Integrand size = 32, antiderivative size = 194 \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} (c-i d)^{3/2} f}-\frac {1}{(i c-d) f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(c-3 i d) d \sqrt {a+i a \tan (e+f x)}}{a (c-i d) (c+i d)^2 f \sqrt {c+d \tan (e+f x)}} \]
-1/2*I*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))/(c-I*d)^(3/2)/f*2^(1/2)/a^(1/2)-1/(I*c-d)/f/(a+I*a*tan (f*x+e))^(1/2)/(c+d*tan(f*x+e))^(1/2)+(c-3*I*d)*d*(a+I*a*tan(f*x+e))^(1/2) /a/(c-I*d)/(c+I*d)^2/f/(c+d*tan(f*x+e))^(1/2)
Time = 1.23 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\frac {i a \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {2} (-a (c-i d))^{3/2} f}+\frac {i c^2+c d-2 i d^2+d (i c+3 d) \tan (e+f x)}{(c-i d) (c+i d)^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \]
(I*a*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*S qrt[c + d*Tan[e + f*x]])])/(Sqrt[2]*(-(a*(c - I*d)))^(3/2)*f) + (I*c^2 + c *d - (2*I)*d^2 + d*(I*c + 3*d)*Tan[e + f*x])/((c - I*d)*(c + I*d)^2*f*Sqrt [a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])
Time = 0.85 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {3042, 4042, 27, 3042, 4081, 27, 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.
| \(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4042 |
| \(\displaystyle -\frac {\int -\frac {\sqrt {i \tan (e+f x) a+a} (a (i c-3 d)+2 i a d \tan (e+f x))}{2 (c+d \tan (e+f x))^{3/2}}dx}{a^2 (-d+i c)}-\frac {1}{f (-d+i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {i \tan (e+f x) a+a} (a (i c-3 d)+2 i a d \tan (e+f x))}{(c+d \tan (e+f x))^{3/2}}dx}{2 a^2 (-d+i c)}-\frac {1}{f (-d+i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {i \tan (e+f x) a+a} (a (i c-3 d)+2 i a d \tan (e+f x))}{(c+d \tan (e+f x))^{3/2}}dx}{2 a^2 (-d+i c)}-\frac {1}{f (-d+i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {\frac {2 \int \frac {i a^2 (c+i d)^2 \sqrt {i \tan (e+f x) a+a}}{2 \sqrt {c+d \tan (e+f x)}}dx}{a \left (c^2+d^2\right )}+\frac {2 a d (3 d+i c) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {1}{f (-d+i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {i a (c+i d)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a d (3 d+i c) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {1}{f (-d+i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {i a (c+i d)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a d (3 d+i c) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {1}{f (-d+i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {\frac {2 a^3 (c+i d)^2 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f \left (c^2+d^2\right )}+\frac {2 a d (3 d+i c) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {1}{f (-d+i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\sqrt {2} a^{3/2} (c+i d)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f \sqrt {c-i d} \left (c^2+d^2\right )}+\frac {2 a d (3 d+i c) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {1}{f (-d+i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
-(1/((I*c - d)*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])) + ( (Sqrt[2]*a^(3/2)*(c + I*d)^2*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f *x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[c - I*d]*(c^2 + d ^2)*f) + (2*a*d*(I*c + 3*d)*Sqrt[a + I*a*Tan[e + f*x]])/((c^2 + d^2)*f*Sqr t[c + d*Tan[e + f*x]]))/(2*a^2*(I*c - d))
3.12.64.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) In t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3195 vs. \(2 (160 ) = 320\).
Time = 1.34 (sec) , antiderivative size = 3196, normalized size of antiderivative = 16.47
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(3196\) |
| default | \(\text {Expression too large to display}\) | \(3196\) |
-1/4/f*(-2^(1/2)*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))*d^5*(-a*(I*d-c))^(1/2)*tan(f*x+e)^3-2^(1/2)*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))*c^5*(-a*(I*d-c))^(1/2)*tan(f* x+e)^2+2^(1/2)*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))*d^5*(-a*(I*d-c))^(1/2)*tan(f*x+e)-4*I*c^4*d*(a*(1+I*tan(f*x+e))*(c+ d*tan(f*x+e)))^(1/2)*tan(f*x+e)^2-16*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e) ))^(1/2)*c^2*d^3*tan(f*x+e)^2+8*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1 /2)*c^3*d^2*tan(f*x+e)+12*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c* d^4*tan(f*x+e)-6*2^(1/2)*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))*c^3*d^2*(-a*(I*d-c))^(1/2)+2^(1/2)*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))*c*d^4*(-a*(I*d-c))^(1/2)-8*(a* (1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^3*d^2*tan(f*x+e)^2-8*(a*(1+I*ta n(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c*d^4*tan(f*x+e)^2-4*(a*(1+I*tan(f*x+e)) *(c+d*tan(f*x+e)))^(1/2)*c^4*d*tan(f*x+e)-24*(a*(1+I*tan(f*x+e))*(c+d*tan( f*x+e)))^(1/2)*c^2*d^3*tan(f*x+e)-4*I*c^5*(a*(1+I*tan(f*x+e))*(c+d*tan(...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 673 vs. \(2 (150) = 300\).
Time = 0.29 (sec) , antiderivative size = 673, normalized size of antiderivative = 3.47 \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {2 \, \sqrt {2} {\left (-i \, c^{2} - i \, d^{2} + {\left (-i \, c^{2} - 2 \, c d + 5 i \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-i \, c^{2} - c d + 2 i \, d^{2}\right )} e^{\left (2 i \, f x + 2 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}} - {\left ({\left (a c^{4} + 2 \, a c^{2} d^{2} + a d^{4}\right )} f e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (a c^{4} + 2 i \, a c^{3} d + 2 i \, a c d^{3} - a d^{4}\right )} f e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {2 i}{{\left (-i \, a c^{3} - 3 \, a c^{2} d + 3 i \, a c d^{2} + a d^{3}\right )} f^{2}}} \log \left ({\left (i \, a c^{2} + 2 \, a c d - i \, a d^{2}\right )} f \sqrt {\frac {2 i}{{\left (-i \, a c^{3} - 3 \, a c^{2} d + 3 i \, a c d^{2} + a d^{3}\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \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}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) + {\left ({\left (a c^{4} + 2 \, a c^{2} d^{2} + a d^{4}\right )} f e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (a c^{4} + 2 i \, a c^{3} d + 2 i \, a c d^{3} - a d^{4}\right )} f e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {2 i}{{\left (-i \, a c^{3} - 3 \, a c^{2} d + 3 i \, a c d^{2} + a d^{3}\right )} f^{2}}} \log \left ({\left (-i \, a c^{2} - 2 \, a c d + i \, a d^{2}\right )} f \sqrt {\frac {2 i}{{\left (-i \, a c^{3} - 3 \, a c^{2} d + 3 i \, a c d^{2} + a d^{3}\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \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}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right )}{4 \, {\left ({\left (a c^{4} + 2 \, a c^{2} d^{2} + a d^{4}\right )} f e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (a c^{4} + 2 i \, a c^{3} d + 2 i \, a c d^{3} - a d^{4}\right )} f e^{\left (i \, f x + i \, e\right )}\right )}} \]
-1/4*(2*sqrt(2)*(-I*c^2 - I*d^2 + (-I*c^2 - 2*c*d + 5*I*d^2)*e^(4*I*f*x + 4*I*e) + 2*(-I*c^2 - c*d + 2*I*d^2)*e^(2*I*f*x + 2*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)) - ((a*c^4 + 2*a*c^2*d^2 + a*d^4)*f*e^(3*I*f*x + 3*I*e) + (a*c^4 + 2*I*a*c^3*d + 2*I*a*c*d^3 - a*d^4)*f*e^(I*f*x + I*e))*sqrt(2*I/(( -I*a*c^3 - 3*a*c^2*d + 3*I*a*c*d^2 + a*d^3)*f^2))*log((I*a*c^2 + 2*a*c*d - I*a*d^2)*f*sqrt(2*I/((-I*a*c^3 - 3*a*c^2*d + 3*I*a*c*d^2 + a*d^3)*f^2))*e ^(I*f*x + I*e) + sqrt(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^(2*I*f*x + 2 *I*e) + 1)) + ((a*c^4 + 2*a*c^2*d^2 + a*d^4)*f*e^(3*I*f*x + 3*I*e) + (a*c^ 4 + 2*I*a*c^3*d + 2*I*a*c*d^3 - a*d^4)*f*e^(I*f*x + I*e))*sqrt(2*I/((-I*a* c^3 - 3*a*c^2*d + 3*I*a*c*d^2 + a*d^3)*f^2))*log((-I*a*c^2 - 2*a*c*d + I*a *d^2)*f*sqrt(2*I/((-I*a*c^3 - 3*a*c^2*d + 3*I*a*c*d^2 + a*d^3)*f^2))*e^(I* f*x + I*e) + sqrt(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^(2*I*f*x + 2*I*e ) + 1)))/((a*c^4 + 2*a*c^2*d^2 + a*d^4)*f*e^(3*I*f*x + 3*I*e) + (a*c^4 + 2 *I*a*c^3*d + 2*I*a*c*d^3 - a*d^4)*f*e^(I*f*x + I*e))
\[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\int { \frac {1}{\sqrt {i \, a \tan \left (f x + e\right ) + a} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Exception generated. \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Argument Typeindex.cc inde x_m i_lex
Timed out. \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]