Integrand size = 25, antiderivative size = 212 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=-\frac {(i a-b)^{3/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {(i a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d} \]
-(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 ))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+2*b^(3/2)*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)/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))*cot (d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d
Time = 0.65 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.12 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\sqrt [4]{-1} \left (\sqrt {-a+i b} (i a+b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt {a+i b} (-i a+b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right ) \sqrt {a+b \tan (c+d x)}+2 \sqrt {a} b^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}\right )}{d \sqrt {a+b \tan (c+d x)}} \]
(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-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]]] + Sqrt[a + I*b]*((-I)*a + b)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqr t[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])*Sqrt[a + b*Tan[c + d*x]] + 2*S qrt[a]*b^(3/2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*T an[c + d*x])/a]))/(d*Sqrt[a + b*Tan[c + d*x]])
Time = 0.66 (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, 610, 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.
\(\displaystyle \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 4729 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{3/2}}{\sqrt {\tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{3/2}}{\sqrt {\tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4058 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{3/2}}{\sqrt {\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 610 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \left (\frac {b^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {a^2+2 b \tan (c+d x) a-b^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left ((-b+i a)^{3/2} \left (-\arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )+2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(b+i a)^{3/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )}{d}\) |
((-((I*a - b)^(3/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*T an[c + d*x]]]) + 2*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b *Tan[c + d*x]]] - (I*a + b)^(3/2)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x] ])/Sqrt[a + b*Tan[c + d*x]]])*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d
3.9.50.3.1 Defintions of rubi rules used
Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_))/((a_) + (b_.)*(x_)^2), x_S ymbol] :> Simp[e^(m + 1/2) Int[ExpandIntegrand[1/(Sqrt[e*x]*Sqrt[c + d*x] ), x^(m + 1/2)*((c + d*x)^(n + 1/2)/(a + b*x^2)), x], x], x] /; FreeQ[{a, b , c, d, e}, x] && IGtQ[n + 1/2, 0] && ILtQ[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. \(1947\) vs. \(2(172)=344\).
Time = 33.87 (sec) , antiderivative size = 1948, normalized size of antiderivative = 9.19
1/4/d*(-1/(1-cos(d*x+c))*(csc(d*x+c)*(1-cos(d*x+c))^2-sin(d*x+c)))^(1/2)*( (csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)/(csc(d*x+c )^2*(1-cos(d*x+c))^2-1))^(1/2)*(4*b^(3/2)*2^(1/2)*arctanh(1/2/(1-cos(d*x+c ))*sin(d*x+c)*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c )-cot(d*x+c))-a)*(1-cos(d*x+c)))^(1/2)*2^(1/2)/b^(1/2))*(-b+(a^2+b^2)^(1/2 ))^(1/2)+(b+(a^2+b^2)^(1/2))^(1/2)*(a^2+b^2)^(1/2)*ln(1/(1-cos(d*x+c))*(-c sc(d*x+c)*a*(1-cos(d*x+c))^2+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))+2*sin(d*x+c) *(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c)) -a)*(1-cos(d*x+c)))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*b*(1-cos(d*x+c))+sin (d*x+c)*a))*(-b+(a^2+b^2)^(1/2))^(1/2)-(b+(a^2+b^2)^(1/2))^(1/2)*(a^2+b^2) ^(1/2)*ln(-1/(1-cos(d*x+c))*(csc(d*x+c)*a*(1-cos(d*x+c))^2+2*sin(d*x+c)*(- csc(d*x+c)*(csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a) *(1-cos(d*x+c)))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)-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)-2*(b+ (a^2+b^2)^(1/2))^(1/2)*b*ln(1/(1-cos(d*x+c))*(-csc(d*x+c)*a*(1-cos(d*x+c)) ^2+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))+2*sin(d*x+c)*(-csc(d*x+c)*(csc(d*x+c)^ 2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)))^(1/2)* (b+(a^2+b^2)^(1/2))^(1/2)+2*b*(1-cos(d*x+c))+sin(d*x+c)*a))*(-b+(a^2+b^2)^ (1/2))^(1/2)+2*(b+(a^2+b^2)^(1/2))^(1/2)*b*ln(-1/(1-cos(d*x+c))*(csc(d*x+c )*a*(1-cos(d*x+c))^2+2*sin(d*x+c)*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-cos(d...
Leaf count of result is larger than twice the leaf count of optimal. 3715 vs. \(2 (168) = 336\).
Time = 0.68 (sec) , antiderivative size = 7462, normalized size of antiderivative = 35.20 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\text {Too large to display} \]
\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \sqrt {\cot {\left (c + d x \right )}}\, dx \]
\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\cot \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} \, dx=\int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2} \,d x \]