Integrand size = 25, antiderivative size = 286 \[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(i a-b)^{3/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\left (3 a^2-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{4 \sqrt {b} d}+\frac {(i a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {3 a \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}} \]
(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+(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+1/4*(3*a^2-8*b^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/b^(1/2)+3/4*a*(a+b*tan(d*x+c))^(1/ 2)/d/cot(d*x+c)^(1/2)+1/2*(a+b*tan(d*x+c))^(3/2)/d/cot(d*x+c)^(1/2)
Time = 3.80 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\left (3 a^2-8 b^2\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) (a+b \tan (c+d x))+\sqrt {a} \sqrt {b} \sqrt {1+\frac {b \tan (c+d x)}{a}} \left (-4 \sqrt [4]{-1} \sqrt {-a+i b} (i a+b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+4 (-1)^{3/4} (a+i b)^{3/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt {\tan (c+d x)} \left (5 a^2+7 a b \tan (c+d x)+2 b^2 \tan ^2(c+d x)\right )\right )\right )}{4 \sqrt {a} \sqrt {b} d \sqrt {a+b \tan (c+d x)} \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]
(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((3*a^2 - 8*b^2)*ArcSinh[(Sqrt[b]*S qrt[Tan[c + d*x]])/Sqrt[a]]*(a + b*Tan[c + d*x]) + Sqrt[a]*Sqrt[b]*Sqrt[1 + (b*Tan[c + d*x])/a]*(-4*(-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 + b*Tan[c + d*x]] + 4*(-1)^(3/4)*(a + I*b)^(3/2)*ArcTan[((-1)^(1/4)*Sqrt[ a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]] + Sqrt[Tan[c + d*x]]*(5*a^2 + 7*a*b*Tan[c + d*x] + 2*b^2*Tan[c + d*x ]^2))))/(4*Sqrt[a]*Sqrt[b]*d*Sqrt[a + b*Tan[c + d*x]]*Sqrt[1 + (b*Tan[c + d*x])/a])
Time = 1.26 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.86, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4729, 3042, 4053, 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.
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{3/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{3/2}}{\cot (c+d x)^{3/2}}dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \tan (c+d x)^{3/2} (a+b \tan (c+d x))^{3/2}dx\) |
| \(\Big \downarrow \) 4053 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} \int -\frac {\sqrt {a+b \tan (c+d x)} \left (-3 a \tan ^2(c+d x)+4 b \tan (c+d x)+a\right )}{2 \sqrt {\tan (c+d x)}}dx+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {1}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-3 a \tan ^2(c+d x)+4 b \tan (c+d x)+a\right )}{\sqrt {\tan (c+d x)}}dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {1}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-3 a \tan (c+d x)^2+4 b \tan (c+d x)+a\right )}{\sqrt {\tan (c+d x)}}dx\right )\) |
| \(\Big \downarrow \) 4130 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\int \frac {5 a^2+16 b \tan (c+d x) a-\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {1}{2} \int \frac {5 a^2+16 b \tan (c+d x) a-\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {1}{2} \int \frac {5 a^2+16 b \tan (c+d x) a-\left (3 a^2-8 b^2\right ) \tan (c+d x)^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {5 a^2+16 b \tan (c+d x) a-\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {5 a^2+16 b \tan (c+d x) a-\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 2257 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \left (\frac {8 b^2-3 a^2}{\sqrt {a+b \tan (c+d x)}}+\frac {8 \left (a^2+2 b \tan (c+d x) a-b^2\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {-\frac {\left (3 a^2-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}-4 (-b+i a)^{3/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-4 (b+i a)^{3/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/(2*d) + (-((-4*(I*a - b)^(3/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan [c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - ((3*a^2 - 8*b^2)*ArcTanh[(Sqrt[b]* Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] - 4*(I*a + b)^(3/2) *ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) + (3*a*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d)/4)
3.9.52.3.1 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*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^n/(f*(m + n - 1))), x] + Simp[1/(m + n - 1) Int[(a + b*Ta n[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a^2*c*(m + n - 1) - b *(b*c*(m - 1) + a*d*n) + (2*a*b*c + a^2*d - b^2*d)*(m + n - 1)*Tan[e + f*x] + b*(b*c*n + a*d*(2*m + n - 2))*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] && GtQ[m, 1] && GtQ[n, 0] && IntegerQ[2*n]
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]
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. \(2274\) vs. \(2(232)=464\).
Time = 39.34 (sec) , antiderivative size = 2275, normalized size of antiderivative = 7.95
1/8/d*csc(d*x+c)*(2*2^(1/2)*b^(3/2)*(-b+(a^2+b^2)^(1/2))^(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)*cos(d*x+c)*sin(d* x+c)+4*cos(d*x+c)^2*b^(5/2)*arctan((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)*sin(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2) *cos(d*x+c)+(b+(a^2+b^2)^(1/2))^(1/2))/(cos(d*x+c)-1)/(-b+(a^2+b^2)^(1/2)) ^(1/2))+4*cos(d*x+c)^2*b^(5/2)*arctan((2^(1/2)*((sin(d*x+c)*cos(d*x+c)*a-c os(d*x+c)^2*b+b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)+(b+(a^2+b^2)^(1/2))^(1 /2)*cos(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2))/(cos(d*x+c)-1)/(-b+(a^2+b^2)^(1/ 2))^(1/2))+4*cos(d*x+c)^2*b^(3/2)*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)^(1/2)*cos(d*x+c)-2*b*cos(d*x+ c)-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)-4*cos(d*x+c)^2*b^(3/2)*ln((cos(d*x+c)*co t(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)+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)*2^(1/2)*b^(3/2)*(-b+...
Leaf count of result is larger than twice the leaf count of optimal. 3781 vs. \(2 (228) = 456\).
Time = 1.13 (sec) , antiderivative size = 7599, normalized size of antiderivative = 26.57 \[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\cot ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cot \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}} \,d x \]