Integrand size = 25, antiderivative size = 179 \[ \int \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=-\frac {\sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{d} \] Output:
-(I*a-b)^(1/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2 ))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+(I*a+b)^(1/2)*arctanh((I*a+b)^(1/2) *tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2 )/d-2*cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d
Time = 0.10 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.97 \[ \int \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=-\frac {\sqrt {\cot (c+d x)} \left (\sqrt [4]{-1} \sqrt {-a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}+\sqrt [4]{-1} \sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}+2 \sqrt {a+b \tan (c+d x)}\right )}{d} \] Input:
Integrate[Cot[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]],x]
Output:
-((Sqrt[Cot[c + d*x]]*((-1)^(1/4)*Sqrt[-a + I*b]*ArcTan[((-1)^(1/4)*Sqrt[- a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Tan[c + d*x]] + (-1)^(1/4)*Sqrt[a + I*b]*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d *x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Tan[c + d*x]] + 2*Sqrt[a + b*Tan[c + d*x]]))/d)
Time = 0.88 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.89, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4729, 3042, 4051, 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.
\(\displaystyle \int \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cot (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4729 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan (c+d x)^{3/2}}dx\) |
\(\Big \downarrow \) 4051 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-2 \int -\frac {b-a \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\int \frac {b-a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\int \frac {b-a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 4099 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {1}{2} (-b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {1}{2} (-b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 4098 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(b+i a) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {(-b+i a) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {(-b+i a) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {(b+i a) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(b+i a) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {\sqrt {-b+i a} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {\sqrt {-b+i a} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b+i a} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )\) |
Input:
Int[Cot[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]],x]
Output:
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-((Sqrt[I*a - b]*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) + (Sqrt[I*a + b]*Arc Tanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (2* Sqrt[a + b*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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])
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])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^n/(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 - 1)*Simp[a*c *(m + 1) - b*d*n - (b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(m + n + 1)*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] && GtQ[n, 0] && Int egerQ[2*m]
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]
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]
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1054\) vs. \(2(147)=294\).
Time = 4.94 (sec) , antiderivative size = 1055, normalized size of antiderivative = 5.89
Input:
int(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2/d*2^(1/2)/(-b+(a^2+b^2)^(1/2))^(1/2)*cot(d*x+c)^(3/2)*(a+b*tan(d*x+c)) ^(1/2)*((b+(a^2+b^2)^(1/2))^(1/2)*ln(1/(1-cos(d*x+c))*(a*(1-cos(d*x+c))^2* csc(d*x+c)+2*2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1 )^2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)-2*(a^2+b^2)^(1/2)*(1-cos(d *x+c))-2*b*(1-cos(d*x+c))-sin(d*x+c)*a))*(-b+(a^2+b^2)^(1/2))^(1/2)*(-cot( d*x+c)+csc(d*x+c))+(-csc(d*x+c)+cot(d*x+c))*(-b+(a^2+b^2)^(1/2))^(1/2)*ln( 1/(1-cos(d*x+c))*(-a*(1-cos(d*x+c))^2*csc(d*x+c)+2*(a^2+b^2)^(1/2)*(1-cos( d*x+c))+2*2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2 )^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)+2*b*(1-cos(d*x+c))+sin(d*x+c) *a))*(b+(a^2+b^2)^(1/2))^(1/2)+(2*csc(d*x+c)-2*cot(d*x+c))*(a^2+b^2)^(1/2) *arctan(1/(-b+(a^2+b^2)^(1/2))^(1/2)*((b+(a^2+b^2)^(1/2))^(1/2)*(-cot(d*x+ c)+csc(d*x+c))-2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c) +1)^2)^(1/2))/(1-cos(d*x+c))*sin(d*x+c))+(-2*csc(d*x+c)+2*cot(d*x+c))*b*ar ctan(1/(-b+(a^2+b^2)^(1/2))^(1/2)*((b+(a^2+b^2)^(1/2))^(1/2)*(-cot(d*x+c)+ csc(d*x+c))-2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1) ^2)^(1/2))/(1-cos(d*x+c))*sin(d*x+c))+(-2*csc(d*x+c)+2*cot(d*x+c))*(a^2+b^ 2)^(1/2)*arctan(1/(-b+(a^2+b^2)^(1/2))^(1/2)*((b+(a^2+b^2)^(1/2))^(1/2)*(- cot(d*x+c)+csc(d*x+c))+2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(co s(d*x+c)+1)^2)^(1/2))/(1-cos(d*x+c))*sin(d*x+c))+(2*csc(d*x+c)-2*cot(d*x+c ))*b*arctan(1/(-b+(a^2+b^2)^(1/2))^(1/2)*((b+(a^2+b^2)^(1/2))^(1/2)*(-c...
Leaf count of result is larger than twice the leaf count of optimal. 2780 vs. \(2 (143) = 286\).
Time = 0.31 (sec) , antiderivative size = 2780, normalized size of antiderivative = 15.53 \[ \int \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")
Output:
-1/8*(d*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2)*log(((a^5*d - (a^5 + 6*a^3*b^2 + 8*a*b^4)*d*tan(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3)*d*tan(d*x + c) - ((a^3 *b + 4*a*b^3)*d^3*tan(d*x + c)^2 - 2*(a^4 + 3*a^2*b^2 + 4*b^4)*d^3*tan(d*x + c) - (3*a^3*b + 4*a*b^3)*d^3)*sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2) + 2*((a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c)^2 + 2*(a^4*b + 2*a ^2*b^3)*tan(d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c)^2 - (a^4 + 3* a^2*b^2 + 4*b^4)*d^2*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) + a )/sqrt(tan(d*x + c)))/(tan(d*x + c)^2 + 1)) + d*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2)*log(-((a^5*d - (a^5 + 6*a^3*b^2 + 8*a*b^4)*d*tan(d*x + c)^2 - 4*( a^4*b + 2*a^2*b^3)*d*tan(d*x + c) - ((a^3*b + 4*a*b^3)*d^3*tan(d*x + c)^2 - 2*(a^4 + 3*a^2*b^2 + 4*b^4)*d^3*tan(d*x + c) - (3*a^3*b + 4*a*b^3)*d^3)* sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2) + 2*((a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c)^2 + 2*(a^4*b + 2*a^2*b^3)*tan(d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c)^2 - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2*tan(d*x + c) )*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) + a)/sqrt(tan(d*x + c)))/(tan(d*x + c)^2 + 1)) - d*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2)*log(((a^5*d - (a^5 + 6*a ^3*b^2 + 8*a*b^4)*d*tan(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3)*d*tan(d*x + c) - ((a^3*b + 4*a*b^3)*d^3*tan(d*x + c)^2 - 2*(a^4 + 3*a^2*b^2 + 4*b^4)*d^3* tan(d*x + c) - (3*a^3*b + 4*a*b^3)*d^3)*sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^ 2/d^4) + b)/d^2) - 2*((a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c)^2 + 2*(a...
\[ \int \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {a + b \tan {\left (c + d x \right )}} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\, dx \] Input:
integrate(cot(d*x+c)**(3/2)*(a+b*tan(d*x+c))**(1/2),x)
Output:
Integral(sqrt(a + b*tan(c + d*x))*cot(c + d*x)**(3/2), x)
\[ \int \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int { \sqrt {b \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*tan(d*x + c) + a)*cot(d*x + c)^(3/2), x)
Timed out. \[ \int \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \] Input:
int(cot(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(1/2),x)
Output:
int(cot(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(1/2), x)
\[ \int \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {a +\tan \left (d x +c \right ) b}\, \sqrt {\cot \left (d x +c \right )}\, \cot \left (d x +c \right )d x \] Input:
int(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(1/2),x)
Output:
int(sqrt(tan(c + d*x)*b + a)*sqrt(cot(c + d*x))*cot(c + d*x),x)