\(\int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [760]

Optimal result

Integrand size = 43, antiderivative size = 140 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {8 a^3 (i A+B)}{f \sqrt {c-i c \tan (e+f x)}}-\frac {8 a^3 (i A+2 B) \sqrt {c-i c \tan (e+f x)}}{c f}+\frac {2 a^3 (i A+5 B) (c-i c \tan (e+f x))^{3/2}}{3 c^2 f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{5/2}}{5 c^3 f} \] Output:

-8*a^3*(I*A+B)/f/(c-I*c*tan(f*x+e))^(1/2)-8*a^3*(I*A+2*B)*(c-I*c*tan(f*x+e 
))^(1/2)/c/f+2/3*a^3*(I*A+5*B)*(c-I*c*tan(f*x+e))^(3/2)/c^2/f-2/5*a^3*B*(c 
-I*c*tan(f*x+e))^(5/2)/c^3/f
 

Mathematica [A] (verified)

Time = 3.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.59 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i a^3 \left (115 A-158 i B+(-50 i A-79 B) \tan (e+f x)+(5 A-16 i B) \tan ^2(e+f x)+3 B \tan ^3(e+f x)\right )}{15 f \sqrt {c-i c \tan (e+f x)}} \] Input:

Integrate[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan 
[e + f*x]],x]
 

Output:

(((-2*I)/15)*a^3*(115*A - (158*I)*B + ((-50*I)*A - 79*B)*Tan[e + f*x] + (5 
*A - (16*I)*B)*Tan[e + f*x]^2 + 3*B*Tan[e + f*x]^3))/(f*Sqrt[c - I*c*Tan[e 
 + f*x]])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.91, 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.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}}dx\)

\(\Big \downarrow \) 4071

\(\displaystyle \frac {a c \int \frac {a^2 (i \tan (e+f x)+1)^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^3 c \int \frac {(i \tan (e+f x)+1)^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {a^3 c \int \left (\frac {i B (c-i c \tan (e+f x))^{3/2}}{c^3}+\frac {(A-5 i B) \sqrt {c-i c \tan (e+f x)}}{c^2}-\frac {4 (A-2 i B)}{c \sqrt {c-i c \tan (e+f x)}}+\frac {4 (A-i B)}{(c-i c \tan (e+f x))^{3/2}}\right )d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 c \left (\frac {2 (5 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c^3}-\frac {8 (2 B+i A) \sqrt {c-i c \tan (e+f x)}}{c^2}-\frac {8 (B+i A)}{c \sqrt {c-i c \tan (e+f x)}}-\frac {2 B (c-i c \tan (e+f x))^{5/2}}{5 c^4}\right )}{f}\)

Input:

Int[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f 
*x]],x]
 

Output:

(a^3*c*((-8*(I*A + B))/(c*Sqrt[c - I*c*Tan[e + f*x]]) - (8*(I*A + 2*B)*Sqr 
t[c - I*c*Tan[e + f*x]])/c^2 + (2*(I*A + 5*B)*(c - I*c*Tan[e + f*x])^(3/2) 
)/(3*c^3) - (2*B*(c - I*c*Tan[e + f*x])^(5/2))/(5*c^4)))/f
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 86

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]))))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 = 0.60 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {5 i B c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+\frac {A c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+8 i \sqrt {c -i c \tan \left (f x +e \right )}\, B \,c^{2}-4 \sqrt {c -i c \tan \left (f x +e \right )}\, A \,c^{2}-\frac {4 c^{3} \left (-i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{3}}\)	\(135\)
default	\(\frac {2 i a^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {5 i B c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+\frac {A c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+8 i \sqrt {c -i c \tan \left (f x +e \right )}\, B \,c^{2}-4 \sqrt {c -i c \tan \left (f x +e \right )}\, A \,c^{2}-\frac {4 c^{3} \left (-i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{3}}\)	\(135\)
parts	\(\frac {2 i A \,a^{3} c \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}-\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}+\frac {a^{3} \left (3 i A +B \right ) \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}-\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}+\frac {2 B \,a^{3} \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 \sqrt {c -i c \tan \left (f x +e \right )}\, c^{2}+\frac {c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4}-\frac {c^{3}}{2 \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{3}}-\frac {6 i a^{3} \left (-i B +A \right ) \left (\sqrt {c -i c \tan \left (f x +e \right )}-\frac {\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4}+\frac {c}{2 \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}-\frac {2 a^{3} \left (i A +3 B \right ) \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+c \sqrt {c -i c \tan \left (f x +e \right )}+\frac {c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4}+\frac {c^{2}}{2 \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}}\)	\(423\)

Input:

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x,metho 
d=_RETURNVERBOSE)
 

Output:

2*I/f*a^3/c^3*(1/5*I*B*(c-I*c*tan(f*x+e))^(5/2)-5/3*I*B*c*(c-I*c*tan(f*x+e 
))^(3/2)+1/3*A*c*(c-I*c*tan(f*x+e))^(3/2)+8*I*(c-I*c*tan(f*x+e))^(1/2)*B*c 
^2-4*(c-I*c*tan(f*x+e))^(1/2)*A*c^2-4*c^3*(A-I*B)/(c-I*c*tan(f*x+e))^(1/2) 
)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.91 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {4 \, \sqrt {2} {\left (15 \, {\left (i \, A + B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 15 \, {\left (5 i \, A + 7 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 20 \, {\left (5 i \, A + 7 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (5 i \, A + 7 \, B\right )} a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, {\left (c f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )}} \] Input:

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x 
, algorithm="fricas")
 

Output:

-4/15*sqrt(2)*(15*(I*A + B)*a^3*e^(6*I*f*x + 6*I*e) + 15*(5*I*A + 7*B)*a^3 
*e^(4*I*f*x + 4*I*e) + 20*(5*I*A + 7*B)*a^3*e^(2*I*f*x + 2*I*e) + 8*(5*I*A 
 + 7*B)*a^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c*f*e^(4*I*f*x + 4*I*e) + 
2*c*f*e^(2*I*f*x + 2*I*e) + c*f)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=- i a^{3} \left (\int \frac {i A}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 A \tan {\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan ^{3}{\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 B \tan ^{2}{\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {B \tan ^{4}{\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 i A \tan ^{2}{\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {i B \tan {\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 i B \tan ^{3}{\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \] Input:

integrate((a+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(1/2) 
,x)
 

Output:

-I*a**3*(Integral(I*A/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-3*A*tan( 
e + f*x)/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(A*tan(e + f*x)**3/sqrt 
(-I*c*tan(e + f*x) + c), x) + Integral(-3*B*tan(e + f*x)**2/sqrt(-I*c*tan( 
e + f*x) + c), x) + Integral(B*tan(e + f*x)**4/sqrt(-I*c*tan(e + f*x) + c) 
, x) + Integral(-3*I*A*tan(e + f*x)**2/sqrt(-I*c*tan(e + f*x) + c), x) + I 
ntegral(I*B*tan(e + f*x)/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-3*I*B 
*tan(e + f*x)**3/sqrt(-I*c*tan(e + f*x) + c), x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.77 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i \, {\left (\frac {60 \, {\left (A - i \, B\right )} a^{3} c}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}} - \frac {3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} B a^{3} + 5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - 5 i \, B\right )} a^{3} c - 60 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - 2 i \, B\right )} a^{3} c^{2}}{c^{2}}\right )}}{15 \, c f} \] Input:

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x 
, algorithm="maxima")
 

Output:

-2/15*I*(60*(A - I*B)*a^3*c/sqrt(-I*c*tan(f*x + e) + c) - (3*I*(-I*c*tan(f 
*x + e) + c)^(5/2)*B*a^3 + 5*(-I*c*tan(f*x + e) + c)^(3/2)*(A - 5*I*B)*a^3 
*c - 60*sqrt(-I*c*tan(f*x + e) + c)*(A - 2*I*B)*a^3*c^2)/c^2)/(c*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),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 = 8.53 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.51 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c\,f}+\frac {a^3\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (A-B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c\,f}\right )-\left (\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c\,f}+\frac {a^3\,\left (A-B\,3{}\mathrm {i}\right )\,4{}\mathrm {i}}{c\,f}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}-\frac {\left (\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{5\,c\,f}-\frac {a^3\,\left (A+B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{5\,c\,f}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\left (\frac {16\,B\,a^3}{3\,c\,f}+\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{3\,c\,f}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1} \] Input:

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3)/(c - c*tan(e + f*x)*1 
i)^(1/2),x)
 

Output:

(((a^3*(A - B*1i)*4i)/(3*c*f) + (16*B*a^3)/(3*c*f))*(c + (c*(exp(e*2i + f* 
x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2))/(exp(e*2i + f*x*2i) + 
1) - ((a^3*(A - B*1i)*4i)/(c*f) + (a^3*(A - B*3i)*4i)/(c*f))*(c + (c*(exp( 
e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2) - (((a^3*(A - 
B*1i)*4i)/(5*c*f) - (a^3*(A + B*1i)*4i)/(5*c*f))*(c + (c*(exp(e*2i + f*x*2 
i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2))/(exp(e*2i + f*x*2i) + 1)^ 
2 - (c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/ 
2)*((a^3*(A - B*1i)*4i)/(c*f) + (a^3*exp(e*2i + f*x*2i)*(A - B*1i)*4i)/(c* 
f))
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx =\text {Too large to display} \] Input:

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x)
 

Output:

(sqrt(c)*a**3*(2*sqrt( - tan(e + f*x)*i + 1)*a*i + int((sqrt( - tan(e + f* 
x)*i + 1)*tan(e + f*x)**5)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*b*f + 
int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**5)/(tan(e + f*x)**2 + 1),x) 
*b*f + int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**2 
+ 1),x)*tan(e + f*x)**2*a*f - 4*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f 
*x)**4)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*b*f*i + int((sqrt( - tan( 
e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**2 + 1),x)*a*f - 4*int((sqr 
t( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**2 + 1),x)*b*f*i - 
 4*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**3)/(tan(e + f*x)**2 + 1) 
,x)*tan(e + f*x)**2*a*f*i - 6*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x 
)**3)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*b*f - 4*int((sqrt( - tan(e 
+ f*x)*i + 1)*tan(e + f*x)**3)/(tan(e + f*x)**2 + 1),x)*a*f*i - 6*int((sqr 
t( - tan(e + f*x)*i + 1)*tan(e + f*x)**3)/(tan(e + f*x)**2 + 1),x)*b*f - 6 
*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**2 + 1),x 
)*tan(e + f*x)**2*a*f + 4*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2 
)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*b*f*i - 6*int((sqrt( - tan(e + 
f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**2 + 1),x)*a*f + 4*int((sqrt( - 
 tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**2 + 1),x)*b*f*i + 7*i 
nt((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**2 + 1),x)*tan 
(e + f*x)**2*a*f*i + int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(ta...