Integrand size = 25, antiderivative size = 212 \[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a-b} d}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {b} d}-\frac {\text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a+b} d} \] Output:
-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)/(I*a-b)^(1/2)/d+2*arctanh(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)/b^(1/2)/d-arctan h((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)/(I*a+b)^(1/2)/d
Time = 0.73 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left ((-1)^{3/4} \left (\frac {\arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )+\frac {2 \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}\right )}{d} \] Input:
Integrate[1/(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]]*((-1)^(3/4)*(ArcTan[((-1)^(1/4)*Sqr t[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/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[a + I*b]) + (2*Sqrt[a]*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqr t[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b*Tan[c + d*x]])))/d
Time = 0.52 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.79, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4729, 3042, 4058, 614, 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.
\(\displaystyle \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cot (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4729 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan (c+d x)^{3/2}}{\sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4058 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 614 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \left (\frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {\arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}-\frac {\text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}\right )}{d}\) |
Input:
Int[1/(Cot[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]]),x]
Output:
((-(ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/Sq rt[I*a - b]) + (2*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] - ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Ta n[c + d*x]]]/Sqrt[I*a + b])*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d
Int[((e_.)*(x_))^(m_)/(Sqrt[(c_) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^2)), x_Sy mbol] :> Simp[e^(m + 1/2) Int[ExpandIntegrand[1/(Sqrt[e*x]*Sqrt[c + d*x]) , x^(m + 1/2)/(a + b*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[ m - 1/2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), 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/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a *d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^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. \(1025\) vs. \(2(172)=344\).
Time = 3.68 (sec) , antiderivative size = 1026, normalized size of antiderivative = 4.84
Input:
int(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/16/d*2^(1/2)/(a^2+b^2)^(1/2)/(-b+(a^2+b^2)^(1/2))^(1/2)/b^(1/2)*((1-cos (d*x+c))^2*csc(d*x+c)^2-1)^2*(a+b*tan(d*x+c))^(1/2)*(2*b^(3/2)*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*b^(3/2)*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))-b^(1/2)*(b+(a^2+b^2)^(1/2))^(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*si n(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^(1/2)*(b+(a^2+b^2)^(1/2))^(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*(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))-4*2^ (1/2)*arctanh(1/b^(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))*(a^2+b^2)^(1/2)*(-b+(a^2+b^2)^(1/ 2))^(1/2)-2*b^(1/2)*(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*...
Leaf count of result is larger than twice the leaf count of optimal. 4165 vs. \(2 (168) = 336\).
Time = 0.76 (sec) , antiderivative size = 8363, normalized size of antiderivative = 39.45 \[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:
integrate(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt {a + b \tan {\left (c + d x \right )}} \cot ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate(1/cot(d*x+c)**(3/2)/(a+b*tan(d*x+c))**(1/2),x)
Output:
Integral(1/(sqrt(a + b*tan(c + d*x))*cot(c + d*x)**(3/2)), x)
\[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(b*tan(d*x + c) + a)*cot(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \] Input:
int(1/(cot(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(1/2)),x)
Output:
int(1/(cot(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(1/2)), x)
\[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \sqrt {\cot \left (d x +c \right )}}{\cot \left (d x +c \right )^{2} \tan \left (d x +c \right ) b +\cot \left (d x +c \right )^{2} a}d x \] Input:
int(1/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)**2*tan(c + d*x)*b + cot(c + d*x)**2*a),x)