\(\int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx\) [825]

Optimal result

Integrand size = 45, antiderivative size = 102 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac {(i A-8 B) (a+i a \tan (e+f x))^{7/2}}{63 c f (c-i c \tan (e+f x))^{7/2}} \] Output:

-1/9*(I*A+B)*(a+I*a*tan(f*x+e))^(7/2)/f/(c-I*c*tan(f*x+e))^(9/2)-1/63*(I*A 
-8*B)*(a+I*a*tan(f*x+e))^(7/2)/c/f/(c-I*c*tan(f*x+e))^(7/2)
 

Mathematica [A] (verified)

Time = 14.21 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.19 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=\frac {a^3 \cos (e+f x) ((-8 i A+B) \cos (e+f x)-(A+8 i B) \sin (e+f x)) (\cos (8 e+11 f x)+i \sin (8 e+11 f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{63 c^5 f (\cos (f x)+i \sin (f x))^3} \] Input:

Integrate[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan 
[e + f*x])^(9/2),x]
 

Output:

(a^3*Cos[e + f*x]*(((-8*I)*A + B)*Cos[e + f*x] - (A + (8*I)*B)*Sin[e + f*x 
])*(Cos[8*e + 11*f*x] + I*Sin[8*e + 11*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sq 
rt[c - I*c*Tan[e + f*x]])/(63*c^5*f*(Cos[f*x] + I*Sin[f*x])^3)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3042, 4071, 87, 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.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}}dx\)

\(\Big \downarrow \) 4071

\(\displaystyle \frac {a c \int \frac {(i \tan (e+f x) a+a)^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {a c \left (\frac {(A+8 i B) \int \frac {(i \tan (e+f x) a+a)^{5/2}}{(c-i c \tan (e+f x))^{9/2}}d\tan (e+f x)}{9 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{9 a c (c-i c \tan (e+f x))^{9/2}}\right )}{f}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {a c \left (-\frac {i (A+8 i B) (a+i a \tan (e+f x))^{7/2}}{63 a c^2 (c-i c \tan (e+f x))^{7/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{9 a c (c-i c \tan (e+f x))^{9/2}}\right )}{f}\)

Input:

Int[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f 
*x])^(9/2),x]
 

Output:

(a*c*(-1/9*((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(a*c*(c - I*c*Tan[e + 
f*x])^(9/2)) - ((I/63)*(A + (8*I)*B)*(a + I*a*Tan[e + f*x])^(7/2))/(a*c^2* 
(c - I*c*Tan[e + f*x])^(7/2))))/f
 

Defintions of rubi rules used

rule 48

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]
 

rule 87

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]))))
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4071

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]
 

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04

method	result	size
risch	\(-\frac {a^{3} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (7 i A \,{\mathrm e}^{8 i \left (f x +e \right )}+7 B \,{\mathrm e}^{8 i \left (f x +e \right )}+9 i A \,{\mathrm e}^{6 i \left (f x +e \right )}-9 B \,{\mathrm e}^{6 i \left (f x +e \right )}\right )}{126 c^{4} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(106\)
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (8 i B \tan \left (f x +e \right )^{3}+6 i A \tan \left (f x +e \right )^{2}+A \tan \left (f x +e \right )^{3}-6 i B \tan \left (f x +e \right )+15 B \tan \left (f x +e \right )^{2}-8 i A +15 A \tan \left (f x +e \right )+B \right )}{63 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}\)	\(134\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (8 i B \tan \left (f x +e \right )^{3}+6 i A \tan \left (f x +e \right )^{2}+A \tan \left (f x +e \right )^{3}-6 i B \tan \left (f x +e \right )+15 B \tan \left (f x +e \right )^{2}-8 i A +15 A \tan \left (f x +e \right )+B \right )}{63 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}\)	\(134\)
parts	\(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (6 i \tan \left (f x +e \right )^{2}+\tan \left (f x +e \right )^{3}-8 i+15 \tan \left (f x +e \right )\right )}{63 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}+\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (15 i \tan \left (f x +e \right )^{2}-8 \tan \left (f x +e \right )^{3}+i+6 \tan \left (f x +e \right )\right )}{63 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}\)	\(193\)

Input:

int((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(9/2),x,m 
ethod=_RETURNVERBOSE)
 

Output:

-1/126*a^3/c^4*(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*(7*I*A*exp(8*I*(f*x+e))+7*B*exp(8*I*(f*x+e))+9*I*A*e 
xp(6*I*(f*x+e))-9*B*exp(6*I*(f*x+e)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {{\left (7 \, {\left (i \, A + B\right )} a^{3} e^{\left (11 i \, f x + 11 i \, e\right )} + 2 \, {\left (8 i \, A - B\right )} a^{3} e^{\left (9 i \, f x + 9 i \, e\right )} + 9 \, {\left (i \, A - B\right )} a^{3} e^{\left (7 i \, f x + 7 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{126 \, c^{5} f} \] Input:

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(9/ 
2),x, algorithm="fricas")
 

Output:

-1/126*(7*(I*A + B)*a^3*e^(11*I*f*x + 11*I*e) + 2*(8*I*A - B)*a^3*e^(9*I*f 
*x + 9*I*e) + 9*(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^5*f)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(f*x+e))**(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**( 
9/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (78) = 156\).

Time = 0.23 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.63 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {126 \, {\left (7 \, {\left (A - i \, B\right )} a^{3} \cos \left (11 \, f x + 11 \, e\right ) + 2 \, {\left (8 \, A + i \, B\right )} a^{3} \cos \left (9 \, f x + 9 \, e\right ) + 9 \, {\left (A + i \, B\right )} a^{3} \cos \left (7 \, f x + 7 \, e\right ) - 7 \, {\left (-i \, A - B\right )} a^{3} \sin \left (11 \, f x + 11 \, e\right ) - 2 \, {\left (-8 i \, A + B\right )} a^{3} \sin \left (9 \, f x + 9 \, e\right ) - 9 \, {\left (-i \, A + B\right )} a^{3} \sin \left (7 \, f x + 7 \, e\right )\right )} \sqrt {a} \sqrt {c}}{-15876 \, {\left (i \, c^{5} \cos \left (2 \, f x + 2 \, e\right ) - c^{5} \sin \left (2 \, f x + 2 \, e\right ) + i \, c^{5}\right )} f} \] Input:

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(9/ 
2),x, algorithm="maxima")
 

Output:

-126*(7*(A - I*B)*a^3*cos(11*f*x + 11*e) + 2*(8*A + I*B)*a^3*cos(9*f*x + 9 
*e) + 9*(A + I*B)*a^3*cos(7*f*x + 7*e) - 7*(-I*A - B)*a^3*sin(11*f*x + 11* 
e) - 2*(-8*I*A + B)*a^3*sin(9*f*x + 9*e) - 9*(-I*A + B)*a^3*sin(7*f*x + 7* 
e))*sqrt(a)*sqrt(c)/((-15876*I*c^5*cos(2*f*x + 2*e) + 15876*c^5*sin(2*f*x 
+ 2*e) - 15876*I*c^5)*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(9/ 
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 TypeError: Bad Argument TypeError: Bad Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 7.83 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.88 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {a^3\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (6\,e+6\,f\,x\right )\,9{}\mathrm {i}+A\,\cos \left (8\,e+8\,f\,x\right )\,7{}\mathrm {i}-9\,B\,\cos \left (6\,e+6\,f\,x\right )+7\,B\,\cos \left (8\,e+8\,f\,x\right )-9\,A\,\sin \left (6\,e+6\,f\,x\right )-7\,A\,\sin \left (8\,e+8\,f\,x\right )-B\,\sin \left (6\,e+6\,f\,x\right )\,9{}\mathrm {i}+B\,\sin \left (8\,e+8\,f\,x\right )\,7{}\mathrm {i}\right )}{126\,c^4\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \] Input:

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(7/2))/(c - c*tan(e + f* 
x)*1i)^(9/2),x)
 

Output:

-(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)*9i + A*cos(8*e + 8*f*x)*7i - 9*B*cos(6*e + 
 6*f*x) + 7*B*cos(8*e + 8*f*x) - 9*A*sin(6*e + 6*f*x) - 7*A*sin(8*e + 8*f* 
x) - B*sin(6*e + 6*f*x)*9i + B*sin(8*e + 8*f*x)*7i))/(126*c^4*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))
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx =\text {Too large to display} \] Input:

int((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(9/2),x)
 

Output:

(sqrt(c)*sqrt(a)*a**3*( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i 
+ 1)*tan(e + f*x)*a - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1) 
*b - 2*int(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e 
+ f*x)**4)/(tan(e + f*x)**6 + 4*tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 5* 
tan(e + f*x)**2 - 4*tan(e + f*x)*i + 1),x)*tan(e + f*x)**2*a*f + 8*int(( - 
 sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(ta 
n(e + f*x)**6 + 4*tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 5*tan(e + f*x)** 
2 - 4*tan(e + f*x)*i + 1),x)*tan(e + f*x)**2*b*f*i - 2*int(( - sqrt(tan(e 
+ f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)** 
6 + 4*tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 5*tan(e + f*x)**2 - 4*tan(e 
+ f*x)*i + 1),x)*a*f + 8*int(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + 
f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**6 + 4*tan(e + f*x)**5*i - 5*ta 
n(e + f*x)**4 - 5*tan(e + f*x)**2 - 4*tan(e + f*x)*i + 1),x)*b*f*i - 2*int 
(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)** 
6 + 4*tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 5*tan(e + f*x)**2 - 4*tan(e 
+ f*x)*i + 1),x)*tan(e + f*x)**2*a*f - 2*int(( - sqrt(tan(e + f*x)*i + 1)* 
sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**6 + 4*tan(e + f*x)**5*i - 5*ta 
n(e + f*x)**4 - 5*tan(e + f*x)**2 - 4*tan(e + f*x)*i + 1),x)*a*f - 12*int( 
(sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(ta 
n(e + f*x)**6 + 4*tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 5*tan(e + f*x...