Integrand size = 35, antiderivative size = 241 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(i a-b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b^{3/2} (2 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}} \]
(I*a-b)^(5/2)*(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c ))^(1/2))/d+b^(3/2)*(2*A*b+5*B*a)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*ta n(d*x+c))^(1/2))/d-(I*a+b)^(5/2)*(A-I*B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^ (1/2)/(a+b*tan(d*x+c))^(1/2))/d+b*(2*A*a+B*b)*tan(d*x+c)^(1/2)*(a+b*tan(d* x+c))^(1/2)/d-2*a*A*(a+b*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(1/2)
Time = 2.71 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.49 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(-1)^{3/4} (-a+i b)^{5/2} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\sqrt [4]{-1} (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {(a-i b)^2 (A-i B) \sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}}-\frac {(a+i b)^2 (A+i B) \sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{\sqrt {\tan (c+d x)}}+\frac {b (2 A b+5 a B) \sqrt {a+b \tan (c+d x)} \left (\frac {\sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {\tan (c+d x)}}\right )}{\sqrt {1+\frac {b \tan (c+d x)}{a}}}}{d} \]
((-1)^(3/4)*(-a + I*b)^(5/2)*(I*A + 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 + I*b)^(5/2)* (A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b* Tan[c + d*x]]] - ((a - I*b)^2*(A - I*B)*Sqrt[a + b*Tan[c + d*x]])/Sqrt[Tan [c + d*x]] - ((a + I*b)^2*(A + I*B)*Sqrt[a + b*Tan[c + d*x]])/Sqrt[Tan[c + d*x]] + (b*B*(a + b*Tan[c + d*x])^(3/2))/Sqrt[Tan[c + d*x]] + (b*(2*A*b + 5*a*B)*Sqrt[a + b*Tan[c + d*x]]*((Sqrt[b]*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d *x]])/Sqrt[a]])/Sqrt[a] - Sqrt[1 + (b*Tan[c + d*x])/a]/Sqrt[Tan[c + d*x]]) )/Sqrt[1 + (b*Tan[c + d*x])/a])/d
Time = 1.51 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.98, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 4088, 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.
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)^{3/2}}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle 2 \int \frac {\sqrt {a+b \tan (c+d x)} \left (b (2 a A+b B) \tan ^2(c+d x)-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (4 A b+a B)\right )}{2 \sqrt {\tan (c+d x)}}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)} \left (b (2 a A+b B) \tan ^2(c+d x)-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (4 A b+a B)\right )}{\sqrt {\tan (c+d x)}}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)} \left (b (2 a A+b B) \tan (c+d x)^2-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (4 A b+a B)\right )}{\sqrt {\tan (c+d x)}}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 4130 |
| \(\displaystyle \int \frac {b^2 (2 A b+5 a B) \tan ^2(c+d x)-2 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (2 B a^2+6 A b a-b^2 B\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {b^2 (2 A b+5 a B) \tan ^2(c+d x)-2 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (2 B a^2+6 A b a-b^2 B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {b^2 (2 A b+5 a B) \tan (c+d x)^2-2 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (2 B a^2+6 A b a-b^2 B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \frac {\int \frac {b^2 (2 A b+5 a B) \tan ^2(c+d x)-2 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (2 B a^2+6 A b a-b^2 B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {\int \frac {b^2 (2 A b+5 a B) \tan ^2(c+d x)-2 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (2 B a^2+6 A b a-b^2 B\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 2257 |
| \(\displaystyle \frac {\int \left (\frac {(2 A b+5 a B) b^2}{\sqrt {a+b \tan (c+d x)}}+\frac {2 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3-\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(-b+i a)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b^{3/2} (5 a B+2 A b) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(b+i a)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\) |
((I*a - b)^(5/2)*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[ a + b*Tan[c + d*x]]] + b^(3/2)*(2*A*b + 5*a*B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - (I*a + b)^(5/2)*(A - I*B)*ArcTanh[(S qrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (b*(2*a*A + b*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d - (2*a*A*(a + b*Tan[ c + d*x])^(3/2))/(d*Sqrt[Tan[c + d*x]])
3.5.45.3.1 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] :> 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]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.08 (sec) , antiderivative size = 2653772, normalized size of antiderivative = 11011.50
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 17614 vs. \(2 (197) = 394\).
Time = 7.59 (sec) , antiderivative size = 35230, normalized size of antiderivative = 146.18 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \]