Integrand size = 33, antiderivative size = 150 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a^2 f}+\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{8 a^2 f \left (c^2+i c^2 \tan (e+f x)\right )} \] Output:
-1/16*I*c^(3/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1 /2)/a^2/f+1/2*I*c^3*(c-I*c*tan(f*x+e))^(1/2)/a^2/f/(c+I*c*tan(f*x+e))^2-1/ 8*I*c^3*(c-I*c*tan(f*x+e))^(1/2)/a^2/f/(c^2+I*c^2*tan(f*x+e))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.67 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.35 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {i \operatorname {Hypergeometric2F1}\left (\frac {3}{2},3,\frac {5}{2},-\frac {1}{2} i (i+\tan (e+f x))\right ) (c-i c \tan (e+f x))^{3/2}}{12 a^2 f} \] Input:
Integrate[(c - I*c*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^2,x]
Output:
((I/12)*Hypergeometric2F1[3/2, 3, 5/2, (-1/2*I)*(I + Tan[e + f*x])]*(c - I *c*Tan[e + f*x])^(3/2))/(a^2*f)
Time = 0.42 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.85, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3042, 4005, 3042, 3968, 51, 52, 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.
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle \frac {\int \cos ^4(e+f x) (c-i c \tan (e+f x))^{7/2}dx}{a^2 c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(c-i c \tan (e+f x))^{7/2}}{\sec (e+f x)^4}dx}{a^2 c^2}\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i c^3 \int \frac {\sqrt {c-i c \tan (e+f x)}}{(i \tan (e+f x) c+c)^3}d(-i c \tan (e+f x))}{a^2 f}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {i c^3 \left (\frac {\sqrt {c-i c \tan (e+f x)}}{2 (c+i c \tan (e+f x))^2}-\frac {1}{4} \int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))\right )}{a^2 f}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {i c^3 \left (\frac {1}{4} \left (-\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}-\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}\right )+\frac {\sqrt {c-i c \tan (e+f x)}}{2 (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {i c^3 \left (\frac {1}{4} \left (-\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{2 c}-\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}\right )+\frac {\sqrt {c-i c \tan (e+f x)}}{2 (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {i c^3 \left (\frac {1}{4} \left (\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{2 \sqrt {2} c^{3/2}}-\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}\right )+\frac {\sqrt {c-i c \tan (e+f x)}}{2 (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
Input:
Int[(c - I*c*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^2,x]
Output:
(I*c^3*(Sqrt[c - I*c*Tan[e + f*x]]/(2*(c + I*c*Tan[e + f*x])^2) + (((I/2)* ArcTan[(Sqrt[c]*Tan[e + f*x])/Sqrt[2]])/(Sqrt[2]*c^(3/2)) - Sqrt[c - I*c*T an[e + f*x]]/(2*c*(c + I*c*Tan[e + f*x])))/4))/(a^2*f)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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 = 0.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.65
method | result | size |
derivativedivides | \(\frac {2 i c^{3} \left (\frac {\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{16 c}+\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 c^{\frac {3}{2}}}\right )}{f \,a^{2}}\) | \(97\) |
default | \(\frac {2 i c^{3} \left (\frac {\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{16 c}+\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 c^{\frac {3}{2}}}\right )}{f \,a^{2}}\) | \(97\) |
Input:
int((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
Output:
2*I/f/a^2*c^3*(4*(1/64/c*(c-I*c*tan(f*x+e))^(3/2)+1/32*(c-I*c*tan(f*x+e))^ (1/2))/(c+I*c*tan(f*x+e))^2-1/32/c^(3/2)*2^(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. 291 vs. \(2 (115) = 230\).
Time = 0.07 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.94 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {{\left (\sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c^{3}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{3}}{a^{4} f^{2}}} + i \, c^{2}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2} f}\right ) - \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c^{3}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{3}}{a^{4} f^{2}}} - i \, c^{2}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2} f}\right ) - \sqrt {2} {\left (i \, c e^{\left (4 i \, f x + 4 i \, e\right )} + 3 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} f} \] Input:
integrate((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="fric as")
Output:
-1/16*(sqrt(1/2)*a^2*f*sqrt(-c^3/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/4*( sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c^3/(a^4*f^2)) + I*c^2)*e^(-I*f*x - I*e)/(a^2*f)) - sq rt(1/2)*a^2*f*sqrt(-c^3/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/4*(sqrt(2)*sq rt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c^3/(a^4*f^2)) - I*c^2)*e^(-I*f*x - I*e)/(a^2*f)) - sqrt(2)*(I*c *e^(4*I*f*x + 4*I*e) + 3*I*c*e^(2*I*f*x + 2*I*e) + 2*I*c)*sqrt(c/(e^(2*I*f *x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*f)
\[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {c \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \left (- \frac {i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\right )\, dx}{a^{2}} \] Input:
integrate((c-I*c*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**2,x)
Output:
-(Integral(c*sqrt(-I*c*tan(e + f*x) + c)/(tan(e + f*x)**2 - 2*I*tan(e + f* x) - 1), x) + Integral(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)/(tan( e + f*x)**2 - 2*I*tan(e + f*x) - 1), x))/a**2
Time = 0.15 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {i \, {\left (\frac {\sqrt {2} c^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}} + \frac {4 \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{3} + 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}}\right )}}{32 \, c f} \] Input:
integrate((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="maxi ma")
Output:
1/32*I*(sqrt(2)*c^(5/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c ))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/a^2 + 4*((-I*c*tan(f*x + e) + c)^(3/2)*c^3 + 2*sqrt(-I*c*tan(f*x + e) + c)*c^4)/((-I*c*tan(f*x + e) + c)^2*a^2 - 4*(-I*c*tan(f*x + e) + c)*a^2*c + 4*a^2*c^2))/(c*f)
Exception generated. \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^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 = 2.04 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.89 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {\frac {c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{4\,a^2\,f}+\frac {c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,1{}\mathrm {i}}{8\,a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}+\frac {\sqrt {2}\,{\left (-c\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{16\,a^2\,f} \] Input:
int((c - c*tan(e + f*x)*1i)^(3/2)/(a + a*tan(e + f*x)*1i)^2,x)
Output:
((c^3*(c - c*tan(e + f*x)*1i)^(1/2)*1i)/(4*a^2*f) + (c^2*(c - c*tan(e + f* x)*1i)^(3/2)*1i)/(8*a^2*f))/((c - c*tan(e + f*x)*1i)^2 - 4*c*(c - c*tan(e + f*x)*1i) + 4*c^2) + (2^(1/2)*(-c)^(3/2)*atan((2^(1/2)*(c - c*tan(e + f*x )*1i)^(1/2))/(2*(-c)^(1/2)))*1i)/(16*a^2*f)
\[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {\sqrt {c}\, c \left (-\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right )+\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right ) i -1}d x \right ) i \right )}{a^{2}} \] Input:
int((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x)
Output:
(sqrt(c)*c*( - int(sqrt( - tan(e + f*x)*i + 1)/(tan(e + f*x)**2 - 2*tan(e + f*x)*i - 1),x) + int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**2 - 2*tan(e + f*x)*i - 1),x)*i))/a**2