3.5.73 \(\int (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}})^3 \, dx\) [473]

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

[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)))

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Rubi [A]
time = 0.27, 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 = {2141, 907} \begin {gather*} -\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} \end {gather*}

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 907

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 2141

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 \text {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 \text {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*}

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Mathematica [A]
time = 5.83, size = 490, normalized size = 1.62 \begin {gather*} \frac {1}{64} \left (32 x \left (2 d^3+3 d^2 e x+e x \left (3 a f^2+2 x \left (b f^2+e^2 x\right )\right )+d \left (6 a f^2+x \left (3 b f^2+4 e^2 x\right )\right )\right )+\frac {4 \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )} \left (3 b^3 f^7-2 b^2 e f^5 (6 d+e x)+4 b e^2 f^3 \left (3 d^2-2 a f^2+2 d e x+2 e^2 x^2\right )+8 e^3 f \left (2 a f^2 (2 d+e x)+e x \left (3 d^2+4 d e x+2 e^2 x^2\right )\right )\right )}{e^4}-\frac {3 \left (-b^2 f^4 \left (e+\sqrt {\frac {e^2}{f^2}} f\right ) \left (-2 d e+b f^2\right )^2+4 a e^2 f^2 \left (4 d^2 e^3-4 b d e f^2 \left (e+\sqrt {\frac {e^2}{f^2}} f\right )+b^2 f^4 \left (e+\sqrt {\frac {e^2}{f^2}} f\right )\right )\right ) \log \left (b f+2 e \sqrt {\frac {e^2}{f^2}} x-2 e \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )}{e^6}-\frac {48 a d^2 f \log \left (e \left (-b f+2 e \left (-\sqrt {\frac {e^2}{f^2}} x+\sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )\right )}{\sqrt {\frac {e^2}{f^2}}}+\frac {3 \left (e-\sqrt {\frac {e^2}{f^2}} f\right ) \left (4 a e^2-b^2 f^2\right ) \left (-2 d e f+b f^3\right )^2 \log \left (b f+2 e \left (-\sqrt {\frac {e^2}{f^2}} x+\sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )}{e^6}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(32*x*(2*d^3 + 3*d^2*e*x + e*x*(3*a*f^2 + 2*x*(b*f^2 + e^2*x)) + d*(6*a*f^2 + x*(3*b*f^2 + 4*e^2*x))) + (4*Sqr
t[a + x*(b + (e^2*x)/f^2)]*(3*b^3*f^7 - 2*b^2*e*f^5*(6*d + e*x) + 4*b*e^2*f^3*(3*d^2 - 2*a*f^2 + 2*d*e*x + 2*e
^2*x^2) + 8*e^3*f*(2*a*f^2*(2*d + e*x) + e*x*(3*d^2 + 4*d*e*x + 2*e^2*x^2))))/e^4 - (3*(-(b^2*f^4*(e + Sqrt[e^
2/f^2]*f)*(-2*d*e + b*f^2)^2) + 4*a*e^2*f^2*(4*d^2*e^3 - 4*b*d*e*f^2*(e + Sqrt[e^2/f^2]*f) + b^2*f^4*(e + Sqrt
[e^2/f^2]*f)))*Log[b*f + 2*e*Sqrt[e^2/f^2]*x - 2*e*Sqrt[a + x*(b + (e^2*x)/f^2)]])/e^6 - (48*a*d^2*f*Log[e*(-(
b*f) + 2*e*(-(Sqrt[e^2/f^2]*x) + Sqrt[a + x*(b + (e^2*x)/f^2)]))])/Sqrt[e^2/f^2] + (3*(e - Sqrt[e^2/f^2]*f)*(4
*a*e^2 - b^2*f^2)*(-2*d*e*f + b*f^3)^2*Log[b*f + 2*e*(-(Sqrt[e^2/f^2]*x) + Sqrt[a + x*(b + (e^2*x)/f^2)])])/e^
6)/64

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(808\) vs. \(2(283)=566\).
time = 0.36, size = 809, normalized size = 2.67

method	result	size
default	\(f^{3} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \left (a +b x +\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 \left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )}{16 e^{2}}\right )+3 f^{2} \left (\frac {e^{3} x^{4}}{4 f^{2}}+\frac {\left (\frac {d \,e^{2}}{f^{2}}+e b \right ) x^{3}}{3}+\frac {\left (a e +b d \right ) x^{2}}{2}+a d x \right )+3 f \left (e^{2} \left (\frac {x \left (a +b x +\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}} f^{2}}{4 e^{2}}-\frac {5 b \,f^{2} \left (\frac {\left (a +b x +\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}} f^{2}}{3 e^{2}}-\frac {b \,f^{2} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )}{2 e^{2}}\right )}{8 e^{2}}-\frac {a \,f^{2} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )}{4 e^{2}}\right )+2 d e \left (\frac {\left (a +b x +\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}} f^{2}}{3 e^{2}}-\frac {b \,f^{2} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )}{2 e^{2}}\right )+d^{2} \left (\frac {\left (b +\frac {2 e^{2} x}{f^{2}}\right ) f^{2} \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}{4 e^{2}}+\frac {\left (\frac {4 e^{2} a}{f^{2}}-b^{2}\right ) f^{2} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}{8 e^{2} \sqrt {\frac {e^{2}}{f^{2}}}}\right )\right )+\frac {\left (e x +d \right )^{4}}{4 e}\)	\(809\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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*%e^2*a>0)', see `ass
ume?` for mo

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Fricas [A]
time = 0.35, size = 326, normalized size = 1.08 \begin {gather*} \frac {1}{32} \, {\left (32 \, x^{4} e^{8} + 64 \, d x^{3} e^{7} + 16 \, {\left (2 \, b f^{2} x^{3} + 3 \, {\left (a f^{2} + d^{2}\right )} x^{2}\right )} e^{6} + 16 \, {\left (3 \, b d f^{2} x^{2} + 2 \, {\left (3 \, a d f^{2} + d^{3}\right )} x\right )} e^{5} + 3 \, {\left (b^{4} f^{8} - 4 \, b^{3} d f^{6} e + 16 \, a b d f^{4} e^{3} - 16 \, a d^{2} f^{2} e^{4} - 4 \, {\left (a b^{2} f^{6} - b^{2} d^{2} f^{4}\right )} e^{2}\right )} \log \left (-b f^{2} + 2 \, f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} e - 2 \, x e^{2}\right ) + 2 \, {\left (3 \, b^{3} f^{7} e - 12 \, b^{2} d f^{5} e^{2} + 16 \, f x^{3} e^{7} + 32 \, d f x^{2} e^{6} + 8 \, {\left (b f^{3} x^{2} + {\left (2 \, a f^{3} + 3 \, d^{2} f\right )} x\right )} e^{5} + 8 \, {\left (b d f^{3} x + 4 \, a d f^{3}\right )} e^{4} - 2 \, {\left (b^{2} f^{5} x + 4 \, a b f^{5} - 6 \, b d^{2} f^{3}\right )} e^{3}\right )} \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} e^{\left (-5\right )} \end {gather*}

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*x^4*e^8 + 64*d*x^3*e^7 + 16*(2*b*f^2*x^3 + 3*(a*f^2 + d^2)*x^2)*e^6 + 16*(3*b*d*f^2*x^2 + 2*(3*a*d*f^
2 + d^3)*x)*e^5 + 3*(b^4*f^8 - 4*b^3*d*f^6*e + 16*a*b*d*f^4*e^3 - 16*a*d^2*f^2*e^4 - 4*(a*b^2*f^6 - b^2*d^2*f^
4)*e^2)*log(-b*f^2 + 2*f*sqrt((b*f^2*x + a*f^2 + x^2*e^2)/f^2)*e - 2*x*e^2) + 2*(3*b^3*f^7*e - 12*b^2*d*f^5*e^
2 + 16*f*x^3*e^7 + 32*d*f*x^2*e^6 + 8*(b*f^3*x^2 + (2*a*f^3 + 3*d^2*f)*x)*e^5 + 8*(b*d*f^3*x + 4*a*d*f^3)*e^4
- 2*(b^2*f^5*x + 4*a*b*f^5 - 6*b*d^2*f^3)*e^3)*sqrt((b*f^2*x + a*f^2 + x^2*e^2)/f^2))*e^(-5)

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

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)

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Giac [A]
time = 3.78, size = 373, normalized size = 1.23 \begin {gather*} 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 )} \end {gather*}

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)

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

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)

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