\(\int \frac {(d \tan (e+f x))^m}{(a+b \sqrt {c \tan (e+f x)})^2} \, dx\) [411]

Optimal result

Integrand size = 29, antiderivative size = 617 \[ \int \frac {(d \tan (e+f x))^m}{\left (a+b \sqrt {c \tan (e+f x)}\right )^2} \, dx=\frac {\left (a^6-3 a^2 b^4 c^2-3 a^4 b^2 \sqrt {-c^2}-b^6 \left (-c^2\right )^{3/2}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 \left (a^4+b^4 c^2\right )^2 f (1+m)}+\frac {\left (a^6-3 a^2 b^4 c^2+3 a^4 b^2 \sqrt {-c^2}+b^6 \left (-c^2\right )^{3/2}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 \left (a^4+b^4 c^2\right )^2 f (1+m)}+\frac {4 a^2 b^4 c^2 \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{\left (a^4+b^4 c^2\right )^2 f (1+m)}+\frac {b^4 c^2 \operatorname {Hypergeometric2F1}\left (2,2 (1+m),3+2 m,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a^2 \left (a^4+b^4 c^2\right ) f (1+m)}-\frac {2 a b \left (a^4-b^4 c^2-2 a^2 b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3+2 m),\frac {1}{2} (5+2 m),-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right )^2 f (3+2 m)}-\frac {2 a b \left (a^4-b^4 c^2+2 a^2 b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3+2 m),\frac {1}{2} (5+2 m),\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right )^2 f (3+2 m)} \] Output:

1/2*(a^6-3*a^2*b^4*c^2-3*a^4*b^2*(-c^2)^(1/2)-b^6*(-c^2)^(3/2))*hypergeom( 
[1, 1+m],[2+m],-c*tan(f*x+e)/(-c^2)^(1/2))*tan(f*x+e)*(d*tan(f*x+e))^m/(b^ 
4*c^2+a^4)^2/f/(1+m)+1/2*(a^6-3*a^2*b^4*c^2+3*a^4*b^2*(-c^2)^(1/2)+b^6*(-c 
^2)^(3/2))*hypergeom([1, 1+m],[2+m],c*tan(f*x+e)/(-c^2)^(1/2))*tan(f*x+e)* 
(d*tan(f*x+e))^m/(b^4*c^2+a^4)^2/f/(1+m)+4*a^2*b^4*c^2*hypergeom([1, 2+2*m 
],[3+2*m],-b*(c*tan(f*x+e))^(1/2)/a)*tan(f*x+e)*(d*tan(f*x+e))^m/(b^4*c^2+ 
a^4)^2/f/(1+m)+b^4*c^2*hypergeom([2, 2+2*m],[3+2*m],-b*(c*tan(f*x+e))^(1/2 
)/a)*tan(f*x+e)*(d*tan(f*x+e))^m/a^2/(b^4*c^2+a^4)/f/(1+m)-2*a*b*(a^4-b^4* 
c^2-2*a^2*b^2*(-c^2)^(1/2))*hypergeom([1, 3/2+m],[5/2+m],-c*tan(f*x+e)/(-c 
^2)^(1/2))*(c*tan(f*x+e))^(3/2)*(d*tan(f*x+e))^m/c/(b^4*c^2+a^4)^2/f/(3+2* 
m)-2*a*b*(a^4-b^4*c^2+2*a^2*b^2*(-c^2)^(1/2))*hypergeom([1, 3/2+m],[5/2+m] 
,c*tan(f*x+e)/(-c^2)^(1/2))*(c*tan(f*x+e))^(3/2)*(d*tan(f*x+e))^m/c/(b^4*c 
^2+a^4)^2/f/(3+2*m)
 

Mathematica [A] (verified)

Time = 3.89 (sec) , antiderivative size = 381, normalized size of antiderivative = 0.62 \[ \int \frac {(d \tan (e+f x))^m}{\left (a+b \sqrt {c \tan (e+f x)}\right )^2} \, dx=\frac {c (d \tan (e+f x))^m \left (\frac {a^2 \left (a^4-3 b^4 c^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{c (1+m)}+\frac {4 a^2 b^4 c \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right ) \tan (e+f x)}{1+m}+\frac {b^4 c \left (a^4+b^4 c^2\right ) \operatorname {Hypergeometric2F1}\left (2,2 (1+m),3+2 m,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right ) \tan (e+f x)}{a^2 (1+m)}+\frac {b^2 \left (3 a^4-b^4 c^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(e+f x)\right ) \tan ^2(e+f x)}{2+m}+\frac {4 a b \left (-a^4+b^4 c^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (3+2 m),\frac {1}{4} (7+2 m),-\tan ^2(e+f x)\right ) (c \tan (e+f x))^{3/2}}{c^2 (3+2 m)}-\frac {8 a^3 b^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (5+2 m),\frac {1}{4} (9+2 m),-\tan ^2(e+f x)\right ) (c \tan (e+f x))^{5/2}}{c^2 (5+2 m)}\right )}{\left (a^4+b^4 c^2\right )^2 f} \] Input:

Integrate[(d*Tan[e + f*x])^m/(a + b*Sqrt[c*Tan[e + f*x]])^2,x]
 

Output:

(c*(d*Tan[e + f*x])^m*((a^2*(a^4 - 3*b^4*c^2)*Hypergeometric2F1[1, (1 + m) 
/2, (3 + m)/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(c*(1 + m)) + (4*a^2*b^4*c*H 
ypergeometric2F1[1, 2*(1 + m), 3 + 2*m, -((b*Sqrt[c*Tan[e + f*x]])/a)]*Tan 
[e + f*x])/(1 + m) + (b^4*c*(a^4 + b^4*c^2)*Hypergeometric2F1[2, 2*(1 + m) 
, 3 + 2*m, -((b*Sqrt[c*Tan[e + f*x]])/a)]*Tan[e + f*x])/(a^2*(1 + m)) + (b 
^2*(3*a^4 - b^4*c^2)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[e + f 
*x]^2]*Tan[e + f*x]^2)/(2 + m) + (4*a*b*(-a^4 + b^4*c^2)*Hypergeometric2F1 
[1, (3 + 2*m)/4, (7 + 2*m)/4, -Tan[e + f*x]^2]*(c*Tan[e + f*x])^(3/2))/(c^ 
2*(3 + 2*m)) - (8*a^3*b^3*Hypergeometric2F1[1, (5 + 2*m)/4, (9 + 2*m)/4, - 
Tan[e + f*x]^2]*(c*Tan[e + f*x])^(5/2))/(c^2*(5 + 2*m))))/((a^4 + b^4*c^2) 
^2*f)
 

Rubi [A] (warning: unable to verify)

Time = 1.52 (sec) , antiderivative size = 625, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 4153, 7267, 30, 7276, 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.

\(\displaystyle \int \frac {(d \tan (e+f x))^m}{\left (a+b \sqrt {c \tan (e+f x)}\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \tan (e+f x))^m}{\left (a+b \sqrt {c \tan (e+f x)}\right )^2}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle \frac {c \int \frac {(d \tan (e+f x))^m}{\left (\tan ^2(e+f x) c^2+c^2\right ) \left (a+b \sqrt {c \tan (e+f x)}\right )^2}d(c \tan (e+f x))}{f}\)

\(\Big \downarrow \) 7267

\(\displaystyle \frac {2 c \int \frac {\sqrt {c \tan (e+f x)} \left (c d \tan ^2(e+f x)\right )^m}{(a+b c \tan (e+f x))^2 \left (c^4 \tan ^4(e+f x)+c^2\right )}d\sqrt {c \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 30

\(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \int \frac {(c \tan (e+f x))^{\frac {1}{2} (2 m+1)}}{\left (c^4 \tan ^4(e+f x)+c^2\right ) \left (a+b \sqrt {c \tan (e+f x)}\right )^2}d\sqrt {c \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 7276

\(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \int \left (\frac {\left (-4 a^3 b^3 c^3 \tan ^3(e+f x)+b^2 c^2 \left (3 a^4-b^4 c^2\right ) \tan ^2(e+f x)+a^2 \left (a^4-3 b^4 c^2\right )-2 a b \left (a^4-b^4 c^2\right ) \sqrt {c \tan (e+f x)}\right ) (c \tan (e+f x))^{\frac {1}{2} (2 m+1)}}{\left (a^4+b^4 c^2\right )^2 \left (c^4 \tan ^4(e+f x)+c^2\right )}+\frac {4 a^3 b^4 (c \tan (e+f x))^{\frac {1}{2} (2 m+1)}}{\left (a^4+b^4 c^2\right )^2 \left (a+b \sqrt {c \tan (e+f x)}\right )}+\frac {b^4 (c \tan (e+f x))^{\frac {1}{2} (2 m+1)}}{\left (a^4+b^4 c^2\right ) \left (a+b \sqrt {c \tan (e+f x)}\right )^2}\right )d\sqrt {c \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \left (\frac {2 a^2 b^4 (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,2 (m+1),2 m+3,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right )}{(m+1) \left (a^4+b^4 c^2\right )^2}+\frac {b^4 (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (2,2 (m+1),2 m+3,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right )}{2 a^2 (m+1) \left (a^4+b^4 c^2\right )}-\frac {a b \left (a^4-2 a^2 b^2 \sqrt {-c^2}-b^4 c^2\right ) (c \tan (e+f x))^{\frac {1}{2} (2 m+3)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+5),-\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{c^2 (2 m+3) \left (a^4+b^4 c^2\right )^2}-\frac {a b \left (a^4+2 a^2 b^2 \sqrt {-c^2}-b^4 c^2\right ) (c \tan (e+f x))^{\frac {1}{2} (2 m+3)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+5),\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{c^2 (2 m+3) \left (a^4+b^4 c^2\right )^2}+\frac {\left (a^6-3 a^4 b^2 \sqrt {-c^2}-3 a^2 b^4 c^2-b^6 \left (-c^2\right )^{3/2}\right ) (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{4 c^2 (m+1) \left (a^4+b^4 c^2\right )^2}+\frac {\left (a^6+3 a^4 b^2 \sqrt {-c^2}-3 a^2 b^4 c^2+b^6 \left (-c^2\right )^{3/2}\right ) (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{4 c^2 (m+1) \left (a^4+b^4 c^2\right )^2}\right )}{f}\)

Input:

Int[(d*Tan[e + f*x])^m/(a + b*Sqrt[c*Tan[e + f*x]])^2,x]
 

Output:

(2*c*(c*d*Tan[e + f*x]^2)^m*(((a^6 - 3*a^2*b^4*c^2 - 3*a^4*b^2*Sqrt[-c^2] 
- b^6*(-c^2)^(3/2))*Hypergeometric2F1[1, 1 + m, 2 + m, -((c^2*Tan[e + f*x] 
^2)/Sqrt[-c^2])]*(c*Tan[e + f*x])^(1 + m))/(4*c^2*(a^4 + b^4*c^2)^2*(1 + m 
)) + ((a^6 - 3*a^2*b^4*c^2 + 3*a^4*b^2*Sqrt[-c^2] + b^6*(-c^2)^(3/2))*Hype 
rgeometric2F1[1, 1 + m, 2 + m, (c^2*Tan[e + f*x]^2)/Sqrt[-c^2]]*(c*Tan[e + 
 f*x])^(1 + m))/(4*c^2*(a^4 + b^4*c^2)^2*(1 + m)) + (2*a^2*b^4*Hypergeomet 
ric2F1[1, 2*(1 + m), 3 + 2*m, -((b*Sqrt[c*Tan[e + f*x]])/a)]*(c*Tan[e + f* 
x])^(1 + m))/((a^4 + b^4*c^2)^2*(1 + m)) + (b^4*Hypergeometric2F1[2, 2*(1 
+ m), 3 + 2*m, -((b*Sqrt[c*Tan[e + f*x]])/a)]*(c*Tan[e + f*x])^(1 + m))/(2 
*a^2*(a^4 + b^4*c^2)*(1 + m)) - (a*b*(a^4 - b^4*c^2 - 2*a^2*b^2*Sqrt[-c^2] 
)*Hypergeometric2F1[1, (3 + 2*m)/2, (5 + 2*m)/2, -((c^2*Tan[e + f*x]^2)/Sq 
rt[-c^2])]*(c*Tan[e + f*x])^((3 + 2*m)/2))/(c^2*(a^4 + b^4*c^2)^2*(3 + 2*m 
)) - (a*b*(a^4 - b^4*c^2 + 2*a^2*b^2*Sqrt[-c^2])*Hypergeometric2F1[1, (3 + 
 2*m)/2, (5 + 2*m)/2, (c^2*Tan[e + f*x]^2)/Sqrt[-c^2]]*(c*Tan[e + f*x])^(( 
3 + 2*m)/2))/(c^2*(a^4 + b^4*c^2)^2*(3 + 2*m))))/(f*(c*Tan[e + f*x])^m)
 

Defintions of rubi rules used

rule 30

Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I 
ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) 
Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & 
&  !IntegerQ[p]
 

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 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

rule 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

rule 7276

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 

Maple [F]

\[\int \frac {\left (d \tan \left (f x +e \right )\right )^{m}}{\left (a +b \sqrt {c \tan \left (f x +e \right )}\right )^{2}}d x\]

Input:

int((d*tan(f*x+e))^m/(a+b*(c*tan(f*x+e))^(1/2))^2,x)
 

Output:

int((d*tan(f*x+e))^m/(a+b*(c*tan(f*x+e))^(1/2))^2,x)
 

Fricas [F]

\[ \int \frac {(d \tan (e+f x))^m}{\left (a+b \sqrt {c \tan (e+f x)}\right )^2} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{m}}{{\left (\sqrt {c \tan \left (f x + e\right )} b + a\right )}^{2}} \,d x } \] Input:

integrate((d*tan(f*x+e))^m/(a+b*(c*tan(f*x+e))^(1/2))^2,x, algorithm="fric 
as")
 

Output:

integral(-(2*sqrt(c*tan(f*x + e))*(d*tan(f*x + e))^m*a*b - (b^2*c*tan(f*x 
+ e) + a^2)*(d*tan(f*x + e))^m)/(b^4*c^2*tan(f*x + e)^2 - 2*a^2*b^2*c*tan( 
f*x + e) + a^4), x)
 

Sympy [F]

\[ \int \frac {(d \tan (e+f x))^m}{\left (a+b \sqrt {c \tan (e+f x)}\right )^2} \, dx=\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{m}}{\left (a + b \sqrt {c \tan {\left (e + f x \right )}}\right )^{2}}\, dx \] Input:

integrate((d*tan(f*x+e))**m/(a+b*(c*tan(f*x+e))**(1/2))**2,x)
 

Output:

Integral((d*tan(e + f*x))**m/(a + b*sqrt(c*tan(e + f*x)))**2, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d \tan (e+f x))^m}{\left (a+b \sqrt {c \tan (e+f x)}\right )^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((d*tan(f*x+e))^m/(a+b*(c*tan(f*x+e))^(1/2))^2,x, algorithm="maxi 
ma")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

\[ \int \frac {(d \tan (e+f x))^m}{\left (a+b \sqrt {c \tan (e+f x)}\right )^2} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{m}}{{\left (\sqrt {c \tan \left (f x + e\right )} b + a\right )}^{2}} \,d x } \] Input:

integrate((d*tan(f*x+e))^m/(a+b*(c*tan(f*x+e))^(1/2))^2,x, algorithm="giac 
")
 

Output:

integrate((d*tan(f*x + e))^m/(sqrt(c*tan(f*x + e))*b + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \tan (e+f x))^m}{\left (a+b \sqrt {c \tan (e+f x)}\right )^2} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^m}{{\left (a+b\,\sqrt {c\,\mathrm {tan}\left (e+f\,x\right )}\right )}^2} \,d x \] Input:

int((d*tan(e + f*x))^m/(a + b*(c*tan(e + f*x))^(1/2))^2,x)
 

Output:

int((d*tan(e + f*x))^m/(a + b*(c*tan(e + f*x))^(1/2))^2, x)
 

Reduce [F]

\[ \int \frac {(d \tan (e+f x))^m}{\left (a+b \sqrt {c \tan (e+f x)}\right )^2} \, dx=\text {too large to display} \] Input:

int((d*tan(f*x+e))^m/(a+b*(c*tan(f*x+e))^(1/2))^2,x)
                                                                                    
                                                                                    
 

Output:

(d**m*(tan(e + f*x)**m*a**4*m + 8*sqrt(c)*int(tan(e + f*x)**((2*m + 1)/2)/ 
(4*tan(e + f*x)**3*b**4*c**2*m**2 - tan(e + f*x)**3*b**4*c**2 - 8*tan(e + 
f*x)**2*a**2*b**2*c*m**2 + 2*tan(e + f*x)**2*a**2*b**2*c + 4*tan(e + f*x)* 
a**4*m**2 - tan(e + f*x)*a**4),x)*tan(e + f*x)*a**5*b**3*c*f*m**4 - 6*sqrt 
(c)*int(tan(e + f*x)**((2*m + 1)/2)/(4*tan(e + f*x)**3*b**4*c**2*m**2 - ta 
n(e + f*x)**3*b**4*c**2 - 8*tan(e + f*x)**2*a**2*b**2*c*m**2 + 2*tan(e + f 
*x)**2*a**2*b**2*c + 4*tan(e + f*x)*a**4*m**2 - tan(e + f*x)*a**4),x)*tan( 
e + f*x)*a**5*b**3*c*f*m**2 - 2*sqrt(c)*int(tan(e + f*x)**((2*m + 1)/2)/(4 
*tan(e + f*x)**3*b**4*c**2*m**2 - tan(e + f*x)**3*b**4*c**2 - 8*tan(e + f* 
x)**2*a**2*b**2*c*m**2 + 2*tan(e + f*x)**2*a**2*b**2*c + 4*tan(e + f*x)*a* 
*4*m**2 - tan(e + f*x)*a**4),x)*tan(e + f*x)*a**5*b**3*c*f*m - 8*sqrt(c)*i 
nt(tan(e + f*x)**((2*m + 1)/2)/(4*tan(e + f*x)**3*b**4*c**2*m**2 - tan(e + 
 f*x)**3*b**4*c**2 - 8*tan(e + f*x)**2*a**2*b**2*c*m**2 + 2*tan(e + f*x)** 
2*a**2*b**2*c + 4*tan(e + f*x)*a**4*m**2 - tan(e + f*x)*a**4),x)*a**7*b*f* 
m**4 + 6*sqrt(c)*int(tan(e + f*x)**((2*m + 1)/2)/(4*tan(e + f*x)**3*b**4*c 
**2*m**2 - tan(e + f*x)**3*b**4*c**2 - 8*tan(e + f*x)**2*a**2*b**2*c*m**2 
+ 2*tan(e + f*x)**2*a**2*b**2*c + 4*tan(e + f*x)*a**4*m**2 - tan(e + f*x)* 
a**4),x)*a**7*b*f*m**2 + 2*sqrt(c)*int(tan(e + f*x)**((2*m + 1)/2)/(4*tan( 
e + f*x)**3*b**4*c**2*m**2 - tan(e + f*x)**3*b**4*c**2 - 8*tan(e + f*x)**2 
*a**2*b**2*c*m**2 + 2*tan(e + f*x)**2*a**2*b**2*c + 4*tan(e + f*x)*a**4...