Integrand size = 43, antiderivative size = 105 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {4 a^2 (i A+B)}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {2 a^2 (i A+3 B)}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^2 B}{3 c^2 f (c-i c \tan (e+f x))^{3/2}} \]
-4/7*a^2*(I*A+B)/f/(c-I*c*tan(f*x+e))^(7/2)+2/5*a^2*(I*A+3*B)/c/f/(c-I*c*t an(f*x+e))^(5/2)-2/3*a^2*B/c^2/f/(c-I*c*tan(f*x+e))^(3/2)
Time = 6.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.75 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\frac {2 a^2 \left (-9 A+2 i B+7 (-3 i A+B) \tan (e+f x)-35 i B \tan ^2(e+f x)\right )}{105 c^3 f (i+\tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \]
(2*a^2*(-9*A + (2*I)*B + 7*((-3*I)*A + B)*Tan[e + f*x] - (35*I)*B*Tan[e + f*x]^2))/(105*c^3*f*(I + Tan[e + f*x])^3*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.38 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3042, 4071, 27, 86, 2009}
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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {a (i \tan (e+f x)+1) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^2 c \int \frac {(i \tan (e+f x)+1) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {a^2 c \int \left (\frac {2 (A-i B)}{(c-i c \tan (e+f x))^{9/2}}-\frac {i B}{c^2 (c-i c \tan (e+f x))^{5/2}}+\frac {3 i B-A}{c (c-i c \tan (e+f x))^{7/2}}\right )d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 c \left (\frac {2 (3 B+i A)}{5 c^2 (c-i c \tan (e+f x))^{5/2}}-\frac {4 (B+i A)}{7 c (c-i c \tan (e+f x))^{7/2}}-\frac {2 B}{3 c^3 (c-i c \tan (e+f x))^{3/2}}\right )}{f}\) |
(a^2*c*((-4*(I*A + B))/(7*c*(c - I*c*Tan[e + f*x])^(7/2)) + (2*(I*A + 3*B) )/(5*c^2*(c - I*c*Tan[e + f*x])^(5/2)) - (2*B)/(3*c^3*(c - I*c*Tan[e + f*x ])^(3/2))))/f
3.8.55.3.1 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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x], x , Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.76
| method | result | size |
| derivativedivides | \(\frac {2 i a^{2} \left (\frac {i B}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2 c^{2} \left (-i B +A \right )}{7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}+\frac {c \left (-3 i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,c^{2}}\) | \(80\) |
| default | \(\frac {2 i a^{2} \left (\frac {i B}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2 c^{2} \left (-i B +A \right )}{7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}+\frac {c \left (-3 i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,c^{2}}\) | \(80\) |
| risch | \(-\frac {a^{2} \left (15 i A \,{\mathrm e}^{6 i \left (f x +e \right )}+15 B \,{\mathrm e}^{6 i \left (f x +e \right )}+24 i A \,{\mathrm e}^{4 i \left (f x +e \right )}-18 B \,{\mathrm e}^{4 i \left (f x +e \right )}+3 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-11 B \,{\mathrm e}^{2 i \left (f x +e \right )}-6 i A +22 B \right ) \sqrt {2}}{420 c^{3} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) | \(115\) |
| parts | \(\frac {2 i A \,a^{2} c \left (-\frac {1}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{24 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{20 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {1}{14 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{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 {9}{2}}}\right )}{f}+\frac {a^{2} \left (2 i A +B \right ) \left (-\frac {1}{7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}+\frac {1}{8 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 {1}{10 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 c^{\frac {7}{2}}}\right )}{f}-\frac {2 B \,a^{2} \left (\frac {7}{24 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{16 c \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {c}{4 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {c^{2}}{14 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{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 \,c^{2}}-\frac {2 i a^{2} \left (-2 i B +A \right ) \left (-\frac {3}{20 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {1}{16 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {1}{24 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {c}{14 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{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 {5}{2}}}\right )}{f c}\) | \(475\) |
2*I/f*a^2/c^2*(1/3*I*B/(c-I*c*tan(f*x+e))^(3/2)-2/7*c^2*(A-I*B)/(c-I*c*tan (f*x+e))^(7/2)+1/5*c*(A-3*I*B)/(c-I*c*tan(f*x+e))^(5/2))
Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.17 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {\sqrt {2} {\left (15 \, {\left (i \, A + B\right )} a^{2} e^{\left (8 i \, f x + 8 i \, e\right )} + 3 \, {\left (13 i \, A - B\right )} a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (-27 i \, A + 29 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (3 i \, A - 11 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (-3 i \, A + 11 \, B\right )} a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{420 \, c^{4} f} \]
integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x , algorithm="fricas")
-1/420*sqrt(2)*(15*(I*A + B)*a^2*e^(8*I*f*x + 8*I*e) + 3*(13*I*A - B)*a^2* e^(6*I*f*x + 6*I*e) - (-27*I*A + 29*B)*a^2*e^(4*I*f*x + 4*I*e) - (3*I*A - 11*B)*a^2*e^(2*I*f*x + 2*I*e) + 2*(-3*I*A + 11*B)*a^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^4*f)
\[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=- a^{2} \left (\int \left (- \frac {A}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan ^{2}{\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {B \tan {\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {B \tan ^{3}{\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {2 i A \tan {\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \left (- \frac {2 i B \tan ^{2}{\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \]
-a**2*(Integral(-A/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3 *c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan (e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)), x) + Inte gral(A*tan(e + f*x)**2/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c *tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(-B*tan(e + f*x)/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)* *3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I *c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(B*tan(e + f*x)**3/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f *x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqr t(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(-2*I*A*tan(e + f*x)/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan (e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c* *3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(-2*I*B*tan(e + f*x)**2/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan( e + f*x) + c)), x))
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.72 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\frac {2 i \, {\left (35 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} B a^{2} + 21 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (A - 3 i \, B\right )} a^{2} c - 30 \, {\left (A - i \, B\right )} a^{2} c^{2}\right )}}{105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} c^{2} f} \]
integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x , algorithm="maxima")
2/105*I*(35*I*(-I*c*tan(f*x + e) + c)^2*B*a^2 + 21*(-I*c*tan(f*x + e) + c) *(A - 3*I*B)*a^2*c - 30*(A - I*B)*a^2*c^2)/((-I*c*tan(f*x + e) + c)^(7/2)* c^2*f)
\[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]
integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x , algorithm="giac")
Time = 10.12 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.59 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\sqrt {c-\frac {c\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}\,\left (-\frac {a^2\,\left (3\,A+B\,11{}\mathrm {i}\right )\,1{}\mathrm {i}}{210\,c^4\,f}-\frac {a^2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (3\,A+B\,11{}\mathrm {i}\right )\,1{}\mathrm {i}}{420\,c^4\,f}+\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (13\,A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{140\,c^4\,f}+\frac {a^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (27\,A+B\,29{}\mathrm {i}\right )\,1{}\mathrm {i}}{420\,c^4\,f}+\frac {a^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\left (A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{28\,c^4\,f}\right ) \]
-(c - (c*sin(e + f*x)*1i)/cos(e + f*x))^(1/2)*((a^2*exp(e*6i + f*x*6i)*(13 *A + B*1i)*1i)/(140*c^4*f) - (a^2*exp(e*2i + f*x*2i)*(3*A + B*11i)*1i)/(42 0*c^4*f) - (a^2*(3*A + B*11i)*1i)/(210*c^4*f) + (a^2*exp(e*4i + f*x*4i)*(2 7*A + B*29i)*1i)/(420*c^4*f) + (a^2*exp(e*8i + f*x*8i)*(A - B*1i)*1i)/(28* c^4*f))