\(\int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [620]

Optimal result

Integrand size = 25, antiderivative size = 173 \[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(i a-b)^{3/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(i a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \] Output:

(I*a-b)^(3/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2) 
)/d+(I*a+b)^(3/2)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^ 
(1/2))/d-2/3*a*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)-8/3*b*(a+b*tan(d* 
x+c))^(1/2)/d/tan(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.02 \[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-3 \sqrt [4]{-1} \sqrt {-a+i b} (i a+b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+3 (-1)^{3/4} (a+i b)^{3/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {2 \sqrt {a+b \tan (c+d x)} (a+4 b \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)}}{3 d} \] Input:

Integrate[(a + b*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(5/2),x]
 

Output:

(-3*(-1)^(1/4)*Sqrt[-a + I*b]*(I*a + b)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]* 
Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 3*(-1)^(3/4)*(a + I*b)^(3/ 
2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + 
 d*x]]] - (2*Sqrt[a + b*Tan[c + d*x]]*(a + 4*b*Tan[c + d*x]))/Tan[c + d*x] 
^(3/2))/(3*d)
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.17, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4050, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 219}

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.

\(\displaystyle \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan (c+d x)^{5/2}}dx\)

\(\Big \downarrow \) 4050

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

\(\displaystyle \frac {1}{3} \left (-\frac {2 \int \frac {3 \left (2 b \tan (c+d x) a^2+\left (a^2-b^2\right ) a\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-\frac {3 \int \frac {2 b \tan (c+d x) a^2+\left (a^2-b^2\right ) a}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-\frac {3 \int \frac {2 b \tan (c+d x) a^2+\left (a^2-b^2\right ) a}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 4099

\(\displaystyle -\frac {2 a \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {8 b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {1}{2} a (a-i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 a \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {8 b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {1}{2} a (a-i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}\right )\)

\(\Big \downarrow \) 4098

\(\displaystyle -\frac {2 a \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {8 b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a-i b)^2 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+\frac {a (a+i b)^2 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a}\right )\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {2 a \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {8 b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a-i b)^2 \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a (a+i b)^2 \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2 a \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {8 b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a-i b)^2 \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a (a+i b)^2 \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {2 a \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {8 b \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {3 \left (\frac {a (a+i b)^2 \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {a (a-i b)^2 \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\right )}{a}\right )\)

Input:

Int[(a + b*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(5/2),x]
 

Output:

(-2*a*Sqrt[a + b*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) + ((-3*((a*(a + I 
*b)^2*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]) 
/(Sqrt[I*a - b]*d) + (a*(a - I*b)^2*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d* 
x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d)))/a - (8*b*Sqrt[a + b*Ta 
n[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/3
 

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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 4050

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

rule 4098

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^2/f   Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), 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] && 
 NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
 

rule 4099

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 + I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T 
an[e + f*x]), x], x] + Simp[(A - I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d 
*Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A 
, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 
0]
 

rule 4132

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

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 0.68 (sec) , antiderivative size = 1343875, normalized size of antiderivative = 7768.06

\[\text {output too large to display}\]

Input:

int((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3507 vs. \(2 (137) = 274\).

Time = 0.36 (sec) , antiderivative size = 3507, normalized size of antiderivative = 20.27 \[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:

integrate((a+b*tan(d*x+c))**(3/2)/tan(d*x+c)**(5/2),x)
 

Output:

Integral((a + b*tan(c + d*x))**(3/2)/tan(c + d*x)**(5/2), x)
 

Maxima [F]

\[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*tan(d*x + c) + a)^(3/2)/tan(d*x + c)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int((a + b*tan(c + d*x))^(3/2)/tan(c + d*x)^(5/2),x)
 

Output:

int((a + b*tan(c + d*x))^(3/2)/tan(c + d*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right ) b -2 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, a +6 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{3} b +\tan \left (d x +c \right )^{2} a}d x \right ) \tan \left (d x +c \right )^{2} a b d -3 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{2} b +\tan \left (d x +c \right ) a}d x \right ) \tan \left (d x +c \right )^{2} a^{2} d +3 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{2} b +\tan \left (d x +c \right ) a}d x \right ) \tan \left (d x +c \right )^{2} b^{2} d}{3 \tan \left (d x +c \right )^{2} d} \] Input:

int((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x)
 

Output:

(4*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)*b - 2*sqrt(tan 
(c + d*x))*sqrt(tan(c + d*x)*b + a)*a + 6*int((sqrt(tan(c + d*x))*sqrt(tan 
(c + d*x)*b + a))/(tan(c + d*x)**3*b + tan(c + d*x)**2*a),x)*tan(c + d*x)* 
*2*a*b*d - 3*int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a))/(tan(c + d* 
x)**2*b + tan(c + d*x)*a),x)*tan(c + d*x)**2*a**2*d + 3*int((sqrt(tan(c + 
d*x))*sqrt(tan(c + d*x)*b + a))/(tan(c + d*x)**2*b + tan(c + d*x)*a),x)*ta 
n(c + d*x)**2*b**2*d)/(3*tan(c + d*x)**2*d)