Integrand size = 33, antiderivative size = 156 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {5 i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a c^{3/2} f}-\frac {5 i}{12 a f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {5 i}{8 a c f \sqrt {c-i c \tan (e+f x)}} \] Output:
5/16*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a/c^( 3/2)/f-5/12*I/a/f/(c-I*c*tan(f*x+e))^(3/2)+1/2*I/a/f/(1+I*tan(f*x+e))/(c-I *c*tan(f*x+e))^(3/2)-5/8*I/a/c/f/(c-I*c*tan(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.69 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.34 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right )}{6 a f (c-i c \tan (e+f x))^{3/2}} \] Input:
Integrate[1/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)),x]
Output:
((-1/6*I)*Hypergeometric2F1[-3/2, 2, -1/2, (-1/2*I)*(I + Tan[e + f*x])])/( a*f*(c - I*c*Tan[e + f*x])^(3/2))
Time = 0.41 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4005, 3042, 3968, 52, 61, 61, 73, 219}
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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4005 |
| \(\displaystyle \frac {\int \frac {\cos ^2(e+f x)}{\sqrt {c-i c \tan (e+f x)}}dx}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{\sec (e+f x)^2 \sqrt {c-i c \tan (e+f x)}}dx}{a c}\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle \frac {i c^2 \int \frac {1}{(c-i c \tan (e+f x))^{5/2} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{a f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {i c^2 \left (\frac {5 \int \frac {1}{(c-i c \tan (e+f x))^{5/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{a f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {i c^2 \left (\frac {5 \left (\frac {\int \frac {1}{(c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{a f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {i c^2 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{a f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {i c^2 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{a f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {i c^2 \left (\frac {5 \left (\frac {-\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{\sqrt {2} c^{3/2}}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{a f}\) |
Input:
Int[1/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)),x]
Output:
(I*c^2*(1/(2*c*(c - I*c*Tan[e + f*x])^(3/2)*(c + I*c*Tan[e + f*x])) + (5*( -1/3*1/(c*(c - I*c*Tan[e + f*x])^(3/2)) + (((-I)*ArcTan[(Sqrt[c]*Tan[e + f *x])/Sqrt[2]])/(Sqrt[2]*c^(3/2)) - 1/(c*Sqrt[c - I*c*Tan[e + f*x]]))/(2*c) ))/(4*c)))/(a*f)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, 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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] && !(IGtQ[n, 0] && (LtQ[ m, 0] || GtQ[m, n]))
Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(\frac {2 i c^{2} \left (-\frac {1}{4 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{12 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{4 c +4 i c \tan \left (f x +e \right )}+\frac {5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}}{4 c^{3}}\right )}{f a}\) | \(121\) |
| default | \(\frac {2 i c^{2} \left (-\frac {1}{4 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{12 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{4 c +4 i c \tan \left (f x +e \right )}+\frac {5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}}{4 c^{3}}\right )}{f a}\) | \(121\) |
Input:
int(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
Output:
2*I/f/a*c^2*(-1/4/c^3/(c-I*c*tan(f*x+e))^(1/2)-1/12/c^2/(c-I*c*tan(f*x+e)) ^(3/2)+1/4/c^3*(1/8*(c-I*c*tan(f*x+e))^(1/2)/(1/2*c+1/2*I*c*tan(f*x+e))+5/ 8*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 297 vs. \(2 (115) = 230\).
Time = 0.09 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.90 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left (-15 i \, \sqrt {\frac {1}{2}} a c^{2} f \sqrt {\frac {1}{a^{2} c^{3} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {5 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a c f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{2} c^{3} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a c f}\right ) + 15 i \, \sqrt {\frac {1}{2}} a c^{2} f \sqrt {\frac {1}{a^{2} c^{3} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {5 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a c f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{2} c^{3} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a c f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-2 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 16 i \, e^{\left (4 i \, f x + 4 i \, e\right )} - 11 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{48 \, a c^{2} f} \] Input:
integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fric as")
Output:
1/48*(-15*I*sqrt(1/2)*a*c^2*f*sqrt(1/(a^2*c^3*f^2))*e^(2*I*f*x + 2*I*e)*lo g(-5/4*(sqrt(2)*sqrt(1/2)*(I*a*c*f*e^(2*I*f*x + 2*I*e) + I*a*c*f)*sqrt(c/( e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^2*c^3*f^2)) - I)*e^(-I*f*x - I*e)/(a*c *f)) + 15*I*sqrt(1/2)*a*c^2*f*sqrt(1/(a^2*c^3*f^2))*e^(2*I*f*x + 2*I*e)*lo g(-5/4*(sqrt(2)*sqrt(1/2)*(-I*a*c*f*e^(2*I*f*x + 2*I*e) - I*a*c*f)*sqrt(c/ (e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^2*c^3*f^2)) - I)*e^(-I*f*x - I*e)/(a* c*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-2*I*e^(6*I*f*x + 6*I*e ) - 16*I*e^(4*I*f*x + 4*I*e) - 11*I*e^(2*I*f*x + 2*I*e) + 3*I))*e^(-2*I*f* x - 2*I*e)/(a*c^2*f)
\[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=- \frac {i \int \frac {1}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - i c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a} \] Input:
integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**(3/2),x)
Output:
-I*Integral(1/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - I*c*sqrt (-I*c*tan(e + f*x) + c)), x)/a
Time = 0.14 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} - 20 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c - 8 \, c^{2}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a - 2 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a c} + \frac {15 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a \sqrt {c}}\right )}}{96 \, c f} \] Input:
integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxi ma")
Output:
-1/96*I*(4*(15*(-I*c*tan(f*x + e) + c)^2 - 20*(-I*c*tan(f*x + e) + c)*c - 8*c^2)/((-I*c*tan(f*x + e) + c)^(5/2)*a - 2*(-I*c*tan(f*x + e) + c)^(3/2)* a*c) + 15*sqrt(2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sq rt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/(a*sqrt(c)))/(c*f)
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac ")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Time = 0.53 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {\frac {c\,1{}\mathrm {i}}{3\,a\,f}+\frac {\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{6\,a\,f}-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,5{}\mathrm {i}}{8\,a\,c\,f}}{2\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,5{}\mathrm {i}}{16\,a\,{\left (-c\right )}^{3/2}\,f} \] Input:
int(1/((a + a*tan(e + f*x)*1i)*(c - c*tan(e + f*x)*1i)^(3/2)),x)
Output:
(2^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*5i)/ (16*a*(-c)^(3/2)*f) - ((c*1i)/(3*a*f) + ((c - c*tan(e + f*x)*1i)*5i)/(6*a* f) - ((c - c*tan(e + f*x)*1i)^2*5i)/(8*a*c*f))/(2*c*(c - c*tan(e + f*x)*1i )^(3/2) - (c - c*tan(e + f*x)*1i)^(5/2))
\[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {2 \sqrt {c}\, i \left (\sqrt {-\tan \left (f x +e \right ) i +1}+4 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{4} f +8 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} f +4 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{2}+1}d x \right ) f \right )}{a \,c^{2} f \left (\tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{2}+1\right )} \] Input:
int(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x)
Output:
(2*sqrt(c)*i*(sqrt( - tan(e + f*x)*i + 1) + 4*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**4 + 2*tan(e + f*x)**2 + 1),x)*tan(e + f* x)**4*f + 8*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)** 4 + 2*tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*f + 4*int((sqrt( - tan(e + f *x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**4 + 2*tan(e + f*x)**2 + 1),x)*f))/ (a*c**2*f*(tan(e + f*x)**4 + 2*tan(e + f*x)**2 + 1))