Integrand size = 35, antiderivative size = 219 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}+\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]
-(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I* a-b)^(3/2)/d+2*B*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/ b^(3/2)/d-(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^ (1/2))/(I*a+b)^(3/2)/d+2*a*(A*b-B*a)*tan(d*x+c)^(1/2)/b/(a^2+b^2)/d/(a+b*t an(d*x+c))^(1/2)
Time = 2.03 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.56 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {2 \sqrt {a} \sqrt {-a+i b} \sqrt {a+i b} \left (a^2+b^2\right ) B \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}+\sqrt {b} \left (\sqrt [4]{-1} (a+i b)^{3/2} b (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt {-a+i b} \left (2 a \sqrt {a+i b} (A b-a B) \sqrt {\tan (c+d x)}+\sqrt [4]{-1} b (i a+b) (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}\right )\right )}{(-a+i b)^{3/2} (a+i b)^{3/2} b^{3/2} d \sqrt {a+b \tan (c+d x)}} \]
-((2*Sqrt[a]*Sqrt[-a + I*b]*Sqrt[a + I*b]*(a^2 + b^2)*B*ArcSinh[(Sqrt[b]*S qrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a] + Sqrt[b]*((-1)^( 1/4)*(a + I*b)^(3/2)*b*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Ta n[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]] + Sqrt[-a + I*b]*(2*a*Sqrt[a + I*b]*(A*b - a*B)*Sqrt[Tan[c + d*x]] + (-1)^(1/4)*b*(I *a + b)*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqr t[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]])))/((-a + I*b)^(3/2)*(a + I*b)^(3/2)*b^(3/2)*d*Sqrt[a + b*Tan[c + d*x]]))
Time = 1.14 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.16, 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.
\(\displaystyle \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^{3/2} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4088 |
\(\displaystyle \frac {2 \int -\frac {-\left (\left (a^2+b^2\right ) B \tan ^2(c+d x)\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}+\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\int \frac {-\left (\left (a^2+b^2\right ) B \tan ^2(c+d x)\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\int \frac {-\left (\left (a^2+b^2\right ) B \tan (c+d x)^2\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 4138 |
\(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\int \frac {-\left (\left (a^2+b^2\right ) B \tan ^2(c+d x)\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{b d \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {2 \int \frac {-\left (\left (a^2+b^2\right ) B \tan ^2(c+d x)\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 2257 |
\(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {2 \int \left (\frac {b (a A+b B)-b (A b-a B) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}-\frac {\left (a^2+b^2\right ) B}{\sqrt {a+b \tan (c+d x)}}\right )d\sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {2 \left (-\frac {B \left (a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}+\frac {b (a-i b) (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 \sqrt {-b+i a}}+\frac {b (a+i b) (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 \sqrt {b+i a}}\right )}{b d \left (a^2+b^2\right )}\) |
(-2*(((a - I*b)*b*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt [a + b*Tan[c + d*x]]])/(2*Sqrt[I*a - b]) - ((a^2 + b^2)*B*ArcTanh[(Sqrt[b] *Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] + ((a + I*b)*b*(A - I*B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]] ])/(2*Sqrt[I*a + b])))/(b*(a^2 + b^2)*d) + (2*a*(A*b - a*B)*Sqrt[Tan[c + d *x]])/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]])
3.5.58.3.1 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_)*((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 = 1.55 (sec) , antiderivative size = 1560668, normalized size of antiderivative = 7126.34
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 18781 vs. \(2 (176) = 352\).
Time = 10.95 (sec) , antiderivative size = 37564, normalized size of antiderivative = 171.53 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]