Integrand size = 30, antiderivative size = 100 \[ \int (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=-\frac {4 i a^2 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f} \]
-4*I*a^2*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c-I*d)^(1/2)/f+4*I *a^2*(c+d*tan(f*x+e))^(1/2)/f-2/3*a^2*(c+d*tan(f*x+e))^(3/2)/d/f
Time = 0.92 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=-\frac {2 a^2 \left (6 i \sqrt {c-i d} d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} (c-6 i d+d \tan (e+f x))\right )}{3 d f} \]
(-2*a^2*((6*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 - (6*I)*d + d*Tan[e + f*x])))/(3*d*f)
Time = 0.60 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4026, 3042, 4011, 3042, 4020, 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.
\(\displaystyle \int (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}dx\) |
\(\Big \downarrow \) 4026 |
\(\displaystyle -\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\int \left (2 i \tan (e+f x) a^2+2 a^2\right ) \sqrt {c+d \tan (e+f x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\int \left (2 i \tan (e+f x) a^2+2 a^2\right ) \sqrt {c+d \tan (e+f x)}dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \frac {2 (c-i d) a^2+2 (i c+d) \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {2 (c-i d) a^2+2 (i c+d) \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {4 i a^4 (c-i d)^2 \int \frac {1}{\sqrt {2} a^2 \sqrt {2 c+2 d \tan (e+f x)} \left (2 (i c+d)^2 a^2+2 (c-i d) (i c+d) \tan (e+f x) a^2\right )}d\left (2 a^2 (i c+d) \tan (e+f x)\right )}{f}-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 i \sqrt {2} a^2 (c-i d)^2 \int \frac {1}{\sqrt {2 c+2 d \tan (e+f x)} \left (2 (i c+d)^2 a^2+2 (c-i d) (i c+d) \tan (e+f x) a^2\right )}d\left (2 a^2 (i c+d) \tan (e+f x)\right )}{f}-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {4 i \sqrt {2} a^4 (c-i d)^2 (d+i c) \int \frac {1}{\frac {4 i (c-i d)^2 (i c+d)^2 \tan ^2(e+f x) a^6}{d}+\frac {2 (i c+d)^3 a^2}{d}}d\sqrt {2 c+2 d \tan (e+f x)}}{d f}-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {4 a^2 (d+i c) \text {arctanh}\left (\frac {\sqrt {2} a^2 (d+i c) \tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}-\frac {2 a^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {4 i a^2 \sqrt {c+d \tan (e+f x)}}{f}\) |
(-4*a^2*(I*c + d)*ArcTanh[(Sqrt[2]*a^2*(I*c + d)*Tan[e + f*x])/Sqrt[c - I* d]])/(Sqrt[c - I*d]*f) + ((4*I)*a^2*Sqrt[c + d*Tan[e + f*x]])/f - (2*a^2*( c + d*Tan[e + f*x])^(3/2))/(3*d*f)
3.12.2.3.1 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[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
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[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !LeQ [m, -1] && !(EqQ[m, 2] && EqQ[a, 0])
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1228 vs. \(2 (84 ) = 168\).
Time = 0.74 (sec) , antiderivative size = 1229, normalized size of antiderivative = 12.29
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1229\) |
default | \(\text {Expression too large to display}\) | \(1229\) |
parts | \(\text {Expression too large to display}\) | \(1317\) |
-2/3*a^2*(c+d*tan(f*x+e))^(3/2)/d/f+4*I*a^2*(c+d*tan(f*x+e))^(1/2)/f-2*I/f *a^2/(4*(c^2+d^2)^(1/2)+4*c)*ln(d*tan(f*x+e)+c+(c+d*tan(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+d^2)^(1/2)+2*I/f*a^2/(4*(c^2+d^2)^(1/2)+4*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^2+d^2)^(1/2)+2/f*a^2*d/(4*(c^2+d^2)^(1/2)+4*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)+2*I/f*a^2/(4*(c^2+d^2)^(1/2)+4*c)*l n((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-8*I/f*a^2*d^2/(4*(c^2+d^2)^(1 /2)+4*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))-8/f*a^2*d/(4*( c^2+d^2)^(1/2)+4*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+d^2)^(1/2)-8/f*a^2*d/(4*(c^2+d^2)^(1/2)+4*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+8*I/f*a^2*d^2/(4*(c^2+d^2)^(1/2)+4*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))-2/f*a^2*d/(4*(c^2+d^2)^(1/2)+4*c) *ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (80) = 160\).
Time = 0.26 (sec) , antiderivative size = 397, normalized size of antiderivative = 3.97 \[ \int (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\frac {3 \, {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {a^{4} c - i \, a^{4} d}{f^{2}}} \log \left (\frac {2 \, {\left (a^{2} c + {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{4} c - i \, a^{4} d}{f^{2}}} + {\left (a^{2} c - i \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{2}}\right ) - 3 \, {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {a^{4} c - i \, a^{4} d}{f^{2}}} \log \left (\frac {2 \, {\left (a^{2} c + {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{4} c - i \, a^{4} d}{f^{2}}} + {\left (a^{2} c - i \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{2}}\right ) - 2 \, {\left (a^{2} c - 5 i \, a^{2} d + {\left (a^{2} c - 7 i \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, {\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )}} \]
1/3*(3*(d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(a^4*c - I*a^4*d)/f^2)*log(2* (a^2*c + (I*f*e^(2*I*f*x + 2*I*e) + I*f)*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^4*c - I*a^4*d)/f^2) + (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) - 3*(d*f*e^ (2*I*f*x + 2*I*e) + d*f)*sqrt(-(a^4*c - I*a^4*d)/f^2)*log(2*(a^2*c + (-I*f *e^(2*I*f*x + 2*I*e) - I*f)*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^4*c - I*a^4*d)/f^2) + (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) - 2*(a^2*c - 5*I*a^2*d + (a^2*c - 7*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)))/(d*f*e^(2*I*f*x + 2*I*e) + d*f)
\[ \int (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=- a^{2} \left (\int \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- 2 i \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- \sqrt {c + d \tan {\left (e + f x \right )}}\right )\, dx\right ) \]
-a**2*(Integral(sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(-2 *I*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(-sqrt(c + d*tan(e + f*x)), x))
\[ \int (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \sqrt {d \tan \left (f x + e\right ) + c} \,d x } \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (80) = 160\).
Time = 0.54 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.23 \[ \int (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=-\frac {8 \, {\left (-i \, a^{2} c - a^{2} d\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 \, {\left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} d^{2} f^{2} - 6 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{2} d^{3} f^{2}\right )}}{3 \, d^{3} f^{3}} \]
-8*(-I*a^2*c - a^2*d)*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)) ))/(sqrt(-2*c + 2*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) - 2 /3*((d*tan(f*x + e) + c)^(3/2)*a^2*d^2*f^2 - 6*I*sqrt(d*tan(f*x + e) + c)* a^2*d^3*f^2)/(d^3*f^3)
Time = 7.91 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.90 \[ \int (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\frac {a^2\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,4{}\mathrm {i}}{f}-\frac {2\,a^2\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,d\,f}-\frac {2\,\sqrt {4{}\mathrm {i}}\,a^2\,\mathrm {atanh}\left (\frac {\sqrt {4{}\mathrm {i}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d+c\,1{}\mathrm {i}}}\right )\,\sqrt {d+c\,1{}\mathrm {i}}}{f} \]