Integrand size = 30, antiderivative size = 92 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {4 i a^2 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}+\frac {2 a^2 (i c-d)}{d (i c+d) f \sqrt {c+d \tan (e+f x)}} \]
-4*I*a^2*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(3/2)/f+2*a ^2*(I*c-d)/d/(I*c+d)/f/(c+d*tan(f*x+e))^(1/2)
Time = 1.42 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {2 a^2 \left (c^2+d^2-2 i \sqrt {c-i d} d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right ) \sqrt {c+d \tan (e+f x)}\right )}{(c-i d)^2 d f \sqrt {c+d \tan (e+f x)}} \]
(2*a^2*(c^2 + d^2 - (2*I)*Sqrt[c - I*d]*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]] /Sqrt[c - I*d]]*Sqrt[c + d*Tan[e + f*x]]))/((c - I*d)^2*d*f*Sqrt[c + d*Tan [e + f*x]])
Time = 0.53 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4025, 27, 3042, 4020, 25, 27, 73, 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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4025 |
| \(\displaystyle \frac {\int \frac {2 \left ((c+i d) a^2+(i c-d) \tan (e+f x) a^2\right )}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^2 (-d+i c)}{d f (d+i c) \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \int \frac {(c+i d) a^2+(i c-d) \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^2 (-d+i c)}{d f (d+i c) \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int \frac {(c+i d) a^2+(i c-d) \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^2 (-d+i c)}{d f (d+i c) \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {2 i a^4 (c+i d)^2 \int -\frac {1}{a^2 (c+i d) \left (a^2 (c+i d)-a^2 (i c-d) \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}d\left (a^2 (i c-d) \tan (e+f x)\right )}{f \left (c^2+d^2\right )}+\frac {2 a^2 (-d+i c)}{d f (d+i c) \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 a^2 (-d+i c)}{d f (d+i c) \sqrt {c+d \tan (e+f x)}}-\frac {2 i a^4 (c+i d)^2 \int \frac {1}{a^2 (c+i d) \left (a^2 (c+i d)-a^2 (i c-d) \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}d\left (a^2 (i c-d) \tan (e+f x)\right )}{f \left (c^2+d^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a^2 (-d+i c)}{d f (d+i c) \sqrt {c+d \tan (e+f x)}}-\frac {2 i a^2 (c+i d) \int \frac {1}{\left (a^2 (c+i d)-a^2 (i c-d) \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}d\left (a^2 (i c-d) \tan (e+f x)\right )}{f \left (c^2+d^2\right )}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 a^2 (-d+i c)}{d f (d+i c) \sqrt {c+d \tan (e+f x)}}-\frac {4 i a^4 (-d+i c) (c+i d) \int \frac {1}{\frac {i a^2 \left (c^2+d^2\right )}{d}-\frac {a^6 (i c-d)^3 \tan ^2(e+f x)}{d}}d\sqrt {c+d \tan (e+f x)}}{d f \left (c^2+d^2\right )}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 a^2 (-d+i c)}{d f (d+i c) \sqrt {c+d \tan (e+f x)}}-\frac {4 a^2 (-d+i c) \sqrt {c-i d} (c+i d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f \left (c^2+d^2\right )^2}\) |
(-4*a^2*(I*c - d)*Sqrt[c - I*d]*(c + I*d)*ArcTanh[Sqrt[c + d*Tan[e + f*x]] /Sqrt[c - I*d]])/((c^2 + d^2)^2*f) + (2*a^2*(I*c - d))/(d*(I*c + d)*f*Sqrt [c + d*Tan[e + f*x]])
3.12.26.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, 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[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*c - a*d)^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2533 vs. \(2 (79 ) = 158\).
Time = 0.77 (sec) , antiderivative size = 2534, normalized size of antiderivative = 27.54
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(2534\) |
| default | \(\text {Expression too large to display}\) | \(2534\) |
| parts | \(\text {Expression too large to display}\) | \(5622\) |
2*I/f*a^2*d^2/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^ (1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*( c^2+d^2)^(1/2)-2*c)^(1/2))*c-2*I/f*a^2*d^2/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2 )+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2 *(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-2/f*a^2*d/(c^2+d ^2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2) ^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)) *c+1/f*a^2*d/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)*ln(d*tan(f*x+e)+c+(c+d*ta n(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2 )^(1/2)+2*c)^(1/2)*c+2/f*a^2*d/(c^2+d^2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^ (1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^ (1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-4/f*a^2*d/(c^2+d^2)^(3/2)/((c^2+d^ 2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^ (1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2-1/f*a^2 *d/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d ^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2* c)^(1/2)*c+4/f*a^2*d/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2 )-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2 ))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2-2*I/f*a^2/(c^2+d^2)^(1/2)/((c^2+d^2) ^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 584 vs. \(2 (74) = 148\).
Time = 0.26 (sec) , antiderivative size = 584, normalized size of antiderivative = 6.35 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {{\left ({\left (c^{2} d - 2 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d + d^{3}\right )} f\right )} \sqrt {\frac {16 i \, a^{4}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac {{\left (4 \, a^{2} c + {\left ({\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} f\right )} \sqrt {\frac {16 i \, a^{4}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{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}} + 4 \, {\left (a^{2} c - i \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - {\left ({\left (c^{2} d - 2 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d + d^{3}\right )} f\right )} \sqrt {\frac {16 i \, a^{4}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac {{\left (4 \, a^{2} c + {\left ({\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} f\right )} \sqrt {\frac {16 i \, a^{4}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{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}} + 4 \, {\left (a^{2} c - i \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) + 8 \, {\left (a^{2} c + i \, a^{2} d + {\left (a^{2} c + i \, a^{2} d\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}}}{4 \, {\left ({\left (c^{2} d - 2 i \, c d^{2} - d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d + d^{3}\right )} f\right )}} \]
1/4*(((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)*s qrt(16*I*a^4/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(1/2*(4*a^2*c + ((I*c^2 + 2*c*d - I*d^2)*f*e^(2*I*f*x + 2*I*e) + (I*c^2 + 2*c*d - I*d^2) *f)*sqrt(16*I*a^4/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^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)) + 4*(a^2*c - I*a^2*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2) - ((c^2*d - 2*I*c *d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)*sqrt(16*I*a^4/((-I*c^ 3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(1/2*(4*a^2*c + ((-I*c^2 - 2*c*d + I*d^2)*f*e^(2*I*f*x + 2*I*e) + (-I*c^2 - 2*c*d + I*d^2)*f)*sqrt(16*I*a^4/ ((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^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)) + 4*(a^2*c - I*a^2*d)*e^(2*I*f *x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2) + 8*(a^2*c + I*a^2*d + (a^2*c + I*a ^2*d)*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)))/((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I* e) + (c^2*d + d^3)*f)
\[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx=- a^{2} \left (\int \frac {\tan ^{2}{\left (e + f x \right )}}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\, dx + \int \left (- \frac {2 i \tan {\left (e + f x \right )}}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\right )\, dx + \int \left (- \frac {1}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\right )\, dx\right ) \]
-a**2*(Integral(tan(e + f*x)**2/(c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d *tan(e + f*x))*tan(e + f*x)), x) + Integral(-2*I*tan(e + f*x)/(c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)), x) + Integral( -1/(c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)), x))
\[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (74) = 148\).
Time = 0.76 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.24 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {8 \, a^{2} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{{\left (-i \, c f - d f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {2 \, {\left (a^{2} c + i \, a^{2} d\right )}}{{\left (c d f - i \, d^{2} f\right )} \sqrt {d \tan \left (f x + e\right ) + c}} \]
8*a^2*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f* x + e) + c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) - I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/((-I*c*f - d* f)*sqrt(-2*c + 2*sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) + 2*(a ^2*c + I*a^2*d)/((c*d*f - I*d^2*f)*sqrt(d*tan(f*x + e) + c))
Time = 8.03 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.54 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {a^2\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\left (2\,c^4\,f^2+4\,c^2\,d^2\,f^2+2\,d^4\,f^2\right )}{2\,f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}\,\left (f\,c^3+1{}\mathrm {i}\,f\,c^2\,d+f\,c\,d^2+1{}\mathrm {i}\,f\,d^3\right )}\right )\,4{}\mathrm {i}}{f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}}+\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{d\,f\,\left (c-d\,1{}\mathrm {i}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \]