Integrand size = 25, antiderivative size = 160 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3 \, dx=\frac {2 \sqrt {2} a^3 d^{3/2} \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f}-\frac {4 a^3 d \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{3 f}+\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+a^3 \tan (e+f x)\right )}{7 d f} \] Output:
2*2^(1/2)*a^3*d^(3/2)*arctanh(1/2*(d^(1/2)+d^(1/2)*tan(f*x+e))*2^(1/2)/(d* tan(f*x+e))^(1/2))/f-4*a^3*d*(d*tan(f*x+e))^(1/2)/f+4/3*a^3*(d*tan(f*x+e)) ^(3/2)/f+32/35*a^3*(d*tan(f*x+e))^(5/2)/d/f+2/7*(d*tan(f*x+e))^(5/2)*(a^3+ a^3*tan(f*x+e))/d/f
Leaf count is larger than twice the leaf count of optimal. \(380\) vs. \(2(160)=320\).
Time = 3.65 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.38 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3 \, dx=\frac {a^3 (d \tan (e+f x))^{3/2} (1+\tan (e+f x))^3 \left (-420 \arctan \left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \cos ^3(e+f x) \sqrt [4]{-\tan (e+f x)}+420 \text {arctanh}\left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \cos ^3(e+f x) \sqrt [4]{-\tan (e+f x)}+\cos (e+f x) \sqrt [4]{\tan (e+f x)} \left (-210 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^2(e+f x)+210 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^2(e+f x)-105 \sqrt {2} \cos ^2(e+f x) \log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )+105 \sqrt {2} \cos ^2(e+f x) \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )-840 \cos ^2(e+f x) \sqrt {\tan (e+f x)}+280 \cos ^2(e+f x) \tan ^{\frac {3}{2}}(e+f x)+60 \sin ^2(e+f x) \tan ^{\frac {3}{2}}(e+f x)+126 \sin (2 (e+f x)) \tan ^{\frac {3}{2}}(e+f x)\right )\right )}{210 f (\cos (e+f x)+\sin (e+f x))^3 \tan ^{\frac {7}{4}}(e+f x)} \] Input:
Integrate[(d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^3,x]
Output:
(a^3*(d*Tan[e + f*x])^(3/2)*(1 + Tan[e + f*x])^3*(-420*ArcTan[(-Tan[e + f* x]^2)^(1/4)]*Cos[e + f*x]^3*(-Tan[e + f*x])^(1/4) + 420*ArcTanh[(-Tan[e + f*x]^2)^(1/4)]*Cos[e + f*x]^3*(-Tan[e + f*x])^(1/4) + Cos[e + f*x]*Tan[e + f*x]^(1/4)*(-210*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f *x]^2 + 210*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^2 - 105*Sqrt[2]*Cos[e + f*x]^2*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] + 105*Sqrt[2]*Cos[e + f*x]^2*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Ta n[e + f*x]] - 840*Cos[e + f*x]^2*Sqrt[Tan[e + f*x]] + 280*Cos[e + f*x]^2*T an[e + f*x]^(3/2) + 60*Sin[e + f*x]^2*Tan[e + f*x]^(3/2) + 126*Sin[2*(e + f*x)]*Tan[e + f*x]^(3/2))))/(210*f*(Cos[e + f*x] + Sin[e + f*x])^3*Tan[e + f*x]^(7/4))
Time = 0.85 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4049, 3042, 4113, 3042, 4011, 3042, 4011, 3042, 4015, 221}
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.
| \(\displaystyle \int (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 4049 |
| \(\displaystyle \frac {2 \int (d \tan (e+f x))^{3/2} \left (8 d \tan ^2(e+f x) a^3+d a^3+7 d \tan (e+f x) a^3\right )dx}{7 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{5/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int (d \tan (e+f x))^{3/2} \left (8 d \tan (e+f x)^2 a^3+d a^3+7 d \tan (e+f x) a^3\right )dx}{7 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{5/2}}{7 d f}\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {2 \left (\int (d \tan (e+f x))^{3/2} \left (7 a^3 d \tan (e+f x)-7 a^3 d\right )dx+\frac {16 a^3 (d \tan (e+f x))^{5/2}}{5 f}\right )}{7 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{5/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\int (d \tan (e+f x))^{3/2} \left (7 a^3 d \tan (e+f x)-7 a^3 d\right )dx+\frac {16 a^3 (d \tan (e+f x))^{5/2}}{5 f}\right )}{7 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{5/2}}{7 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {2 \left (\int \sqrt {d \tan (e+f x)} \left (-7 d^2 a^3-7 d^2 \tan (e+f x) a^3\right )dx+\frac {16 a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac {14 a^3 d (d \tan (e+f x))^{3/2}}{3 f}\right )}{7 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{5/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\int \sqrt {d \tan (e+f x)} \left (-7 d^2 a^3-7 d^2 \tan (e+f x) a^3\right )dx+\frac {16 a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac {14 a^3 d (d \tan (e+f x))^{3/2}}{3 f}\right )}{7 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{5/2}}{7 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {2 \left (\int \frac {7 a^3 d^3-7 a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx-\frac {14 a^3 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {16 a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac {14 a^3 d (d \tan (e+f x))^{3/2}}{3 f}\right )}{7 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{5/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\int \frac {7 a^3 d^3-7 a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx-\frac {14 a^3 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {16 a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac {14 a^3 d (d \tan (e+f x))^{3/2}}{3 f}\right )}{7 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{5/2}}{7 d f}\) |
| \(\Big \downarrow \) 4015 |
| \(\displaystyle \frac {2 \left (-\frac {98 a^6 d^6 \int \frac {1}{49 \cot (e+f x) \left (a^3 d^3+a^3 \tan (e+f x) d^3\right )^2-98 a^6 d^6}d\frac {7 \left (a^3 d^3+a^3 \tan (e+f x) d^3\right )}{\sqrt {d \tan (e+f x)}}}{f}-\frac {14 a^3 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {14 a^3 d (d \tan (e+f x))^{3/2}}{3 f}+\frac {16 a^3 (d \tan (e+f x))^{5/2}}{5 f}\right )}{7 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{5/2}}{7 d f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 \left (\frac {7 \sqrt {2} a^3 d^{5/2} \text {arctanh}\left (\frac {a^3 d^3 \tan (e+f x)+a^3 d^3}{\sqrt {2} a^3 d^{5/2} \sqrt {d \tan (e+f x)}}\right )}{f}-\frac {14 a^3 d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {16 a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac {14 a^3 d (d \tan (e+f x))^{3/2}}{3 f}\right )}{7 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{5/2}}{7 d f}\) |
Input:
Int[(d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^3,x]
Output:
(2*(d*Tan[e + f*x])^(5/2)*(a^3 + a^3*Tan[e + f*x]))/(7*d*f) + (2*((7*Sqrt[ 2]*a^3*d^(5/2)*ArcTanh[(a^3*d^3 + a^3*d^3*Tan[e + f*x])/(Sqrt[2]*a^3*d^(5/ 2)*Sqrt[d*Tan[e + f*x]])])/f - (14*a^3*d^2*Sqrt[d*Tan[e + f*x]])/f + (14*a ^3*d*(d*Tan[e + f*x])^(3/2))/(3*f) + (16*a^3*(d*Tan[e + f*x])^(5/2))/(5*f) ))/(7*d)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[-2*(d^2/f) Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((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 - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 , 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I ntegerQ[m]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) )
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(338\) vs. \(2(135)=270\).
Time = 1.68 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.12
| method | result | size |
| derivativedivides | \(\frac {2 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {3 d \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{3} \sqrt {d \tan \left (f x +e \right )}+2 d^{4} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\) | \(339\) |
| default | \(\frac {2 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {3 d \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{3} \sqrt {d \tan \left (f x +e \right )}+2 d^{4} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\) | \(339\) |
| parts | \(\frac {2 a^{3} d \left (\sqrt {d \tan \left (f x +e \right )}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{f}+\frac {2 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}-\frac {d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+\frac {d^{4} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f \,d^{2}}+\frac {3 a^{3} \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}+\frac {6 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-d^{2} \sqrt {d \tan \left (f x +e \right )}+\frac {d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{f d}\) | \(654\) |
Input:
int((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
Output:
2/f*a^3/d^2*(1/7*(d*tan(f*x+e))^(7/2)+3/5*d*(d*tan(f*x+e))^(5/2)+2/3*d^2*( d*tan(f*x+e))^(3/2)-2*d^3*(d*tan(f*x+e))^(1/2)+2*d^4*(1/8/d*(d^2)^(1/4)*2^ (1/2)*(ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/ 2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2 *arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2 )^(1/4)*(d*tan(f*x+e))^(1/2)+1))-1/8/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e) -(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2) ^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^( 1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^ (1/2)+1))))
Time = 0.09 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.61 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3 \, dx=\left [\frac {105 \, \sqrt {2} a^{3} d^{\frac {3}{2}} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (15 \, a^{3} d \tan \left (f x + e\right )^{3} + 63 \, a^{3} d \tan \left (f x + e\right )^{2} + 70 \, a^{3} d \tan \left (f x + e\right ) - 210 \, a^{3} d\right )} \sqrt {d \tan \left (f x + e\right )}}{105 \, f}, -\frac {2 \, {\left (105 \, \sqrt {2} a^{3} \sqrt {-d} d \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) - {\left (15 \, a^{3} d \tan \left (f x + e\right )^{3} + 63 \, a^{3} d \tan \left (f x + e\right )^{2} + 70 \, a^{3} d \tan \left (f x + e\right ) - 210 \, a^{3} d\right )} \sqrt {d \tan \left (f x + e\right )}\right )}}{105 \, f}\right ] \] Input:
integrate((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^3,x, algorithm="fricas")
Output:
[1/105*(105*sqrt(2)*a^3*d^(3/2)*log((d*tan(f*x + e)^2 + 2*sqrt(2)*sqrt(d*t an(f*x + e))*sqrt(d)*(tan(f*x + e) + 1) + 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) + 2*(15*a^3*d*tan(f*x + e)^3 + 63*a^3*d*tan(f*x + e)^2 + 70*a^ 3*d*tan(f*x + e) - 210*a^3*d)*sqrt(d*tan(f*x + e)))/f, -2/105*(105*sqrt(2) *a^3*sqrt(-d)*d*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan(f*x + e) + 1)/(d*tan(f*x + e))) - (15*a^3*d*tan(f*x + e)^3 + 63*a^3*d*tan(f*x + e)^2 + 70*a^3*d*tan(f*x + e) - 210*a^3*d)*sqrt(d*tan(f*x + e)))/f]
\[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3 \, dx=a^{3} \left (\int \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx + \int 3 \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}\, dx + \int 3 \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{3}{\left (e + f x \right )}\, dx\right ) \] Input:
integrate((d*tan(f*x+e))**(3/2)*(a+a*tan(f*x+e))**3,x)
Output:
a**3*(Integral((d*tan(e + f*x))**(3/2), x) + Integral(3*(d*tan(e + f*x))** (3/2)*tan(e + f*x), x) + Integral(3*(d*tan(e + f*x))**(3/2)*tan(e + f*x)** 2, x) + Integral((d*tan(e + f*x))**(3/2)*tan(e + f*x)**3, x))
Time = 0.12 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.02 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3 \, dx=\frac {105 \, a^{3} d^{3} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + \frac {2 \, {\left (15 \, \left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} a^{3} + 63 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} d + 70 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d^{2} - 210 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{3}\right )}}{d}}{105 \, d f} \] Input:
integrate((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^3,x, algorithm="maxima")
Output:
1/105*(105*a^3*d^3*(sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) - sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan (f*x + e))*sqrt(d) + d)/sqrt(d)) + 2*(15*(d*tan(f*x + e))^(7/2)*a^3 + 63*( d*tan(f*x + e))^(5/2)*a^3*d + 70*(d*tan(f*x + e))^(3/2)*a^3*d^2 - 210*sqrt (d*tan(f*x + e))*a^3*d^3)/d)/(d*f)
Exception generated. \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3 \, dx=\text {Exception raised: TypeError} \] Input:
integrate((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^3,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{%%%{1,[6,24]%%%}+%%%{10,[6,22]%%%}+%%%{45,[6,20]%%%}+%%%{1 20,[6,18]
Time = 2.45 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.89 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3 \, dx=\frac {4\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}+\frac {6\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,d\,f}+\frac {2\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}}{7\,d^2\,f}-\frac {4\,a^3\,d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}-\frac {\sqrt {2}\,a^3\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,a^6\,d^{9/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,32{}\mathrm {i}}{32\,a^6\,d^5+32\,a^6\,d^5\,\mathrm {tan}\left (e+f\,x\right )}\right )\,2{}\mathrm {i}}{f} \] Input:
int((d*tan(e + f*x))^(3/2)*(a + a*tan(e + f*x))^3,x)
Output:
(4*a^3*(d*tan(e + f*x))^(3/2))/(3*f) + (6*a^3*(d*tan(e + f*x))^(5/2))/(5*d *f) + (2*a^3*(d*tan(e + f*x))^(7/2))/(7*d^2*f) - (4*a^3*d*(d*tan(e + f*x)) ^(1/2))/f - (2^(1/2)*a^3*d^(3/2)*atan((2^(1/2)*a^6*d^(9/2)*(d*tan(e + f*x) )^(1/2)*32i)/(32*a^6*d^5 + 32*a^6*d^5*tan(e + f*x)))*2i)/f
\[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3 \, dx=\frac {\sqrt {d}\, a^{3} d \left (6 \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}-20 \sqrt {\tan \left (f x +e \right )}+10 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )}d x \right ) f +5 \left (\int \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{4}d x \right ) f +15 \left (\int \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}d x \right ) f \right )}{5 f} \] Input:
int((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^3,x)
Output:
(sqrt(d)*a**3*d*(6*sqrt(tan(e + f*x))*tan(e + f*x)**2 - 20*sqrt(tan(e + f* x)) + 10*int(sqrt(tan(e + f*x))/tan(e + f*x),x)*f + 5*int(sqrt(tan(e + f*x ))*tan(e + f*x)**4,x)*f + 15*int(sqrt(tan(e + f*x))*tan(e + f*x)**2,x)*f)) /(5*f)