Integrand size = 25, antiderivative size = 248 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\frac {i (i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {i (i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \]
I*(I*a-b)^(5/2)*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+5*a*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*( I*a+b)^(5/2)*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+b^2*(a+b*tan(d*x+c))^(1/2)/d/cot(d*x +c)^(1/2)
Time = 1.16 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.97 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left ((-1)^{3/4} (-a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(-1)^{3/4} (a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {5 a^{3/2} b^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]
(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-1)^(3/4)*(-a + I*b)^(5/2)*ArcTan [((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - (-1)^(3/4)*(a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + b^2*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[ c + d*x]] + (5*a^(3/2)*b^(3/2)*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]]))/d
Time = 1.10 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.83, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4729, 3042, 4049, 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.
| \(\displaystyle \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{5/2}}{\sqrt {\tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{5/2}}{\sqrt {\tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4049 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\int \frac {5 a b^2 \tan ^2(c+d x)+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (2 a^2-b^2\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} \int \frac {5 a b^2 \tan ^2(c+d x)+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (2 a^2-b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} \int \frac {5 a b^2 \tan (c+d x)^2+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (2 a^2-b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {5 a b^2 \tan ^2(c+d x)+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (2 a^2-b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {5 a b^2 \tan ^2(c+d x)+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (2 a^2-b^2\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 2257 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \left (\frac {5 a b^2}{\sqrt {a+b \tan (c+d x)}}+\frac {2 \left (a^3-3 b^2 a+b \left (3 a^2-b^2\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {i (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+i (b+i a)^{5/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((I*(I*a - b)^(5/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 5*a*b^(3/2)*ArcTanh[ (Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + I*(I*a + b)^(5/2) *ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (b^2*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d)
3.9.58.3.1 Defintions of rubi rules used
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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((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 - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 , 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I ntegerQ[m]) && !(IGtQ[n, 2] && ( !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]
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. \(2949\) vs. \(2(202)=404\).
Time = 36.74 (sec) , antiderivative size = 2950, normalized size of antiderivative = 11.90
1/4/d*(a+b*tan(d*x+c))^(1/2)*cot(d*x+c)^(1/2)/(cos(d*x+c)+1)/((sin(d*x+c)* cos(d*x+c)*a-cos(d*x+c)^2*b+b)/(cos(d*x+c)+1)^2)^(1/2)*(10*sin(d*x+c)*2^(1 /2)*b^(3/2)*arctanh(((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1) ^2)^(1/2)*(csc(d*x+c)+cot(d*x+c))/b^(1/2))*(-b+(a^2+b^2)^(1/2))^(1/2)*a^2+ 2*sin(d*x+c)*2^(1/2)*((sin(d*x+c)*cos(d*x+c)*a-cos(d*x+c)^2*b+b)/(cos(d*x+ c)+1)^2)^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*a*b^2-sin(d*x+c)*(a^2+b^2)^(1/2) *ln((cos(d*x+c)*cot(d*x+c)*a-2*a*cot(d*x+c)+2*sin(d*x+c)*(csc(d*x+c)*(cot( d*x+c)^2*a-2*cot(d*x+c)*csc(d*x+c)*a+csc(d*x+c)^2*a-2*b*csc(d*x+c)+2*cot(d *x+c)*b-a)*(cos(d*x+c)-1))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*(a^2+b^2)^(1/ 2)*cos(d*x+c)+2*b*cos(d*x+c)-sin(d*x+c)*a+csc(d*x+c)*a-2*(a^2+b^2)^(1/2)-2 *b)/(cos(d*x+c)-1))*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*a ^2+sin(d*x+c)*(a^2+b^2)^(1/2)*ln((cos(d*x+c)*cot(d*x+c)*a-2*a*cot(d*x+c)+2 *sin(d*x+c)*(csc(d*x+c)*(cot(d*x+c)^2*a-2*cot(d*x+c)*csc(d*x+c)*a+csc(d*x+ c)^2*a-2*b*csc(d*x+c)+2*cot(d*x+c)*b-a)*(cos(d*x+c)-1))^(1/2)*(b+(a^2+b^2) ^(1/2))^(1/2)+2*(a^2+b^2)^(1/2)*cos(d*x+c)+2*b*cos(d*x+c)-sin(d*x+c)*a+csc (d*x+c)*a-2*(a^2+b^2)^(1/2)-2*b)/(cos(d*x+c)-1))*(b+(a^2+b^2)^(1/2))^(1/2) *(-b+(a^2+b^2)^(1/2))^(1/2)*b^2+sin(d*x+c)*(a^2+b^2)^(1/2)*ln(-(-cos(d*x+c )*cot(d*x+c)*a+2*sin(d*x+c)*(csc(d*x+c)*(cot(d*x+c)^2*a-2*cot(d*x+c)*csc(d *x+c)*a+csc(d*x+c)^2*a-2*b*csc(d*x+c)+2*cot(d*x+c)*b-a)*(cos(d*x+c)-1))^(1 /2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*a*cot(d*x+c)+sin(d*x+c)*a-2*(a^2+b^2)^(...
Leaf count of result is larger than twice the leaf count of optimal. 4793 vs. \(2 (196) = 392\).
Time = 1.34 (sec) , antiderivative size = 9618, normalized size of antiderivative = 38.78 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]
Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]
\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\cot \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \]