Integrand size = 32, antiderivative size = 181 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{5/2} f}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {4 i a^2 \sqrt {a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}} \]
-4*I*a^(5/2)*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/ (a+I*a*tan(f*x+e))^(1/2))*2^(1/2)/(c-I*d)^(5/2)/f+4*I*a^2*(a+I*a*tan(f*x+e ))^(1/2)/(c-I*d)^2/f/(c+d*tan(f*x+e))^(1/2)-2/3*a*(a+I*a*tan(f*x+e))^(3/2) /(I*c+d)/f/(c+d*tan(f*x+e))^(3/2)
Time = 9.47 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.56 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {(a+i a \tan (e+f x))^{5/2} \left (-\frac {4 i \sqrt {2} e^{-3 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \log \left (2 e^{-i e} \left (\sqrt {c-i d} e^{i (e+f x)}+\sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{(c-i d)^{5/2}}-\frac {2 \sqrt {\sec (e+f x)} (\cos (2 (e+f x))-i \sin (2 (e+f x))) (-7 i c-d+(c-7 i d) \tan (e+f x))}{3 (c-i d)^2 (c+d \tan (e+f x))^{3/2}}\right )}{f \sec ^{\frac {5}{2}}(e+f x)} \]
((a + I*a*Tan[e + f*x])^(5/2)*(((-4*I)*Sqrt[2]*Sqrt[E^(I*(e + f*x))/(1 + E ^((2*I)*(e + f*x)))]*Sqrt[1 + E^((2*I)*(e + f*x))]*Log[(2*(Sqrt[c - I*d]*E ^(I*(e + f*x)) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[c - (I*d*(-1 + E^((2*I )*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))]))/E^(I*e)])/((c - I*d)^(5/2)*E^( (3*I)*(e + f*x))) - (2*Sqrt[Sec[e + f*x]]*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)])*((-7*I)*c - d + (c - (7*I)*d)*Tan[e + f*x]))/(3*(c - I*d)^2*(c + d *Tan[e + f*x])^(3/2))))/(f*Sec[e + f*x]^(5/2))
Time = 0.74 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3042, 4028, 3042, 4028, 3042, 4027, 221}
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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4028 |
| \(\displaystyle \frac {2 a \int \frac {(i \tan (e+f x) a+a)^{3/2}}{(c+d \tan (e+f x))^{3/2}}dx}{c-i d}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \int \frac {(i \tan (e+f x) a+a)^{3/2}}{(c+d \tan (e+f x))^{3/2}}dx}{c-i d}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4028 |
| \(\displaystyle \frac {2 a \left (\frac {2 a \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c-i d}-\frac {2 a \sqrt {a+i a \tan (e+f x)}}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\right )}{c-i d}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (\frac {2 a \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c-i d}-\frac {2 a \sqrt {a+i a \tan (e+f x)}}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\right )}{c-i d}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {2 a \left (-\frac {4 i a^3 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f (c-i d)}-\frac {2 a \sqrt {a+i a \tan (e+f x)}}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\right )}{c-i d}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 a \left (-\frac {2 i \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f (c-i d)^{3/2}}-\frac {2 a \sqrt {a+i a \tan (e+f x)}}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\right )}{c-i d}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}\) |
(-2*a*(a + I*a*Tan[e + f*x])^(3/2))/(3*(I*c + d)*f*(c + d*Tan[e + f*x])^(3 /2)) + (2*a*(((-2*I)*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*T an[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/((c - I*d)^(3/2 )*f) - (2*a*Sqrt[a + I*a*Tan[e + f*x]])/((I*c + d)*f*Sqrt[c + d*Tan[e + f* x]])))/(c - I*d)
3.12.67.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*b*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m - 1)*(a*c - b*d))), x] + Simp[2*(a^2/(a*c - b*d)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0 ] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2967 vs. \(2 (146 ) = 292\).
Time = 1.47 (sec) , antiderivative size = 2968, normalized size of antiderivative = 16.40
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(2968\) |
| default | \(\text {Expression too large to display}\) | \(2968\) |
1/3/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(-3*2^(1/2)*(-a*(I*d-c))^(1/2)*ln (1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/ 2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^2+6*I*2^(1/2)*ln(1/2*(2*I*a*d*tan (f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+ a*d)/(I*a*d)^(1/2))*a*c^3*d^2*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2-18*I*2^(1/2) *ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^ (1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^4*(-a*(I*d-c))^(1/2)*tan(f*x +e)^2+12*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))* (c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^4*d*(-a*(I*d -c))^(1/2)*tan(f*x+e)-36*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*( 1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))* a*c^2*d^3*(-a*(I*d-c))^(1/2)*tan(f*x+e)-3*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a *d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d* tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*a*c^2*(I*a*d)^(1/2)-7*2^(1/2)*(-a*(I*d -c))^(1/2)*(I*a*d)^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^4+2 ^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e )))^(1/2)*d^4-3*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2 )*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x +e)+I))*a*d^2*(I*a*d)^(1/2)*tan(f*x+e)^2-6*ln((3*a*c+I*a*tan(f*x+e)*c-I*a* d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 849 vs. \(2 (137) = 274\).
Time = 0.25 (sec) , antiderivative size = 849, normalized size of antiderivative = 4.69 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {8 \, \sqrt {2} {\left (4 \, {\left (-i \, a^{2} c - a^{2} d\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-7 i \, a^{2} c - a^{2} d\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + 3 \, {\left (-i \, a^{2} c + a^{2} d\right )} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - 3 \, {\left ({\left (c^{4} - 4 i \, c^{3} d - 6 \, c^{2} d^{2} + 4 i \, c d^{3} + d^{4}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} - 2 i \, c^{3} d - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} + 2 \, c^{2} d^{2} + d^{4}\right )} f\right )} \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} \log \left (\frac {{\left ({\left (i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}\right )} \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} f e^{\left (i \, f x + i \, e\right )} + 4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2}}\right ) + 3 \, {\left ({\left (c^{4} - 4 i \, c^{3} d - 6 \, c^{2} d^{2} + 4 i \, c d^{3} + d^{4}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} - 2 i \, c^{3} d - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} + 2 \, c^{2} d^{2} + d^{4}\right )} f\right )} \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} \log \left (\frac {{\left ({\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} f e^{\left (i \, f x + i \, e\right )} + 4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2}}\right )}{6 \, {\left ({\left (c^{4} - 4 i \, c^{3} d - 6 \, c^{2} d^{2} + 4 i \, c d^{3} + d^{4}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} - 2 i \, c^{3} d - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} + 2 \, c^{2} d^{2} + d^{4}\right )} f\right )}} \]
-1/6*(8*sqrt(2)*(4*(-I*a^2*c - a^2*d)*e^(5*I*f*x + 5*I*e) + (-7*I*a^2*c - a^2*d)*e^(3*I*f*x + 3*I*e) + 3*(-I*a^2*c + a^2*d)*e^(I*f*x + I*e))*sqrt((( c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/ (e^(2*I*f*x + 2*I*e) + 1)) - 3*((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f*x + 4*I*e) + 2*(c^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2 *I*f*x + 2*I*e) + (c^4 + 2*c^2*d^2 + d^4)*f)*sqrt(-32*I*a^5/((I*c^5 + 5*c^ 4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(1/4*((I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*sqrt(-32*I*a^5/((I*c^5 + 5*c^4*d - 10*I*c^3*d^ 2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*f*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2 *e^(2*I*f*x + 2*I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d) /(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2) + 3*((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f *x + 4*I*e) + 2*(c^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c^4 + 2*c^2*d^2 + d^4)*f)*sqrt(-32*I*a^5/((I*c^5 + 5*c^4*d - 10*I*c^3*d ^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(1/4*((-I*c^3 - 3*c^2*d + 3*I* c*d^2 + d^3)*sqrt(-32*I*a^5/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*f*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2 *I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2))/((c ^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f*x + 4*I*e) + 2...
\[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Non regular value [0,0] was discard ed and replaced randomly by 0=[72,-20]Warning, replacing 72 by 50, a subst itution v
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \]