Integrand size = 32, antiderivative size = 193 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} \sqrt {c-i d} f}-\frac {\sqrt {c+d \tan (e+f x)}}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2}}+\frac {(3 i c-7 d) \sqrt {c+d \tan (e+f x)}}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)}} \]
-1/4*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))/a^(3/2)/f*2^(1/2)/(c-I*d)^(1/2)+1/6*(3*I*c-7*d)*(c+d*t an(f*x+e))^(1/2)/a/(c+I*d)^2/f/(a+I*a*tan(f*x+e))^(1/2)-1/3*(c+d*tan(f*x+e ))^(1/2)/(I*c-d)/f/(a+I*a*tan(f*x+e))^(3/2)
Time = 1.44 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {\frac {3 i \sqrt {2} a^2 (c+i d)^2 \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a (c-i d)}}+\frac {4 a^3 (-i c+d) \sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^{3/2}}+\frac {2 a^2 (-3 i c+7 d) \sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}}}{12 a^3 (c+i d)^2 f} \]
-1/12*(((3*I)*Sqrt[2]*a^2*(c + I*d)^2*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[-(a*(c - I*d))] + (4*a^3*((-I)*c + d)*Sqrt[c + d*Tan[e + f*x]])/(a + I*a*Tan[e + f* x])^(3/2) + (2*a^2*((-3*I)*c + 7*d)*Sqrt[c + d*Tan[e + f*x]])/Sqrt[a + I*a *Tan[e + f*x]])/(a^3*(c + I*d)^2*f)
Time = 0.84 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.17, 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, 4079, 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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 4042 |
| \(\displaystyle -\frac {\int -\frac {a (3 i c-5 d)+2 i a d \tan (e+f x)}{2 \sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)}}dx}{3 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (3 i c-5 d)+2 i a d \tan (e+f x)}{\sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)}}dx}{6 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (3 i c-5 d)+2 i a d \tan (e+f x)}{\sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)}}dx}{6 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {-\frac {\int \frac {3 a^2 (c+i d)^2 \sqrt {i \tan (e+f x) a+a}}{2 \sqrt {c+d \tan (e+f x)}}dx}{a^2 (-d+i c)}-\frac {a (3 c+7 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) \sqrt {a+i a \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {3 (c+i d)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{2 (-d+i c)}-\frac {a (3 c+7 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) \sqrt {a+i a \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {3 (c+i d)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{2 (-d+i c)}-\frac {a (3 c+7 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) \sqrt {a+i a \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {\frac {3 i a^2 (c+i d)^2 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f (-d+i c)}-\frac {a (3 c+7 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) \sqrt {a+i a \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {3 i \sqrt {a} (c+i d)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} f (-d+i c) \sqrt {c-i d}}-\frac {a (3 c+7 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) \sqrt {a+i a \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2}}\) |
-1/3*Sqrt[c + d*Tan[e + f*x]]/((I*c - d)*f*(a + I*a*Tan[e + f*x])^(3/2)) + (((3*I)*Sqrt[a]*(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[2]*(I*c - d)*Sqrt [c - I*d]*f) - (a*(3*c + (7*I)*d)*Sqrt[c + d*Tan[e + f*x]])/((c + I*d)*f*S qrt[a + I*a*Tan[e + f*x]]))/(6*a^2*(I*c - d))
3.12.59.3.1 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*A + b*B)*(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)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2945 vs. \(2 (154 ) = 308\).
Time = 1.67 (sec) , antiderivative size = 2946, normalized size of antiderivative = 15.26
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(2946\) |
| default | \(\text {Expression too large to display}\) | \(2946\) |
-1/24/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^2*(-3*2^(1/2)* (-a*(I*d-c))^(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^4*tan(f*x+e)^3-3*2^(1/2)*(-a*(I*d-c))^(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^4*tan(f*x+e)^3+9*2^(1/2 )*(-a*(I*d-c))^(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))/(ta n(f*x+e)+I))*c^4*tan(f*x+e)+9*2^(1/2)*(-a*(I*d-c))^(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^4*tan(f*x+e)+64*(a*(1+I *tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*d^4*tan(f*x+e)-56*(a*(1+I*tan(f*x+e)) *(c+d*tan(f*x+e)))^(1/2)*c^3*d-56*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1 /2)*c*d^3-32*c^4*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*tan(f*x+e)-12 *I*c^4*(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^2+28*I*(a*(1+I*tan(f*x+e))*(c +d*tan(f*x+e)))^(1/2)*d^4*tan(f*x+e)^2+20*I*c^4*(a*(1+I*tan(f*x+e))*(c+d*t an(f*x+e)))^(1/2)-36*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*d^4+12* 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))*2^(1/...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (145) = 290\).
Time = 0.27 (sec) , antiderivative size = 488, normalized size of antiderivative = 2.53 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {{\left (3 \, {\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}\right )} f \sqrt {\frac {i}{2 \, {\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (-2 \, {\left (i \, a^{2} c + a^{2} d\right )} f \sqrt {\frac {i}{2 \, {\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) - 3 \, {\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}\right )} f \sqrt {\frac {i}{2 \, {\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (-2 \, {\left (-i \, a^{2} c - a^{2} d\right )} f \sqrt {\frac {i}{2 \, {\left (-i \, a^{3} c - a^{3} d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) - \sqrt {2} {\left (4 \, {\left (i \, c - 2 \, d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-5 i \, c + 9 \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{12 \, {\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}\right )} f} \]
-1/12*(3*(a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d )*f^2))*e^(3*I*f*x + 3*I*e)*log(-2*(I*a^2*c + a^2*d)*f*sqrt(1/2*I/((-I*a^3 *c - a^3*d)*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)) - 3*(a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*sqr t(1/2*I/((-I*a^3*c - a^3*d)*f^2))*e^(3*I*f*x + 3*I*e)*log(-2*(-I*a^2*c - a ^2*d)*f*sqrt(1/2*I/((-I*a^3*c - a^3*d)*f^2))*e^(I*f*x + I*e) + sqrt(2)*sqr t(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqr t(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1)) - sqrt(2)*(4*(I* c - 2*d)*e^(4*I*f*x + 4*I*e) - (-5*I*c + 9*d)*e^(2*I*f*x + 2*I*e) + I*c - d)*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^(-3*I*f*x - 3*I*e)/((a^2*c^2 + 2*I *a^2*c*d - a^2*d^2)*f)
\[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=\text {Exception raised: RuntimeError} \]
Timed out. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]