Integrand size = 35, antiderivative size = 260 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {(i a-b)^{5/2} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b} \left (20 a A b+15 a^2 B-8 b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 d}+\frac {(i a+b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b (4 A b+7 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d} \] Output:
(I*a-b)^(5/2)*(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c ))^(1/2))/d+1/4*b^(1/2)*(20*A*a*b+15*B*a^2-8*B*b^2)*arctanh(b^(1/2)*tan(d* x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+(I*a+b)^(5/2)*(I*A+B)*arctanh((I*a+b) ^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+1/4*b*(4*A*b+7*B*a)*tan( d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d+1/2*b*B*tan(d*x+c)^(1/2)*(a+b*tan(d* x+c))^(3/2)/d
Time = 1.82 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {4 \sqrt [4]{-1} (-a+i b)^{5/2} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-4 (-1)^{3/4} (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b (4 A b+7 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+\frac {\sqrt {a} \sqrt {b} \left (20 a A b+15 a^2 B-8 b^2 B\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}}{4 d} \] Input:
Integrate[((a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Sqrt[Tan[c + d *x]],x]
Output:
(4*(-1)^(1/4)*(-a + I*b)^(5/2)*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b] *Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - 4*(-1)^(3/4)*(a + I*b)^(5 /2)*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + b*(4*A*b + 7*a*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + 2*b*B*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2) + (Sqrt[a]* Sqrt[b]*(20*a*A*b + 15*a^2*B - 8*b^2*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x] ])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/Sqrt[a + b*Tan[c + d*x]])/(4*d)
Time = 1.54 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 4090, 27, 3042, 4130, 27, 3042, 4138, 2035, 2257, 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.
\(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4090 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt {a+b \tan (c+d x)} \left (b (4 A b+7 a B) \tan ^2(c+d x)+4 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (4 a A-b B)\right )}{2 \sqrt {\tan (c+d x)}}dx+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (b (4 A b+7 a B) \tan ^2(c+d x)+4 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (4 a A-b B)\right )}{\sqrt {\tan (c+d x)}}dx+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (b (4 A b+7 a B) \tan (c+d x)^2+4 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (4 a A-b B)\right )}{\sqrt {\tan (c+d x)}}dx+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 4130 |
\(\displaystyle \frac {1}{4} \left (\int \frac {b \left (15 B a^2+20 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-9 b B a-4 A b^2\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {b \left (15 B a^2+20 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-9 b B a-4 A b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {b \left (15 B a^2+20 A b a-8 b^2 B\right ) \tan (c+d x)^2+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-9 b B a-4 A b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 4138 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {b \left (15 B a^2+20 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-9 b B a-4 A b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {b \left (15 B a^2+20 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-9 b B a-4 A b^2\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 2257 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \left (\frac {b \left (15 B a^2+20 A b a-8 b^2 B\right )}{\sqrt {a+b \tan (c+d x)}}+\frac {8 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}+\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (\frac {b (7 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\sqrt {b} \left (15 a^2 B+20 a A b-8 b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+4 (-b+i a)^{5/2} (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+4 (b+i a)^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )\) |
Input:
Int[((a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Sqrt[Tan[c + d*x]],x ]
Output:
(b*B*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/(2*d) + ((4*(I*a - b)^ (5/2)*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + Sqrt[b]*(20*a*A*b + 15*a^2*B - 8*b^2*B)*ArcTanh[(Sqrt[b]*Sqrt[ Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 4*(I*a + b)^(5/2)*(I*A + B)*Arc Tanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (b* (4*A*b + 7*a*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol ] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a , c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
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[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 - 2)*(c + d*Ta n[e + f*x])^n*Simp[a^2*A*d*(m + n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m - 1) - b *(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 , 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
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[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C *(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[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] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ c, 0] && NeQ[a, 0])))
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] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[ff/f Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f , A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 2.23 (sec) , antiderivative size = 2652267, normalized size of antiderivative = 10201.03
\[\text {output too large to display}\]
Input:
int((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x)
Output:
result too large to display
Leaf count of result is larger than twice the leaf count of optimal. 17603 vs. \(2 (210) = 420\).
Time = 6.71 (sec) , antiderivative size = 35212, normalized size of antiderivative = 135.43 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Too large to display} \] Input:
integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algo rithm="fricas")
Output:
Too large to include
\[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx \] Input:
integrate((a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(1/2),x)
Output:
Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**(5/2)/sqrt(tan(c + d*x )), x)
\[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \] Input:
integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algo rithm="maxima")
Output:
integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(5/2)/sqrt(tan(d*x + c )), x)
Exception generated. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algo rithm="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
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}} \,d x \] Input:
int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2))/tan(c + d*x)^(1/2),x )
Output:
int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2))/tan(c + d*x)^(1/2), x)
\[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {2 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, a^{3}-2 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{a +\tan \left (d x +c \right ) b}d x \right ) a^{3} b d -\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )}{a +\tan \left (d x +c \right ) b}d x \right ) a^{4} d +\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{2} b +\tan \left (d x +c \right ) a}d x \right ) a^{4} d +2 \left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}d x \right ) b^{3} d +6 \left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )d x \right ) a \,b^{2} d +6 \left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}d x \right ) a^{2} b d}{2 d} \] Input:
int((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x)
Output:
(2*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*a**3 - 2*int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2)/(tan(c + d*x)*b + a),x)*a* *3*b*d - int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x))/(t an(c + d*x)*b + a),x)*a**4*d + 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)*a**4*d + 2*int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2,x)*b**3*d + 6*int(sqrt(tan (c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x),x)*a*b**2*d + 6*int(sqrt( tan(c + d*x))*sqrt(tan(c + d*x)*b + a),x)*a**2*b*d)/(2*d)