Integrand size = 35, antiderivative size = 209 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {(a+i b)^2 (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {2 b^{3/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \]
2*b^(3/2)*B*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d-(I* a+b)^(3/2)*(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c)) ^(1/2))/d-(a+I*b)^2*(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan (d*x+c))^(1/2))/d/(I*a-b)^(1/2)-2*a*A*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^ (1/2)
Time = 4.46 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.99 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt [4]{-1} (-a+i b)^{3/2} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\sqrt [4]{-1} a A \sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(-1)^{3/4} A \sqrt {a+i b} b \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(-1)^{3/4} a \sqrt {a+i b} B \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt [4]{-1} \sqrt {a+i b} b B \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}}+\frac {2 \sqrt {a} b^{3/2} B \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}}{d} \]
((-1)^(1/4)*(-a + I*b)^(3/2)*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*S qrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - (-1)^(1/4)*a*A*Sqrt[a + I*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*Sqrt[a + I*b]*b*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqr t[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - (-1)^(3/4)*a*Sqrt[a + I*b]*B* ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d* x]]] + (-1)^(1/4)*Sqrt[a + I*b]*b*B*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[ Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - (2*a*A*Sqrt[a + b*Tan[c + d*x]] )/Sqrt[Tan[c + d*x]] + (2*Sqrt[a]*b^(3/2)*B*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 = 1.05 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 4088, 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.
\(\displaystyle \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan (c+d x)^{3/2}}dx\) |
\(\Big \downarrow \) 4088 |
\(\displaystyle 2 \int \frac {b^2 B \tan ^2(c+d x)-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (2 A b+a B)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {b^2 B \tan ^2(c+d x)-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (2 A b+a B)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {b^2 B \tan (c+d x)^2-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (2 A b+a B)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 4138 |
\(\displaystyle \frac {\int \frac {b^2 B \tan ^2(c+d x)-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (2 A b+a B)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {2 \int \frac {b^2 B \tan ^2(c+d x)-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (2 A b+a B)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 2257 |
\(\displaystyle \frac {2 \int \left (\frac {B b^2}{\sqrt {a+b \tan (c+d x)}}+\frac {B a^2+2 A b a-b^2 B-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}-\frac {2 a A \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a A \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {2 \left (\frac {1}{2} (-b+i a)^{3/2} (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {1}{2} (b+i a)^{3/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b^{3/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )}{d}\) |
(2*(((I*a - b)^(3/2)*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/S qrt[a + b*Tan[c + d*x]]])/2 + b^(3/2)*B*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x] ])/Sqrt[a + b*Tan[c + d*x]]] - ((I*a + b)^(3/2)*(I*A + B)*ArcTanh[(Sqrt[I* a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/2))/d - (2*a*A*Sqrt[ a + b*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]])
3.5.37.3.1 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*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x ])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d* (b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1) + a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[ e + f*x] - b*(d*(A*b*c + a*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] & & LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 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] :> 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]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.91 (sec) , antiderivative size = 2394883, normalized size of antiderivative = 11458.77
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 12148 vs. \(2 (167) = 334\).
Time = 4.52 (sec) , antiderivative size = 24297, normalized size of antiderivative = 116.25 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \]