3.5.54 \(\int (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}})^3 \, dx\) [454]

Optimal. Leaf size=175 \[ -\frac {a d^3 f^2}{2 e \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}+\frac {a d f^2 \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{e}+\frac {a f^2 \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^2}{4 e}+\frac {\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^4}{8 e}+\frac {3 a d^2 f^2 \log \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{2 e} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2142, 907} \begin {gather*} -\frac {a d^3 f^2}{2 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}+\frac {3 a d^2 f^2 \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}{2 e}+\frac {\left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^4}{8 e}+\frac {a f^2 \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^2}{4 e}+\frac {a d f^2 \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a*d^3*f^2)/(e*(e*x + f*Sqrt[a + (e^2*x^2)/f^2])) + (a*d*f^2*(e*x + f*Sqrt[a + (e^2*x^2)/f^2]))/e + (a*f^
2*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^2)/(4*e) + (d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^4/(8*e) + (3*a*d^2*f^
2*Log[e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(2*e)

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 2142

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[(g + h*x^n)^p*((d^2 + a*f^2 - 2*d*x + x^2)/(d - x)^2), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, 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+\frac {e^2 x^2}{f^2}}\right )^3 \, dx &=\frac {\text {Subst}\left (\int \frac {x^3 \left (d^2+a f^2-2 d x+x^2\right )}{(d-x)^2} \, dx,x,d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{2 e}\\ &=\frac {\text {Subst}\left (\int \left (2 a d f^2+\frac {a d^3 f^2}{(d-x)^2}-\frac {3 a d^2 f^2}{d-x}+a f^2 x+x^3\right ) \, dx,x,d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{2 e}\\ &=-\frac {a d^3 f^2}{2 e \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}+\frac {a d f^2 \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{e}+\frac {a f^2 \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^2}{4 e}+\frac {\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^4}{8 e}+\frac {3 a d^2 f^2 \log \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{2 e}\\ \end {align*}

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Mathematica [A]
time = 0.78, size = 159, normalized size = 0.91 \begin {gather*} \frac {1}{2} \left (x \left (2 d^3+6 a d f^2+3 d^2 e x+3 a e f^2 x+4 d e^2 x^2+2 e^3 x^3\right )+\frac {\sqrt {a+\frac {e^2 x^2}{f^2}} \left (2 a f^3 (2 d+e x)+e f x \left (3 d^2+4 d e x+2 e^2 x^2\right )\right )}{e}-\frac {3 a d^2 f \log \left (-\sqrt {\frac {e^2}{f^2}} x+\sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{\sqrt {\frac {e^2}{f^2}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(2*d^3 + 6*a*d*f^2 + 3*d^2*e*x + 3*a*e*f^2*x + 4*d*e^2*x^2 + 2*e^3*x^3) + (Sqrt[a + (e^2*x^2)/f^2]*(2*a*f^3
*(2*d + e*x) + e*f*x*(3*d^2 + 4*d*e*x + 2*e^2*x^2)))/e - (3*a*d^2*f*Log[-(Sqrt[e^2/f^2]*x) + Sqrt[a + (e^2*x^2
)/f^2]])/Sqrt[e^2/f^2])/2

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(340\) vs. \(2(157)=314\).
time = 0.21, size = 341, normalized size = 1.95

method	result	size
default	\(f^{3} \left (\frac {x \left (a +\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}}{2}+\frac {a \ln \left (\frac {e^{2} x}{f^{2} \sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}\right )}{2 \sqrt {\frac {e^{2}}{f^{2}}}}\right )}{4}\right )+3 f^{2} \left (\frac {e^{3} x^{4}}{4 f^{2}}+\frac {d \,e^{2} x^{3}}{3 f^{2}}+\frac {a e \,x^{2}}{2}+a d x \right )+3 f \left (e^{2} \left (\frac {x \left (a +\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}} f^{2}}{4 e^{2}}-\frac {a \,f^{2} \left (\frac {x \sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}}{2}+\frac {a \ln \left (\frac {e^{2} x}{f^{2} \sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}\right )}{2 \sqrt {\frac {e^{2}}{f^{2}}}}\right )}{4 e^{2}}\right )+\frac {2 d \,f^{2} \left (\frac {e^{2} x^{2}+a \,f^{2}}{f^{2}}\right )^{\frac {3}{2}}}{3 e}+d^{2} \left (\frac {x \sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}}{2}+\frac {a \ln \left (\frac {e^{2} x}{f^{2} \sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}\right )}{2 \sqrt {\frac {e^{2}}{f^{2}}}}\right )\right )+\frac {\left (e x +d \right )^{4}}{4 e}\)	\(341\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f^3*(1/4*x*(a+1/f^2*e^2*x^2)^(3/2)+3/4*a*(1/2*x*(a+1/f^2*e^2*x^2)^(1/2)+1/2*a*ln(1/f^2*e^2*x/(1/f^2*e^2)^(1/2)
+(a+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*x^3+1/2*a*e*x^2+a*d*x)+3*f*
(e^2*(1/4*x*(a+1/f^2*e^2*x^2)^(3/2)*f^2/e^2-1/4*a*f^2/e^2*(1/2*x*(a+1/f^2*e^2*x^2)^(1/2)+1/2*a*ln(1/f^2*e^2*x/
(1/f^2*e^2)^(1/2)+(a+1/f^2*e^2*x^2)^(1/2))/(1/f^2*e^2)^(1/2)))+2/3*d/e*f^2*((e^2*x^2+a*f^2)/f^2)^(3/2)+d^2*(1/
2*x*(a+1/f^2*e^2*x^2)^(1/2)+1/2*a*ln(1/f^2*e^2*x/(1/f^2*e^2)^(1/2)+(a+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 [A]
time = 0.28, size = 252, normalized size = 1.44 \begin {gather*} \frac {3}{4} \, {\left (a + \frac {x^{2} e^{2}}{f^{2}}\right )}^{2} f^{4} e^{\left (-1\right )} + \frac {1}{4} \, x^{4} e^{3} + \frac {1}{8} \, {\left (3 \, a^{2} f \operatorname {arsinh}\left (\frac {x e}{\sqrt {a} f}\right ) e^{\left (-1\right )} + 2 \, {\left (a + \frac {x^{2} e^{2}}{f^{2}}\right )}^{\frac {3}{2}} x + 3 \, \sqrt {a + \frac {x^{2} e^{2}}{f^{2}}} a x\right )} f^{3} + d^{3} x + \frac {3}{2} \, {\left (x^{2} e + {\left (a f \operatorname {arsinh}\left (\frac {x e}{\sqrt {a} f}\right ) e^{\left (-1\right )} + \sqrt {a + \frac {x^{2} e^{2}}{f^{2}}} x\right )} f\right )} d^{2} - \frac {3}{8} \, {\left (a^{2} f^{3} \operatorname {arsinh}\left (\frac {x e}{\sqrt {a} f}\right ) e^{\left (-3\right )} - 2 \, {\left (a + \frac {x^{2} e^{2}}{f^{2}}\right )}^{\frac {3}{2}} f^{2} x e^{\left (-2\right )} + \sqrt {a + \frac {x^{2} e^{2}}{f^{2}}} a f^{2} x e^{\left (-2\right )}\right )} f e^{2} + {\left (2 \, {\left (a + \frac {x^{2} e^{2}}{f^{2}}\right )}^{\frac {3}{2}} f^{3} e^{\left (-1\right )} + x^{3} e^{2} + {\left (3 \, a x + \frac {x^{3} e^{2}}{f^{2}}\right )} f^{2}\right )} d \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*(a + x^2*e^2/f^2)^2*f^4*e^(-1) + 1/4*x^4*e^3 + 1/8*(3*a^2*f*arcsinh(x*e/(sqrt(a)*f))*e^(-1) + 2*(a + x^2*e
^2/f^2)^(3/2)*x + 3*sqrt(a + x^2*e^2/f^2)*a*x)*f^3 + d^3*x + 3/2*(x^2*e + (a*f*arcsinh(x*e/(sqrt(a)*f))*e^(-1)
+ sqrt(a + x^2*e^2/f^2)*x)*f)*d^2 - 3/8*(a^2*f^3*arcsinh(x*e/(sqrt(a)*f))*e^(-3) - 2*(a + x^2*e^2/f^2)^(3/2)*
f^2*x*e^(-2) + sqrt(a + x^2*e^2/f^2)*a*f^2*x*e^(-2))*f*e^2 + (2*(a + x^2*e^2/f^2)^(3/2)*f^3*e^(-1) + x^3*e^2 +
(3*a*x + x^3*e^2/f^2)*f^2)*d

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Fricas [A]
time = 0.36, size = 150, normalized size = 0.86 \begin {gather*} -\frac {1}{2} \, {\left (3 \, a d^{2} f^{2} \log \left (-x e + f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}}\right ) - 2 \, x^{4} e^{4} - 4 \, d x^{3} e^{3} - 3 \, {\left (a f^{2} + d^{2}\right )} x^{2} e^{2} - 2 \, {\left (3 \, a d f^{2} + d^{3}\right )} x e - {\left (4 \, a d f^{3} + 2 \, f x^{3} e^{3} + 4 \, d f x^{2} e^{2} + {\left (2 \, a f^{3} + 3 \, d^{2} f\right )} x e\right )} \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*a*d^2*f^2*log(-x*e + f*sqrt((a*f^2 + x^2*e^2)/f^2)) - 2*x^4*e^4 - 4*d*x^3*e^3 - 3*(a*f^2 + d^2)*x^2*e^
2 - 2*(3*a*d*f^2 + d^3)*x*e - (4*a*d*f^3 + 2*f*x^3*e^3 + 4*d*f*x^2*e^2 + (2*a*f^3 + 3*d^2*f)*x*e)*sqrt((a*f^2
+ x^2*e^2)/f^2))*e^(-1)

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Sympy [A]
time = 5.47, size = 279, normalized size = 1.59 \begin {gather*} \frac {a^{\frac {3}{2}} f^{3} x \sqrt {1 + \frac {e^{2} x^{2}}{a f^{2}}}}{2} + \frac {a^{\frac {3}{2}} f^{3} x}{2 \sqrt {1 + \frac {e^{2} x^{2}}{a f^{2}}}} + \frac {3 \sqrt {a} d^{2} f x \sqrt {1 + \frac {e^{2} x^{2}}{a f^{2}}}}{2} + \frac {3 \sqrt {a} e^{2} f x^{3}}{2 \sqrt {1 + \frac {e^{2} x^{2}}{a f^{2}}}} + \frac {3 a d^{2} f^{2} \operatorname {asinh}{\left (\frac {e x}{\sqrt {a} f} \right )}}{2 e} + 3 a d f^{2} x + \frac {3 a e f^{2} x^{2}}{2} + d^{3} x + \frac {3 d^{2} e x^{2}}{2} + 2 d e^{2} x^{3} + 6 d e f \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: e^{2} = 0 \\\frac {f^{2} \left (a + \frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}}}{3 e^{2}} & \text {otherwise} \end {cases}\right ) + e^{3} x^{4} + \frac {e^{4} x^{5}}{\sqrt {a} f \sqrt {1 + \frac {e^{2} x^{2}}{a f^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*f**3*x*sqrt(1 + e**2*x**2/(a*f**2))/2 + a**(3/2)*f**3*x/(2*sqrt(1 + e**2*x**2/(a*f**2))) + 3*sqrt(a)*
d**2*f*x*sqrt(1 + e**2*x**2/(a*f**2))/2 + 3*sqrt(a)*e**2*f*x**3/(2*sqrt(1 + e**2*x**2/(a*f**2))) + 3*a*d**2*f*
*2*asinh(e*x/(sqrt(a)*f))/(2*e) + 3*a*d*f**2*x + 3*a*e*f**2*x**2/2 + d**3*x + 3*d**2*e*x**2/2 + 2*d*e**2*x**3
+ 6*d*e*f*Piecewise((sqrt(a)*x**2/2, Eq(e**2, 0)), (f**2*(a + e**2*x**2/f**2)**(3/2)/(3*e**2), True)) + e**3*x
**4 + e**4*x**5/(sqrt(a)*f*sqrt(1 + e**2*x**2/(a*f**2)))

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Giac [A]
time = 3.64, size = 163, normalized size = 0.93 \begin {gather*} -\frac {3}{2} \, a d^{2} f {\left | f \right |} e^{\left (-1\right )} \log \left ({\left | -x e + \sqrt {a f^{2} + x^{2} e^{2}} \right |}\right ) + \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 {1}{2} \, {\left (4 \, a d f {\left | f \right |} e^{\left (-1\right )} + {\left (2 \, {\left (\frac {x {\left | f \right |} e^{2}}{f} + \frac {2 \, d {\left | f \right |} e}{f}\right )} x + \frac {{\left (2 \, a f^{4} {\left | f \right |} e^{4} + 3 \, d^{2} f^{2} {\left | f \right |} e^{4}\right )} e^{\left (-4\right )}}{f^{3}}\right )} x\right )} \sqrt {a f^{2} + x^{2} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*a*d^2*f*abs(f)*e^(-1)*log(abs(-x*e + sqrt(a*f^2 + x^2*e^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 + 1/2*(4*a*d*f*abs(f)*e^(-1) + (2*(x*abs(f)*e^2/f + 2*d*abs(f)*e/f)*x + (2
*a*f^4*abs(f)*e^4 + 3*d^2*f^2*abs(f)*e^4)*e^(-4)/f^3)*x)*sqrt(a*f^2 + x^2*e^2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d+e\,x+f\,\sqrt {a+\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 + (e^2*x^2)/f^2)^(1/2))^3,x)

[Out]

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

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