Integrand size = 35, antiderivative size = 264 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {(i a-b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\sqrt {b} (2 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {(i a+b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {b B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \]
-(I*a-b)^(3/2)*(A+I*B)*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)*(A-I*B)*arcta nh((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*A*b+3*B*a)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*ta n(d*x+c))^(1/2))*b^(1/2)*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+b*B*(a+b*tan( d*x+c))^(1/2)/d/cot(d*x+c)^(1/2)
Time = 1.41 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.00 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-\sqrt [4]{-1} (-a+i b)^{3/2} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(-1)^{3/4} (a+i b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {\sqrt {a} \sqrt {b} (2 A b+3 a B) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]
(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-((-1)^(1/4)*(-a + I*b)^(3/2)*(I*A + B)*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)^(3/2)*(A + I*B)*ArcTan[((-1)^(1/4)*Sqr t[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + b*B*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + (Sqrt[a]*Sqrt[b]*(2*A*b + 3*a*B)*ArcSin h[(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 = 1.29 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.84, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4729, 3042, 4090, 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.
| \(\displaystyle \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4090 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\int \frac {b (2 A b+3 a B) \tan ^2(c+d x)+2 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (2 a A-b B)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} \int \frac {b (2 A b+3 a B) \tan ^2(c+d x)+2 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (2 a A-b B)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} \int \frac {b (2 A b+3 a B) \tan (c+d x)^2+2 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (2 a A-b B)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {b (2 A b+3 a B) \tan ^2(c+d x)+2 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (2 a A-b B)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}+\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {b (2 A b+3 a B) \tan ^2(c+d x)+2 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (2 a A-b B)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}+\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 2257 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \left (\frac {b (2 A b+3 a B)}{\sqrt {a+b \tan (c+d x)}}+\frac {2 \left (A a^2-2 b B a-A b^2+\left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}+\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {-\left ((-b+i a)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )+\sqrt {b} (3 a B+2 A b) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(b+i a)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-((I*a - b)^(3/2)*(A + I*B)*ArcTan [(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]) + Sqrt[b]*( 2*A*b + 3*a*B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x ]]] - (I*a + b)^(3/2)*(A - I*B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]]) /Sqrt[a + b*Tan[c + d*x]]])/d + (b*B*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d)
3.7.24.3.1 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_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Ta n[e + f*x])^n*Simp[a^2*A*d*(m + n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m - 1) - b *(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 , 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 1] && ( !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]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.50 (sec) , antiderivative size = 2398750, normalized size of antiderivative = 9086.17
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 12219 vs. \(2 (212) = 424\).
Time = 4.89 (sec) , antiderivative size = 24471, normalized size of antiderivative = 92.69 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \sqrt {\cot {\left (c + d x \right )}}\, dx \]
\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\cot \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2} \,d x \]