3.473 \(\int (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}})^3 \, dx\)

Optimal. Leaf size=303 \[ -\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^3}{32 e^5 \left (2 e \left (f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac {3 f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^2 \log \left (2 e \left (f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}{32 e^5}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+e x\right )}{8 e^4}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^2}{16 e^3}+\frac {\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^4}{8 e} \]

[Out]

3/32*f^2*(-b*f^2+2*d*e)^2*(-b^2*f^2+4*a*e^2)*ln(b*f^2+2*e*(e*x+f*(a+x*(b*f^2+e^2*x)/f^2)^(1/2)))/e^5+1/8*f^2*(
-b*f^2+2*d*e)*(-b^2*f^2+4*a*e^2)*(e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))/e^4+1/16*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))^2/e^3+1/8*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^4/e-1/32*f^2*(-b*f^2+2*d*e)^3*(-b^2*f^2
+4*a*e^2)/e^5/(b*f^2+2*e*(e*x+f*(a+x*(b*f^2+e^2*x)/f^2)^(1/2)))

________________________________________________________________________________________

Rubi [A]  time = 0.38, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2116, 893} \[ \frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^2}{16 e^3}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+e x\right )}{8 e^4}-\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^3}{32 e^5 \left (2 e \left (f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac {3 f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^2 \log \left (2 e \left (f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}{32 e^5}+\frac {\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^4}{8 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^3,x]

[Out]

(f^2*(2*d*e - b*f^2)*(4*a*e^2 - b^2*f^2)*(e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]))/(8*e^4) + (f^2*(4*a*e^2 - b^
2*f^2)*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^2)/(16*e^3) + (d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^4
/(8*e) - (f^2*(2*d*e - b*f^2)^3*(4*a*e^2 - b^2*f^2))/(32*e^5*(b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x)
)/f^2]))) + (3*f^2*(2*d*e - b*f^2)^2*(4*a*e^2 - b^2*f^2)*Log[b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))
/f^2])])/(32*e^5)

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2116

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol]
:> Dist[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]

Rubi steps

\begin {align*} \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3 \, dx &=2 \operatorname {Subst}\left (\int \frac {x^3 \left (d^2 e-(b d-a e) f^2-\left (2 d e-b f^2\right ) x+e x^2\right )}{\left (-2 d e+b f^2+2 e x\right )^2} \, dx,x,d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right )}{16 e^4}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) x}{16 e^3}+\frac {x^3}{4 e}+\frac {f^2 \left (2 d e-b f^2\right )^3 \left (4 a e^2-b^2 f^2\right )}{32 e^4 \left (2 d e-b f^2-2 e x\right )^2}-\frac {3 \left (4 a e^2-b^2 f^2\right ) \left (2 d e f-b f^3\right )^2}{32 e^4 \left (2 d e-b f^2-2 e x\right )}\right ) \, dx,x,d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )\\ &=\frac {f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \left (e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )}{8 e^4}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^2}{16 e^3}+\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^4}{8 e}-\frac {f^2 \left (2 d e-b f^2\right )^3 \left (4 a e^2-b^2 f^2\right )}{32 e^5 \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 \left (2 d e-b f^2\right )^2 \left (4 a e^2-b^2 f^2\right ) \log \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 )}{32 e^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.55, size = 276, normalized size = 0.91 \[ \frac {2 e^2 f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+d+e x\right )^2+4 e f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+e x\right )-\frac {f^2 \left (b^2 f^2-4 a e^2\right ) \left (b f^2-2 d e\right )^3}{2 e \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+e x\right )+b f^2}-3 \left (b^2 f^2-4 a e^2\right ) \left (b f^3-2 d e f\right )^2 \log \left (-2 e \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+e x\right )-b f^2\right )+4 e^4 \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+d+e x\right )^4}{32 e^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^3,x]

[Out]

(4*e*f^2*(2*d*e - b*f^2)*(4*a*e^2 - b^2*f^2)*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + 2*e^2*f^2*(4*a*e^2 - b^
2*f^2)*(d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])^2 + 4*e^4*(d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])^4 - (
f^2*(-2*d*e + b*f^2)^3*(-4*a*e^2 + b^2*f^2))/(b*f^2 + 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])) - 3*(-4*a*e
^2 + b^2*f^2)*(-2*d*e*f + b*f^3)^2*Log[-(b*f^2) - 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])])/(32*e^5)

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 345, normalized size = 1.14 \[ \frac {32 \, e^{8} x^{4} + 32 \, {\left (b e^{6} f^{2} + 2 \, d e^{7}\right )} x^{3} + 48 \, {\left (d^{2} e^{6} + {\left (b d e^{5} + a e^{6}\right )} f^{2}\right )} x^{2} + 32 \, {\left (3 \, a d e^{5} f^{2} + d^{3} e^{5}\right )} x + 3 \, {\left (b^{4} f^{8} - 16 \, a d^{2} e^{4} f^{2} - 4 \, {\left (b^{3} d e + a b^{2} e^{2}\right )} f^{6} + 4 \, {\left (b^{2} d^{2} e^{2} + 4 \, a b d e^{3}\right )} f^{4}\right )} \log \left (-b f^{2} - 2 \, e^{2} x + 2 \, e f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + 2 \, {\left (3 \, b^{3} e f^{7} + 16 \, e^{7} f x^{3} - 4 \, {\left (3 \, b^{2} d e^{2} + 2 \, a b e^{3}\right )} f^{5} + 4 \, {\left (3 \, b d^{2} e^{3} + 8 \, a d e^{4}\right )} f^{3} + 8 \, {\left (b e^{5} f^{3} + 4 \, d e^{6} f\right )} x^{2} - 2 \, {\left (b^{2} e^{3} f^{5} - 12 \, d^{2} e^{5} f - 4 \, {\left (b d e^{4} + 2 \, a e^{5}\right )} f^{3}\right )} x\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}}{32 \, e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^3,x, algorithm="fricas")

[Out]

1/32*(32*e^8*x^4 + 32*(b*e^6*f^2 + 2*d*e^7)*x^3 + 48*(d^2*e^6 + (b*d*e^5 + a*e^6)*f^2)*x^2 + 32*(3*a*d*e^5*f^2
+ d^3*e^5)*x + 3*(b^4*f^8 - 16*a*d^2*e^4*f^2 - 4*(b^3*d*e + a*b^2*e^2)*f^6 + 4*(b^2*d^2*e^2 + 4*a*b*d*e^3)*f^
4)*log(-b*f^2 - 2*e^2*x + 2*e*f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2)) + 2*(3*b^3*e*f^7 + 16*e^7*f*x^3 - 4*(3*
b^2*d*e^2 + 2*a*b*e^3)*f^5 + 4*(3*b*d^2*e^3 + 8*a*d*e^4)*f^3 + 8*(b*e^5*f^3 + 4*d*e^6*f)*x^2 - 2*(b^2*e^3*f^5
- 12*d^2*e^5*f - 4*(b*d*e^4 + 2*a*e^5)*f^3)*x)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))/e^5

________________________________________________________________________________________

giac [A]  time = 0.37, size = 373, normalized size = 1.23 \[ b f^{2} x^{3} e + \frac {3}{2} \, b d f^{2} x^{2} + \frac {3}{2} \, a f^{2} x^{2} e + 3 \, a d f^{2} x + x^{4} e^{3} + 2 \, d x^{3} e^{2} + \frac {3}{2} \, d^{2} x^{2} e + d^{3} x + \frac {3}{32} \, {\left (b^{4} f^{7} {\left | f \right |} - 4 \, b^{3} d f^{5} {\left | f \right |} e - 4 \, a b^{2} f^{5} {\left | f \right |} e^{2} + 4 \, b^{2} d^{2} f^{3} {\left | f \right |} e^{2} + 16 \, a b d f^{3} {\left | f \right |} e^{3} - 16 \, a d^{2} f {\left | f \right |} e^{4}\right )} e^{\left (-5\right )} \log \left ({\left | -b f^{2} - 2 \, {\left (x e - \sqrt {b f^{2} x + a f^{2} + x^{2} e^{2}}\right )} e \right |}\right ) + \frac {1}{16} \, \sqrt {b f^{2} x + a f^{2} + x^{2} e^{2}} {\left (2 \, {\left (4 \, {\left (\frac {2 \, x {\left | f \right |} e^{2}}{f} + \frac {{\left (b f^{4} {\left | f \right |} e^{6} + 4 \, d f^{2} {\left | f \right |} e^{7}\right )} e^{\left (-6\right )}}{f^{3}}\right )} x - \frac {{\left (b^{2} f^{6} {\left | f \right |} e^{4} - 4 \, b d f^{4} {\left | f \right |} e^{5} - 8 \, a f^{4} {\left | f \right |} e^{6} - 12 \, d^{2} f^{2} {\left | f \right |} e^{6}\right )} e^{\left (-6\right )}}{f^{3}}\right )} x + \frac {{\left (3 \, b^{3} f^{8} {\left | f \right |} e^{2} - 12 \, b^{2} d f^{6} {\left | f \right |} e^{3} - 8 \, a b f^{6} {\left | f \right |} e^{4} + 12 \, b d^{2} f^{4} {\left | f \right |} e^{4} + 32 \, a d f^{4} {\left | f \right |} e^{5}\right )} e^{\left (-6\right )}}{f^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^3,x, algorithm="giac")

[Out]

b*f^2*x^3*e + 3/2*b*d*f^2*x^2 + 3/2*a*f^2*x^2*e + 3*a*d*f^2*x + x^4*e^3 + 2*d*x^3*e^2 + 3/2*d^2*x^2*e + d^3*x
+ 3/32*(b^4*f^7*abs(f) - 4*b^3*d*f^5*abs(f)*e - 4*a*b^2*f^5*abs(f)*e^2 + 4*b^2*d^2*f^3*abs(f)*e^2 + 16*a*b*d*f
^3*abs(f)*e^3 - 16*a*d^2*f*abs(f)*e^4)*e^(-5)*log(abs(-b*f^2 - 2*(x*e - sqrt(b*f^2*x + a*f^2 + x^2*e^2))*e)) +
1/16*sqrt(b*f^2*x + a*f^2 + x^2*e^2)*(2*(4*(2*x*abs(f)*e^2/f + (b*f^4*abs(f)*e^6 + 4*d*f^2*abs(f)*e^7)*e^(-6)
/f^3)*x - (b^2*f^6*abs(f)*e^4 - 4*b*d*f^4*abs(f)*e^5 - 8*a*f^4*abs(f)*e^6 - 12*d^2*f^2*abs(f)*e^6)*e^(-6)/f^3)
*x + (3*b^3*f^8*abs(f)*e^2 - 12*b^2*d*f^6*abs(f)*e^3 - 8*a*b*f^6*abs(f)*e^4 + 12*b*d^2*f^4*abs(f)*e^4 + 32*a*d
*f^4*abs(f)*e^5)*e^(-6)/f^3)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 685, normalized size = 2.26 \[ -\frac {3 b^{4} f^{7} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\right )}{32 \sqrt {\frac {e^{2}}{f^{2}}}\, e^{4}}+\frac {3 a \,b^{2} f^{5} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\right )}{8 \sqrt {\frac {e^{2}}{f^{2}}}\, e^{2}}+\frac {3 b^{3} d \,f^{5} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\right )}{8 \sqrt {\frac {e^{2}}{f^{2}}}\, e^{3}}+b e \,f^{2} x^{3}+e^{3} x^{4}-\frac {3 a b d \,f^{3} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\right )}{2 \sqrt {\frac {e^{2}}{f^{2}}}\, e}+\frac {3 a e \,f^{2} x^{2}}{2}+\frac {3 \sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\, b^{3} f^{7}}{16 e^{4}}-\frac {3 b^{2} d^{2} f^{3} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\right )}{8 \sqrt {\frac {e^{2}}{f^{2}}}\, e^{2}}+\frac {3 \sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\, b^{2} f^{5} x}{8 e^{2}}+\frac {3 b d \,f^{2} x^{2}}{2}+2 d \,e^{2} x^{3}+\frac {3 a \,d^{2} f \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\right )}{2 \sqrt {\frac {e^{2}}{f^{2}}}}+3 a d \,f^{2} x -\frac {3 \sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\, b^{2} d \,f^{5}}{4 e^{3}}-\frac {3 \sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\, b d \,f^{3} x}{2 e}+\frac {3 d^{2} e \,x^{2}}{2}+\frac {3 \sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\, b \,d^{2} f^{3}}{4 e^{2}}-\frac {\left (b x +\frac {e^{2} x^{2}}{f^{2}}+a \right )^{\frac {3}{2}} b \,f^{5}}{2 e^{2}}+d^{3} x +\frac {3 \sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\, d^{2} f x}{2}+\left (b x +\frac {e^{2} x^{2}}{f^{2}}+a \right )^{\frac {3}{2}} f^{3} x +\frac {d^{4}}{4 e}+\frac {2 \left (b x +\frac {e^{2} x^{2}}{f^{2}}+a \right )^{\frac {3}{2}} d \,f^{3}}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d+(b*x+e^2/f^2*x^2+a)^(1/2)*f)^3,x)

[Out]

3/2*d^2*e*x^2+d^3*x+e^3*x^4-3/2*d/e*f^3*b*ln((1/2*b+e^2/f^2*x)/(e^2/f^2)^(1/2)+(b*x+e^2/f^2*x^2+a)^(1/2))/(e^2
/f^2)^(1/2)*a+1/4*d^4/e+2*d/e*f^3*(b*x+e^2/f^2*x^2+a)^(3/2)+3/2*f*d^2*(b*x+e^2/f^2*x^2+a)^(1/2)*x+3/8*f^5/e^2*
a*ln((1/2*b+e^2/f^2*x)/(e^2/f^2)^(1/2)+(b*x+e^2/f^2*x^2+a)^(1/2))/(e^2/f^2)^(1/2)*b^2-1/2*f^5/e^2*(b*x+e^2/f^2
*x^2+a)^(3/2)*b+3/16*f^7/e^4*b^3*(b*x+e^2/f^2*x^2+a)^(1/2)+f^2*x^3*b*e+3/2*f^2*x^2*b*d+2*d*e^2*x^3+3/2*a*e*f^2
*x^2+3*a*d*f^2*x+f^3*(b*x+e^2/f^2*x^2+a)^(3/2)*x-3/4*d/e^3*f^5*b^2*(b*x+e^2/f^2*x^2+a)^(1/2)+3/4*d^2/e^2*f^3*(
b*x+e^2/f^2*x^2+a)^(1/2)*b+3/2*f*d^2*ln((1/2*b+e^2/f^2*x)/(e^2/f^2)^(1/2)+(b*x+e^2/f^2*x^2+a)^(1/2))/(e^2/f^2)
^(1/2)*a-3/2*d/e*f^3*b*(b*x+e^2/f^2*x^2+a)^(1/2)*x+3/8*d/e^3*f^5*b^3*ln((1/2*b+e^2/f^2*x)/(e^2/f^2)^(1/2)+(b*x
+e^2/f^2*x^2+a)^(1/2))/(e^2/f^2)^(1/2)-3/8*d^2/e^2*f^3*ln((1/2*b+e^2/f^2*x)/(e^2/f^2)^(1/2)+(b*x+e^2/f^2*x^2+a
)^(1/2))/(e^2/f^2)^(1/2)*b^2+3/8*f^5/e^2*b^2*(b*x+e^2/f^2*x^2+a)^(1/2)*x-3/32*f^7/e^4*b^4*ln((1/2*b+e^2/f^2*x)
/(e^2/f^2)^(1/2)+(b*x+e^2/f^2*x^2+a)^(1/2))/(e^2/f^2)^(1/2)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b^2*f^2-4*a*e^2>0)', see `assu
me?` for more details)Is b^2*f^2-4*a*e^2    positive, negative or zero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (d+e\,x+f\,\sqrt {a+b\,x+\frac {e^2\,x^2}{f^2}}\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^3,x)

[Out]

int((d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^3, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d + e x + f \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}}\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x+f*(a+b*x+e**2*x**2/f**2)**(1/2))**3,x)

[Out]

Integral((d + e*x + f*sqrt(a + b*x + e**2*x**2/f**2))**3, x)

________________________________________________________________________________________