Integrand size = 30, antiderivative size = 302 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{4 e^3}+\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}}{5 e}-\frac {f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{8 e^3 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}-\frac {3 f^2 \sqrt {2 d e-b f^2} \left (4 a e^2-b^2 f^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\sqrt {2 d e-b f^2}}\right )}{8 \sqrt {2} e^{7/2}} \]
-3/16*f^2*(-b^2*f^2+4*a*e^2)*arctanh(2^(1/2)*e^(1/2)*(d+e*x+f*(a+b*x+e^2*x ^2/f^2)^(1/2))^(1/2)/(-b*f^2+2*d*e)^(1/2))*(-b*f^2+2*d*e)^(1/2)/e^(7/2)*2^ (1/2)+1/5*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(5/2)/e+1/4*f^2*(-b^2*f^2+4* a*e^2)*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2)/e^3-1/8*f^2*(-b*f^2+2*d*e )*(-b^2*f^2+4*a*e^2)*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2)/e^3/(b*f^2+ 2*e*(e*x+f*(a+x*(b*f^2+e^2*x)/f^2)^(1/2)))
Time = 2.57 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.47 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\frac {\sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}} \left (-15 b^3 f^6-2 b^2 e f^4 \left (-5 d+6 e x+10 f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+4 b e^2 f^2 \left (2 d^2+17 a f^2+8 e x \left (2 e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+4 d \left (3 e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )+8 e^3 \left (2 (d+2 e x)^2 \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+a f^2 \left (-d+16 e x+12 f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )\right )}{40 e^3 \left (b f^2+2 e \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )}-\frac {3 a f^2 \sqrt {-d e+\frac {b f^2}{2}} \arctan \left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{2 e^{3/2}}+\frac {3 b^2 f^4 \sqrt {-d e+\frac {b f^2}{2}} \arctan \left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{8 e^{7/2}} \]
(Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]]*(-15*b^3*f^6 - 2*b^2*e*f^ 4*(-5*d + 6*e*x + 10*f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + 4*b*e^2*f^2*(2*d^2 + 17*a*f^2 + 8*e*x*(2*e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + 4*d*(3*e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])) + 8*e^3*(2*(d + 2*e*x)^2*(e*x + f*Sqr t[a + x*(b + (e^2*x)/f^2)]) + a*f^2*(-d + 16*e*x + 12*f*Sqrt[a + x*(b + (e ^2*x)/f^2)]))))/(40*e^3*(b*f^2 + 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2) ]))) - (3*a*f^2*Sqrt[-(d*e) + (b*f^2)/2]*ArcTan[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]])/Sqrt[-2*d*e + b*f^2]])/(2*e^(3/2)) + (3*b^2*f^4*Sqrt[-(d*e) + (b*f^2)/2]*ArcTan[(Sqrt[2]*Sqrt[e]*Sqrt[d + e* x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]])/Sqrt[-2*d*e + b*f^2]])/(8*e^(7/2))
Time = 0.65 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2541, 1192, 1580, 2341, 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.
| \(\displaystyle \int \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 2541 |
| \(\displaystyle 2 \int \frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^{3/2} \left (e d^2-b f^2 d+a e f^2+e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2-\left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}{\left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle 4 \int \frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2 \left (e d^2-b f^2 d+a e f^2+e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2-\left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}{\left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}\) |
| \(\Big \downarrow \) 1580 |
| \(\displaystyle 4 \left (\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{32 e^3 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}-\frac {\int \frac {16 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^3 e^4-8 \left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2 e^3+4 f^2 \left (4 a e^2-b^2 f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right ) e^2+f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) e}{-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}}{32 e^4}\right )\) |
| \(\Big \downarrow \) 2341 |
| \(\displaystyle 4 \left (\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{32 e^3 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}-\frac {\int \left (-8 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2 e^3-2 f^2 \left (4 a e^2-b^2 f^2\right ) e+\frac {3 \left (b^3 e f^6-4 a b e^3 f^4-2 b^2 d e^2 f^4+8 a d e^4 f^2\right )}{-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}\right )d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}}{32 e^4}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 4 \left (\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{32 e^3 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}-\frac {\frac {3 \sqrt {e} f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {2 d e-b f^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {2 d e-b f^2}}\right )}{\sqrt {2}}-2 e f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}-\frac {8}{5} e^3 \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{5/2}}{32 e^4}\right )\) |
4*((f^2*(2*d*e - b*f^2)*(4*a*e^2 - b^2*f^2)*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/(32*e^3*(2*d*e - b*f^2 - 2*e*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]))) - (-2*e*f^2*(4*a*e^2 - b^2*f^2)*Sqrt[d + e*x + f*Sqrt[ a + b*x + (e^2*x^2)/f^2]] - (8*e^3*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f ^2])^(5/2))/5 + (3*Sqrt[e]*f^2*Sqrt[2*d*e - b*f^2]*(4*a*e^2 - b^2*f^2)*Arc Tanh[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/Sqr t[2*d*e - b*f^2]])/Sqrt[2])/(32*e^4))
3.5.80.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^( 2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_) ^4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[1/(2*e^(2*p + m/2)* (q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2* e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b *d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b, c, d, e }, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c _.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Simp[2 Subst[Int[(g + h*x^n)^p*((d ^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2)/(-2*d*e + b*f^2 + 2*e*x )^2), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e , f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
\[\int {\left (d +e x +f \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}^{\frac {3}{2}}d x\]
Time = 0.44 (sec) , antiderivative size = 657, normalized size of antiderivative = 2.18 \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\left [-\frac {15 \, \sqrt {\frac {1}{2}} {\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt {-\frac {b f^{2} - 2 \, d e}{e}} \log \left (-b^{2} f^{4} + 4 \, {\left (b d e - a e^{2}\right )} f^{2} - 4 \, {\left (b e^{2} f^{2} - 2 \, d e^{3}\right )} x + 4 \, {\left (2 \, \sqrt {\frac {1}{2}} e^{2} f \sqrt {-\frac {b f^{2} - 2 \, d e}{e}} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} - \sqrt {\frac {1}{2}} {\left (b e f^{2} + 2 \, e^{3} x\right )} \sqrt {-\frac {b f^{2} - 2 \, d e}{e}}\right )} \sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d} + 4 \, {\left (b e f^{3} - 2 \, d e^{2} f\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + 2 \, {\left (15 \, b^{2} f^{4} - 16 \, e^{4} x^{2} - 8 \, d^{2} e^{2} - 2 \, {\left (5 \, b d e + 24 \, a e^{2}\right )} f^{2} + 2 \, {\left (b e^{2} f^{2} - 18 \, d e^{3}\right )} x - 2 \, {\left (5 \, b e f^{3} + 8 \, e^{3} f x - 2 \, d e^{2} f\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{80 \, e^{3}}, \frac {15 \, \sqrt {\frac {1}{2}} {\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt {\frac {b f^{2} - 2 \, d e}{e}} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d} e \sqrt {\frac {b f^{2} - 2 \, d e}{e}}}{b f^{2} - 2 \, d e}\right ) - {\left (15 \, b^{2} f^{4} - 16 \, e^{4} x^{2} - 8 \, d^{2} e^{2} - 2 \, {\left (5 \, b d e + 24 \, a e^{2}\right )} f^{2} + 2 \, {\left (b e^{2} f^{2} - 18 \, d e^{3}\right )} x - 2 \, {\left (5 \, b e f^{3} + 8 \, e^{3} f x - 2 \, d e^{2} f\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{40 \, e^{3}}\right ] \]
[-1/80*(15*sqrt(1/2)*(b^2*f^4 - 4*a*e^2*f^2)*sqrt(-(b*f^2 - 2*d*e)/e)*log( -b^2*f^4 + 4*(b*d*e - a*e^2)*f^2 - 4*(b*e^2*f^2 - 2*d*e^3)*x + 4*(2*sqrt(1 /2)*e^2*f*sqrt(-(b*f^2 - 2*d*e)/e)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) - sqrt(1/2)*(b*e*f^2 + 2*e^3*x)*sqrt(-(b*f^2 - 2*d*e)/e))*sqrt(e*x + f*sqrt ((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d) + 4*(b*e*f^3 - 2*d*e^2*f)*sqrt((b*f ^2*x + e^2*x^2 + a*f^2)/f^2)) + 2*(15*b^2*f^4 - 16*e^4*x^2 - 8*d^2*e^2 - 2 *(5*b*d*e + 24*a*e^2)*f^2 + 2*(b*e^2*f^2 - 18*d*e^3)*x - 2*(5*b*e*f^3 + 8* e^3*f*x - 2*d*e^2*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*s qrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/e^3, 1/40*(15*sqrt(1/2)*(b^2*f^ 4 - 4*a*e^2*f^2)*sqrt((b*f^2 - 2*d*e)/e)*arctan(2*sqrt(1/2)*sqrt(e*x + f*s qrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d)*e*sqrt((b*f^2 - 2*d*e)/e)/(b*f^2 - 2*d*e)) - (15*b^2*f^4 - 16*e^4*x^2 - 8*d^2*e^2 - 2*(5*b*d*e + 24*a*e^2) *f^2 + 2*(b*e^2*f^2 - 18*d*e^3)*x - 2*(5*b*e*f^3 + 8*e^3*f*x - 2*d*e^2*f)* sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^ 2 + a*f^2)/f^2) + d))/e^3]
\[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\int \left (d + e x + f \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {3}{2}}\, dx \]
\[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\int { {\left (e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {3}{2}} \,d x } \]
\[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\int { {\left (e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\int {\left (d+e\,x+f\,\sqrt {a+b\,x+\frac {e^2\,x^2}{f^2}}\right )}^{3/2} \,d x \]