \right )}}\, dx + \int \left (- 6 \sqrt {e \sec {\left (c + d x \right )}} \tan ^{2}{\left (c + d x \right )}\right )\, dx + \int \sqrt {e \sec {\left (c + d x \right )}} \tan ^{4}{\left (c + d x \right )}\, dx + \int 4 i \sqrt {e \sec {\left (c + d x \right )}} \tan {\left (c + d x \right )}\, dx + \int \left (- 4 i \sqrt {e \sec {\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )}\right )\, dx\right ) \] Input:
integrate((e*sec(d*x+c))**(1/2)*(a+I*a*tan(d*x+c))**4,x)
Output:
a**4*(Integral(sqrt(e*sec(c + d*x)), x) + Integral(-6*sqrt(e*sec(c + d*x)) *tan(c + d*x)**2, x) + Integral(sqrt(e*sec(c + d*x))*tan(c + d*x)**4, x) + Integral(4*I*sqrt(e*sec(c + d*x))*tan(c + d*x), x) + Integral(-4*I*sqrt(e *sec(c + d*x))*tan(c + d*x)**3, x))
\[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^4 \, dx=\int { \sqrt {e \sec \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} \,d x } \] Input:
integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
Output:
integrate(sqrt(e*sec(d*x + c))*(I*a*tan(d*x + c) + a)^4, x)
Exception generated. \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^4 \, dx=\text {Exception raised: TypeError} \] Input:
integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{-1,[2,0]%%%}+%%%{%%{[-2,0]:[1,0,%%%{1,[1]%%%}]%%},[1,0]%%% }+%%%{%%%
Timed out. \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^4 \, dx=\int \sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4 \,d x \] Input:
int((e/cos(c + d*x))^(1/2)*(a + a*tan(c + d*x)*1i)^4,x)
Output:
int((e/cos(c + d*x))^(1/2)*(a + a*tan(c + d*x)*1i)^4, x)
\[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^4 \, dx=\frac {\sqrt {e}\, a^{4} \left (-8 \sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{2} i +72 \sqrt {\sec \left (d x +c \right )}\, i +5 \left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) d +5 \left (\int \sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{4}d x \right ) d -30 \left (\int \sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}d x \right ) d \right )}{5 d} \] Input:
int((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^4,x)
Output:
(sqrt(e)*a**4*( - 8*sqrt(sec(c + d*x))*tan(c + d*x)**2*i + 72*sqrt(sec(c + d*x))*i + 5*int(sqrt(sec(c + d*x)),x)*d + 5*int(sqrt(sec(c + d*x))*tan(c + d*x)**4,x)*d - 30*int(sqrt(sec(c + d*x))*tan(c + d*x)**2,x)*d))/(5*d)