\(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [213]

Optimal result

Integrand size = 36, antiderivative size = 224 \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {2 a (i A+B) \operatorname {AppellF1}\left (\frac {1}{2},-m,1,\frac {3}{2},1+i \tan (c+d x),\frac {1}{2} (1+i \tan (c+d x))\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (B+(i A+B) (3+2 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+i \tan (c+d x)\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m)}+\frac {2 i a B \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m)} \] Output:

-2*a*(I*A+B)*AppellF1(1/2,-m,1,3/2,1+I*tan(d*x+c),1/2+1/2*I*tan(d*x+c))*ta 
n(d*x+c)^m*(a+I*a*tan(d*x+c))^(1/2)/d/((-I*tan(d*x+c))^m)+2*a*(B+(I*A+B)*( 
3+2*m))*hypergeom([1/2, -m],[3/2],1+I*tan(d*x+c))*tan(d*x+c)^m*(a+I*a*tan( 
d*x+c))^(1/2)/d/(3+2*m)/((-I*tan(d*x+c))^m)+2*I*a*B*tan(d*x+c)^(1+m)*(a+I* 
a*tan(d*x+c))^(1/2)/d/(3+2*m)
 

Mathematica [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx \] Input:

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]) 
,x]
 

Output:

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]) 
, x]
 

Rubi [A] (warning: unable to verify)

Time = 1.13 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3042, 4077, 27, 3042, 4084, 3042, 4047, 25, 27, 152, 150, 4082, 77, 75}

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.

\(\displaystyle \int (a+i a \tan (c+d x))^{3/2} \tan ^m(c+d x) (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+i a \tan (c+d x))^{3/2} \tan (c+d x)^m (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4077

\(\displaystyle \frac {2 \int -\frac {1}{2} \tan ^m(c+d x) \sqrt {i \tan (c+d x) a+a} (a (2 i B (m+1)-A (2 m+3))-a (B+(i A+B) (2 m+3)) \tan (c+d x))dx}{2 m+3}+\frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {\int \tan ^m(c+d x) \sqrt {i \tan (c+d x) a+a} (a (2 i B (m+1)-A (2 m+3))-a (B+(i A+B) (2 m+3)) \tan (c+d x))dx}{2 m+3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {\int \tan (c+d x)^m \sqrt {i \tan (c+d x) a+a} (a (2 i B (m+1)-A (2 m+3))-a (B+(i A+B) (2 m+3)) \tan (c+d x))dx}{2 m+3}\)

\(\Big \downarrow \) 4084

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {-2 a (2 m+3) (A-i B) \int \tan ^m(c+d x) \sqrt {i \tan (c+d x) a+a}dx-i (B+(2 m+3) (B+i A)) \int \tan ^m(c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 m+3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {-2 a (2 m+3) (A-i B) \int \tan (c+d x)^m \sqrt {i \tan (c+d x) a+a}dx-i (B+(2 m+3) (B+i A)) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 m+3}\)

\(\Big \downarrow \) 4047

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {-\frac {2 i a^3 (2 m+3) (A-i B) \int -\frac {\tan ^m(c+d x)}{a (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{d}-i (B+(2 m+3) (B+i A)) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 m+3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {\frac {2 i a^3 (2 m+3) (A-i B) \int \frac {\tan ^m(c+d x)}{a (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{d}-i (B+(2 m+3) (B+i A)) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 m+3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {\frac {2 i a^2 (2 m+3) (A-i B) \int \frac {\tan ^m(c+d x)}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{d}-i (B+(2 m+3) (B+i A)) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 m+3}\)

\(\Big \downarrow \) 152

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {\frac {2 i a^2 (2 m+3) (A-i B) \sqrt {1+i \tan (c+d x)} \int \frac {\tan ^m(c+d x)}{\sqrt {i \tan (c+d x)+1} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d \sqrt {a+i a \tan (c+d x)}}-i (B+(2 m+3) (B+i A)) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 m+3}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {-i (B+(2 m+3) (B+i A)) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx-\frac {2 a^2 (2 m+3) (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}}{2 m+3}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {-\frac {i a^2 (B+(2 m+3) (B+i A)) \int \frac {\tan ^m(c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {2 a^2 (2 m+3) (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}}{2 m+3}\)

\(\Big \downarrow \) 77

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {-\frac {i a^2 (B+(2 m+3) (B+i A)) \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \int \frac {(-i \tan (c+d x))^m}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {2 a^2 (2 m+3) (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}}{2 m+3}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)}-\frac {-\frac {2 a^2 (2 m+3) (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}-\frac {2 a (B+(2 m+3) (B+i A)) \sqrt {a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},i \tan (c+d x)+1\right )}{d}}{2 m+3}\)

Input:

Int[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]
 

Output:

((2*I)*a*B*Tan[c + d*x]^(1 + m)*Sqrt[a + I*a*Tan[c + d*x]])/(d*(3 + 2*m)) 
- ((-2*a^2*(A - I*B)*(3 + 2*m)*AppellF1[1 + m, 1/2, 1, 2 + m, (-I)*Tan[c + 
 d*x], I*Tan[c + d*x]]*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m))/(d*( 
1 + m)*Sqrt[a + I*a*Tan[c + d*x]]) - (2*a*(B + (I*A + B)*(3 + 2*m))*Hyperg 
eometric2F1[1/2, -m, 3/2, 1 + I*Tan[c + d*x]]*Tan[c + d*x]^m*Sqrt[a + I*a* 
Tan[c + d*x]])/(d*((-I)*Tan[c + d*x])^m))/(3 + 2*m)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 77

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((-b)*(c/ 
d))^IntPart[m]*((b*x)^FracPart[m]/((-d)*(x/c))^FracPart[m])   Int[((-d)*(x/ 
c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && 
 !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0]
 

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 152

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) 
Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, 
 n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]
 

rule 3042

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

rule 4047

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(b/f)   Subst[Int[(a + x)^(m - 1)*(( 
c + (d/b)*x)^n/(b^2 + a*x)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, c, 
d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d 
^2, 0]
 

rule 4077

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + 
n))), x] + Simp[1/(d*(m + n))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan 
[e + f*x])^n*Simp[a*A*d*(m + n) + B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - 
 a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a 
, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && 
GtQ[m, 1] &&  !LtQ[n, -1]
 

rule 4082

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(B/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], 
x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
 

rule 4084

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*b + a*B)/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] 
 - Simp[B/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ 
e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - 
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
 

Maple [F]

\[\int \tan \left (d x +c \right )^{m} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} \left (A +B \tan \left (d x +c \right )\right )d x\]

Input:

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)
 

Output:

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)
 

Fricas [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{m} \,d x } \] Input:

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algori 
thm="fricas")
 

Output:

integral(2*sqrt(2)*((A - I*B)*a*e^(5*I*d*x + 5*I*c) + (A + I*B)*a*e^(3*I*d 
*x + 3*I*c))*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^m*sq 
rt(a/(e^(2*I*d*x + 2*I*c) + 1))/(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*I* 
c) + 1), x)
 

Sympy [F]

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

integrate(tan(d*x+c)**m*(a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)
 

Output:

Integral((I*a*(tan(c + d*x) - I))**(3/2)*(A + B*tan(c + d*x))*tan(c + d*x) 
**m, x)
 

Maxima [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{m} \,d x } \] Input:

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algori 
thm="maxima")
 

Output:

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(3/2)*tan(d*x + c)^m 
, x)
 

Giac [F(-2)]

Exception generated. \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algori 
thm="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 [F(-1)]

Timed out. \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \] Input:

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2),x)
 

Output:

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2), x)
 

Reduce [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a}\, a \left (-2 \tan \left (d x +c \right )^{m} \sqrt {\tan \left (d x +c \right ) i +1}\, a i +2 \left (\int \frac {\tan \left (d x +c \right )^{m} \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )}d x \right ) a d i m +\left (\int \tan \left (d x +c \right )^{m} \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}d x \right ) b d i +2 \left (\int \tan \left (d x +c \right )^{m} \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )d x \right ) a d i m +2 \left (\int \tan \left (d x +c \right )^{m} \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )d x \right ) a d i +\left (\int \tan \left (d x +c \right )^{m} \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )d x \right ) b d \right )}{d} \] Input:

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)
 

Output:

(sqrt(a)*a*( - 2*tan(c + d*x)**m*sqrt(tan(c + d*x)*i + 1)*a*i + 2*int((tan 
(c + d*x)**m*sqrt(tan(c + d*x)*i + 1))/tan(c + d*x),x)*a*d*i*m + int(tan(c 
 + d*x)**m*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2,x)*b*d*i + 2*int(tan(c 
 + d*x)**m*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)*a*d*i*m + 2*int(tan(c 
+ d*x)**m*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)*a*d*i + int(tan(c + d*x 
)**m*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)*b*d))/d