Integrand size = 45, antiderivative size = 155 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac {(2 i A-9 B) (a+i a \tan (e+f x))^{7/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac {(2 i A-9 B) (a+i a \tan (e+f x))^{7/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}} \]
-1/11*(I*A+B)*(a+I*a*tan(f*x+e))^(7/2)/f/(c-I*c*tan(f*x+e))^(11/2)-1/99*(2 *I*A-9*B)*(a+I*a*tan(f*x+e))^(7/2)/c/f/(c-I*c*tan(f*x+e))^(9/2)-1/693*(2*I *A-9*B)*(a+I*a*tan(f*x+e))^(7/2)/c^2/f/(c-I*c*tan(f*x+e))^(7/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(417\) vs. \(2(155)=310\).
Time = 18.82 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.69 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx=\frac {\cos ^4(e+f x) \left ((-i A+B) \cos (6 f x) \left (\frac {\cos (3 e)}{56 c^6}+\frac {i \sin (3 e)}{56 c^6}\right )+(-23 i A+9 B) \cos (8 f x) \left (\frac {\cos (5 e)}{504 c^6}+\frac {i \sin (5 e)}{504 c^6}\right )+(31 A-9 i B) \cos (10 f x) \left (-\frac {i \cos (7 e)}{792 c^6}+\frac {\sin (7 e)}{792 c^6}\right )+(A-i B) \cos (12 f x) \left (-\frac {i \cos (9 e)}{88 c^6}+\frac {\sin (9 e)}{88 c^6}\right )+(A+i B) \left (\frac {\cos (3 e)}{56 c^6}+\frac {i \sin (3 e)}{56 c^6}\right ) \sin (6 f x)+(23 A+9 i B) \left (\frac {\cos (5 e)}{504 c^6}+\frac {i \sin (5 e)}{504 c^6}\right ) \sin (8 f x)+(31 A-9 i B) \left (\frac {\cos (7 e)}{792 c^6}+\frac {i \sin (7 e)}{792 c^6}\right ) \sin (10 f x)+(A-i B) \left (\frac {\cos (9 e)}{88 c^6}+\frac {i \sin (9 e)}{88 c^6}\right ) \sin (12 f x)\right ) \sqrt {\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]
Integrate[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan [e + f*x])^(11/2),x]
(Cos[e + f*x]^4*(((-I)*A + B)*Cos[6*f*x]*(Cos[3*e]/(56*c^6) + ((I/56)*Sin[ 3*e])/c^6) + ((-23*I)*A + 9*B)*Cos[8*f*x]*(Cos[5*e]/(504*c^6) + ((I/504)*S in[5*e])/c^6) + (31*A - (9*I)*B)*Cos[10*f*x]*(((-1/792*I)*Cos[7*e])/c^6 + Sin[7*e]/(792*c^6)) + (A - I*B)*Cos[12*f*x]*(((-1/88*I)*Cos[9*e])/c^6 + Si n[9*e]/(88*c^6)) + (A + I*B)*(Cos[3*e]/(56*c^6) + ((I/56)*Sin[3*e])/c^6)*S in[6*f*x] + (23*A + (9*I)*B)*(Cos[5*e]/(504*c^6) + ((I/504)*Sin[5*e])/c^6) *Sin[8*f*x] + (31*A - (9*I)*B)*(Cos[7*e]/(792*c^6) + ((I/792)*Sin[7*e])/c^ 6)*Sin[10*f*x] + (A - I*B)*(Cos[9*e]/(88*c^6) + ((I/88)*Sin[9*e])/c^6)*Sin [12*f*x])*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a* Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A* Cos[e + f*x] + B*Sin[e + f*x]))
Time = 0.41 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 4071, 87, 55, 48}
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.
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}}dx\) |
\(\Big \downarrow \) 4071 |
\(\displaystyle \frac {a c \int \frac {(i \tan (e+f x) a+a)^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {a c \left (\frac {(2 A+9 i B) \int \frac {(i \tan (e+f x) a+a)^{5/2}}{(c-i c \tan (e+f x))^{11/2}}d\tan (e+f x)}{11 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{11 a c (c-i c \tan (e+f x))^{11/2}}\right )}{f}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {a c \left (\frac {(2 A+9 i B) \left (\frac {\int \frac {(i \tan (e+f x) a+a)^{5/2}}{(c-i c \tan (e+f x))^{9/2}}d\tan (e+f x)}{9 c}-\frac {i (a+i a \tan (e+f x))^{7/2}}{9 a c (c-i c \tan (e+f x))^{9/2}}\right )}{11 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{11 a c (c-i c \tan (e+f x))^{11/2}}\right )}{f}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {a c \left (\frac {(2 A+9 i B) \left (-\frac {i (a+i a \tan (e+f x))^{7/2}}{63 a c^2 (c-i c \tan (e+f x))^{7/2}}-\frac {i (a+i a \tan (e+f x))^{7/2}}{9 a c (c-i c \tan (e+f x))^{9/2}}\right )}{11 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{11 a c (c-i c \tan (e+f x))^{11/2}}\right )}{f}\) |
(a*c*(-1/11*((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(a*c*(c - I*c*Tan[e + f*x])^(11/2)) + ((2*A + (9*I)*B)*(((-1/9*I)*(a + I*a*Tan[e + f*x])^(7/2)) /(a*c*(c - I*c*Tan[e + f*x])^(9/2)) - ((I/63)*(a + I*a*Tan[e + f*x])^(7/2) )/(a*c^2*(c - I*c*Tan[e + f*x])^(7/2))))/(11*c)))/f
3.9.26.3.1 Defintions of rubi rules used
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] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
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*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.37 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.77
method | result | size |
risch | \(-\frac {a^{3} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (63 i A \,{\mathrm e}^{10 i \left (f x +e \right )}+63 B \,{\mathrm e}^{10 i \left (f x +e \right )}+154 i A \,{\mathrm e}^{8 i \left (f x +e \right )}+99 i A \,{\mathrm e}^{6 i \left (f x +e \right )}-99 B \,{\mathrm e}^{6 i \left (f x +e \right )}\right )}{2772 c^{5} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) | \(119\) |
derivativedivides | \(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (2 i A \tan \left (f x +e \right )^{4}-63 i B \tan \left (f x +e \right )^{3}-9 B \tan \left (f x +e \right )^{4}-45 i A \tan \left (f x +e \right )^{2}-14 A \tan \left (f x +e \right )^{3}+63 i \tan \left (f x +e \right ) B -144 B \tan \left (f x +e \right )^{2}+79 i A -140 A \tan \left (f x +e \right )-9 B \right )}{693 f \,c^{6} \left (i+\tan \left (f x +e \right )\right )^{7}}\) | \(161\) |
default | \(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (2 i A \tan \left (f x +e \right )^{4}-63 i B \tan \left (f x +e \right )^{3}-9 B \tan \left (f x +e \right )^{4}-45 i A \tan \left (f x +e \right )^{2}-14 A \tan \left (f x +e \right )^{3}+63 i \tan \left (f x +e \right ) B -144 B \tan \left (f x +e \right )^{2}+79 i A -140 A \tan \left (f x +e \right )-9 B \right )}{693 f \,c^{6} \left (i+\tan \left (f x +e \right )\right )^{7}}\) | \(161\) |
parts | \(\frac {i A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (2 i \tan \left (f x +e \right )^{4}-45 i \tan \left (f x +e \right )^{2}-14 \tan \left (f x +e \right )^{3}+79 i-140 \tan \left (f x +e \right )\right )}{693 f \,c^{6} \left (i+\tan \left (f x +e \right )\right )^{7}}-\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (7 i \tan \left (f x +e \right )^{3}+\tan \left (f x +e \right )^{4}-7 i \tan \left (f x +e \right )+16 \tan \left (f x +e \right )^{2}+1\right )}{77 f \,c^{6} \left (i+\tan \left (f x +e \right )\right )^{7}}\) | \(215\) |
int((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x, method=_RETURNVERBOSE)
-1/2772*a^3/c^5*(a*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^(1/2)/(c/(exp(2* I*(f*x+e))+1))^(1/2)/f*(63*I*A*exp(10*I*(f*x+e))+63*B*exp(10*I*(f*x+e))+15 4*I*A*exp(8*I*(f*x+e))+99*I*A*exp(6*I*(f*x+e))-99*B*exp(6*I*(f*x+e)))
Time = 0.25 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.81 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx=-\frac {{\left (63 \, {\left (i \, A + B\right )} a^{3} e^{\left (13 i \, f x + 13 i \, e\right )} + 7 \, {\left (31 i \, A + 9 \, B\right )} a^{3} e^{\left (11 i \, f x + 11 i \, e\right )} + 11 \, {\left (23 i \, A - 9 \, B\right )} a^{3} e^{\left (9 i \, f x + 9 i \, e\right )} + 99 \, {\left (i \, A - B\right )} a^{3} e^{\left (7 i \, f x + 7 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{2772 \, c^{6} f} \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11 /2),x, algorithm="fricas")
-1/2772*(63*(I*A + B)*a^3*e^(13*I*f*x + 13*I*e) + 7*(31*I*A + 9*B)*a^3*e^( 11*I*f*x + 11*I*e) + 11*(23*I*A - 9*B)*a^3*e^(9*I*f*x + 9*I*e) + 99*(I*A - B)*a^3*e^(7*I*f*x + 7*I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^( 2*I*f*x + 2*I*e) + 1))/(c^6*f)
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx=\text {Timed out} \]
Time = 0.63 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.28 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx=\frac {{\left (63 \, {\left (-i \, A - B\right )} a^{3} \cos \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 154 i \, A a^{3} \cos \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 99 \, {\left (-i \, A + B\right )} a^{3} \cos \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 63 \, {\left (A - i \, B\right )} a^{3} \sin \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 154 \, A a^{3} \sin \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 99 \, {\left (A + i \, B\right )} a^{3} \sin \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{2772 \, c^{\frac {11}{2}} f} \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11 /2),x, algorithm="maxima")
1/2772*(63*(-I*A - B)*a^3*cos(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2 *e))) - 154*I*A*a^3*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 99*(-I*A + B)*a^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 63*(A - I*B)*a^3*sin(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1 54*A*a^3*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 99*(A + I* B)*a^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(1 1/2)*f)
\[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {11}{2}}} \,d x } \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11 /2),x, algorithm="giac")
integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(7/2)/(-I*c*tan(f*x + e) + c)^(11/2), x)
Time = 10.82 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.40 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx=-\frac {a^3\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (6\,e+6\,f\,x\right )\,99{}\mathrm {i}+A\,\cos \left (8\,e+8\,f\,x\right )\,154{}\mathrm {i}+A\,\cos \left (10\,e+10\,f\,x\right )\,63{}\mathrm {i}-99\,B\,\cos \left (6\,e+6\,f\,x\right )+63\,B\,\cos \left (10\,e+10\,f\,x\right )-99\,A\,\sin \left (6\,e+6\,f\,x\right )-154\,A\,\sin \left (8\,e+8\,f\,x\right )-63\,A\,\sin \left (10\,e+10\,f\,x\right )-B\,\sin \left (6\,e+6\,f\,x\right )\,99{}\mathrm {i}+B\,\sin \left (10\,e+10\,f\,x\right )\,63{}\mathrm {i}\right )}{2772\,c^5\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]
-(a^3*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*cos(6*e + 6*f*x)*99i + A*cos(8*e + 8*f*x)*154i + A*cos(10*e + 10*f*x)*63i - 99*B*cos(6*e + 6*f*x) + 63*B*cos(10*e + 10*f*x) - 99*A*si n(6*e + 6*f*x) - 154*A*sin(8*e + 8*f*x) - 63*A*sin(10*e + 10*f*x) - B*sin( 6*e + 6*f*x)*99i + B*sin(10*e + 10*f*x)*63i))/(2772*c^5*f*((c*(cos(2*e + 2 *f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))