3.5.83 \(\int \frac {1}{(d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}})^{3/2}} \, dx\) [483]

Optimal. Leaf size=269 \[ -\frac {4 \left (d^2 e-b d f^2+a e f^2\right )}{\left (2 d e-b f^2\right )^2 \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}-\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\left (2 d e-b f^2\right )^2 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac {3 f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\sqrt {2 d e-b f^2}}\right )}{\sqrt {2} \sqrt {e} \left (2 d e-b f^2\right )^{5/2}} \]

[Out]

3/2*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)^(5/2)*2^(1/2)/e^(1/2)-4*(a*e*f^2-b*d*f^2+d^2*e)/(-b*f^2+2*d*e)^2/(d+e*x+f*(a+b*x+e^2*x^2/f
^2)^(1/2))^(1/2)-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)/(-b*f^2+2*d*e)^2/(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 = 0.28, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2141, 911, 1273, 464, 214} \begin {gather*} -\frac {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}}{\left (2 d e-b f^2\right )^2 \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 ) \tanh ^{-1}\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} \sqrt {e} \left (2 d e-b f^2\right )^{5/2}}-\frac {4 \left (a e f^2-b d f^2+d^2 e\right )}{\left (2 d e-b f^2\right )^2 \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 214

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]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 911

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - (2*c*d - b*e)*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], 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] && IntegersQ[n,
p] && FractionQ[m]

Rule 1273

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] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*
p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(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] && ILtQ[m/2, 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 \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx &=2 \text {Subst}\left (\int \frac {d^2 e-(b d-a e) f^2-\left (2 d e-b f^2\right ) x+e x^2}{x^{3/2} \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 )\\ &=4 \text {Subst}\left (\int \frac {d^2 e-(b d-a e) f^2+\left (-2 d e+b f^2\right ) x^2+e x^4}{x^2 \left (-2 d e+b f^2+2 e x^2\right )^2} \, dx,x,\sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}\right )\\ &=-\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}}}}{\left (2 d e-b f^2\right )^2 \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 {\text {Subst}\left (\int \frac {8 e^2 \left (2 d e-b f^2\right ) \left (d^2 e-b d f^2+a e f^2\right )-2 e^2 \left (8 d^2 e^2-8 b d e f^2-4 a e^2 f^2+3 b^2 f^4\right ) x^2}{x^2 \left (-2 d e+b f^2+2 e x^2\right )} \, dx,x,\sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}\right )}{2 e^2 \left (2 d e-b f^2\right )^2}\\ &=-\frac {4 \left (d^2 e-b d f^2+a e f^2\right )}{\left (2 d e-b f^2\right )^2 \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}-\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}}}}{\left (2 d e-b f^2\right )^2 \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 {\left (3 f^2 \left (4 a e^2-b^2 f^2\right )\right ) \text {Subst}\left (\int \frac {1}{-2 d e+b f^2+2 e x^2} \, dx,x,\sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}\right )}{\left (2 d e-b f^2\right )^2}\\ &=-\frac {4 \left (d^2 e-b d f^2+a e f^2\right )}{\left (2 d e-b f^2\right )^2 \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}-\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}}}}{\left (2 d e-b f^2\right )^2 \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 (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\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 )}{\sqrt {2} \sqrt {e} \left (2 d e-b f^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 2.57, size = 395, normalized size = 1.47 \begin {gather*} \frac {b^2 f^4 \left (5 d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )-4 b e f^2 \left (d^2+a f^2-2 d \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )-4 e^2 \left (2 d^2 \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+a f^2 \left (d+3 e x+3 f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )}{\left (-2 d e+b f^2\right )^2 \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}} \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 {6 \sqrt {2} a e^{3/2} f^2 \tan ^{-1}\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 )}{\left (-2 d e+b f^2\right )^{5/2}}+\frac {3 b^2 f^4 \tan ^{-1}\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 )}{\sqrt {2} \sqrt {e} \left (-2 d e+b f^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*f^4*(5*d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) - 4*b*e*f^2*(d^2 + a*f^2 - 2*d*(e*x + f*Sqrt[a + x*(b +
(e^2*x)/f^2)])) - 4*e^2*(2*d^2*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + a*f^2*(d + 3*e*x + 3*f*Sqrt[a + x*(b
+ (e^2*x)/f^2)])))/((-2*d*e + b*f^2)^2*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]]*(b*f^2 + 2*e*(e*x + f*
Sqrt[a + x*(b + (e^2*x)/f^2)]))) - (6*Sqrt[2]*a*e^(3/2)*f^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*d*e + b*f^2)^(5/2) + (3*b^2*f^4*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]])/(Sqrt[2]*Sqrt[e]*(-2*d*e + b*f^2)^(5/2))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (d +e x +f \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x*e + sqrt(b*x + a + x^2*e^2/f^2)*f + d)^(-3/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 593 vs. \(2 (242) = 484\).
time = 0.65, size = 1312, normalized size = 4.88 \begin {gather*} \left [\frac {3 \, {\left (b^{3} f^{6} x + a b^{2} f^{6} - 2 \, b^{2} d f^{4} x e - b^{2} d^{2} f^{4} + 8 \, a d f^{2} x e^{3} - 4 \, {\left (a b f^{4} x + a^{2} f^{4} - a d^{2} f^{2}\right )} e^{2}\right )} \sqrt {-2 \, b f^{2} e + 4 \, d e^{2}} \log \left (b^{2} f^{4} - 4 \, b d f^{2} e - 8 \, d x e^{3} + 4 \, {\left (b f^{2} x + a f^{2}\right )} e^{2} + 2 \, {\left (2 \, \sqrt {-2 \, b f^{2} e + 4 \, d e^{2}} f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} e - \sqrt {-2 \, b f^{2} e + 4 \, d e^{2}} {\left (b f^{2} + 2 \, x e^{2}\right )}\right )} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d} - 4 \, {\left (b f^{3} e - 2 \, d f e^{2}\right )} \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}}\right ) + 4 \, {\left (8 \, d^{2} x^{2} e^{5} - 4 \, {\left (2 \, b d f^{2} x^{2} + {\left (3 \, a d f^{2} + d^{3}\right )} x\right )} e^{4} + 2 \, {\left (b^{2} f^{4} x^{2} - 4 \, a d^{2} f^{2} - 4 \, d^{4} + {\left (3 \, a b f^{4} + 7 \, b d^{2} f^{2}\right )} x\right )} e^{3} - 2 \, {\left (4 \, b^{2} d f^{4} x - a b d f^{4} - 7 \, b d^{3} f^{2}\right )} e^{2} + {\left (b^{3} f^{6} x + a b^{2} f^{6} - 5 \, b^{2} d^{2} f^{4}\right )} e + 2 \, {\left (2 \, b^{2} d f^{5} e - 4 \, d^{2} f x e^{4} + 2 \, {\left (2 \, b d f^{3} x + 3 \, a d f^{3} + d^{3} f\right )} e^{3} - {\left (b^{2} f^{5} x + 3 \, a b f^{5} + 5 \, b d^{2} f^{3}\right )} e^{2}\right )} \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d}}{4 \, {\left (16 \, d^{4} x e^{5} - 8 \, {\left (4 \, b d^{3} f^{2} x + a d^{3} f^{2} - d^{5}\right )} e^{4} + 12 \, {\left (2 \, b^{2} d^{2} f^{4} x + a b d^{2} f^{4} - b d^{4} f^{2}\right )} e^{3} - 2 \, {\left (4 \, b^{3} d f^{6} x + 3 \, a b^{2} d f^{6} - 3 \, b^{2} d^{3} f^{4}\right )} e^{2} + {\left (b^{4} f^{8} x + a b^{3} f^{8} - b^{3} d^{2} f^{6}\right )} e\right )}}, -\frac {3 \, {\left (b^{3} f^{6} x + a b^{2} f^{6} - 2 \, b^{2} d f^{4} x e - b^{2} d^{2} f^{4} + 8 \, a d f^{2} x e^{3} - 4 \, {\left (a b f^{4} x + a^{2} f^{4} - a d^{2} f^{2}\right )} e^{2}\right )} \sqrt {2 \, b f^{2} e - 4 \, d e^{2}} \arctan \left (-\frac {\sqrt {2 \, b f^{2} e - 4 \, d e^{2}} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d}}{b f^{2} - 2 \, d e}\right ) - 2 \, {\left (8 \, d^{2} x^{2} e^{5} - 4 \, {\left (2 \, b d f^{2} x^{2} + {\left (3 \, a d f^{2} + d^{3}\right )} x\right )} e^{4} + 2 \, {\left (b^{2} f^{4} x^{2} - 4 \, a d^{2} f^{2} - 4 \, d^{4} + {\left (3 \, a b f^{4} + 7 \, b d^{2} f^{2}\right )} x\right )} e^{3} - 2 \, {\left (4 \, b^{2} d f^{4} x - a b d f^{4} - 7 \, b d^{3} f^{2}\right )} e^{2} + {\left (b^{3} f^{6} x + a b^{2} f^{6} - 5 \, b^{2} d^{2} f^{4}\right )} e + 2 \, {\left (2 \, b^{2} d f^{5} e - 4 \, d^{2} f x e^{4} + 2 \, {\left (2 \, b d f^{3} x + 3 \, a d f^{3} + d^{3} f\right )} e^{3} - {\left (b^{2} f^{5} x + 3 \, a b f^{5} + 5 \, b d^{2} f^{3}\right )} e^{2}\right )} \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d}}{2 \, {\left (16 \, d^{4} x e^{5} - 8 \, {\left (4 \, b d^{3} f^{2} x + a d^{3} f^{2} - d^{5}\right )} e^{4} + 12 \, {\left (2 \, b^{2} d^{2} f^{4} x + a b d^{2} f^{4} - b d^{4} f^{2}\right )} e^{3} - 2 \, {\left (4 \, b^{3} d f^{6} x + 3 \, a b^{2} d f^{6} - 3 \, b^{2} d^{3} f^{4}\right )} e^{2} + {\left (b^{4} f^{8} x + a b^{3} f^{8} - b^{3} d^{2} f^{6}\right )} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*(b^3*f^6*x + a*b^2*f^6 - 2*b^2*d*f^4*x*e - b^2*d^2*f^4 + 8*a*d*f^2*x*e^3 - 4*(a*b*f^4*x + a^2*f^4 - a*
d^2*f^2)*e^2)*sqrt(-2*b*f^2*e + 4*d*e^2)*log(b^2*f^4 - 4*b*d*f^2*e - 8*d*x*e^3 + 4*(b*f^2*x + a*f^2)*e^2 + 2*(
2*sqrt(-2*b*f^2*e + 4*d*e^2)*f*sqrt((b*f^2*x + a*f^2 + x^2*e^2)/f^2)*e - sqrt(-2*b*f^2*e + 4*d*e^2)*(b*f^2 + 2
*x*e^2))*sqrt(x*e + f*sqrt((b*f^2*x + a*f^2 + x^2*e^2)/f^2) + d) - 4*(b*f^3*e - 2*d*f*e^2)*sqrt((b*f^2*x + a*f
^2 + x^2*e^2)/f^2)) + 4*(8*d^2*x^2*e^5 - 4*(2*b*d*f^2*x^2 + (3*a*d*f^2 + d^3)*x)*e^4 + 2*(b^2*f^4*x^2 - 4*a*d^
2*f^2 - 4*d^4 + (3*a*b*f^4 + 7*b*d^2*f^2)*x)*e^3 - 2*(4*b^2*d*f^4*x - a*b*d*f^4 - 7*b*d^3*f^2)*e^2 + (b^3*f^6*
x + a*b^2*f^6 - 5*b^2*d^2*f^4)*e + 2*(2*b^2*d*f^5*e - 4*d^2*f*x*e^4 + 2*(2*b*d*f^3*x + 3*a*d*f^3 + d^3*f)*e^3
- (b^2*f^5*x + 3*a*b*f^5 + 5*b*d^2*f^3)*e^2)*sqrt((b*f^2*x + a*f^2 + x^2*e^2)/f^2))*sqrt(x*e + f*sqrt((b*f^2*x
+ a*f^2 + x^2*e^2)/f^2) + d))/(16*d^4*x*e^5 - 8*(4*b*d^3*f^2*x + a*d^3*f^2 - d^5)*e^4 + 12*(2*b^2*d^2*f^4*x +
a*b*d^2*f^4 - b*d^4*f^2)*e^3 - 2*(4*b^3*d*f^6*x + 3*a*b^2*d*f^6 - 3*b^2*d^3*f^4)*e^2 + (b^4*f^8*x + a*b^3*f^8
- b^3*d^2*f^6)*e), -1/2*(3*(b^3*f^6*x + a*b^2*f^6 - 2*b^2*d*f^4*x*e - b^2*d^2*f^4 + 8*a*d*f^2*x*e^3 - 4*(a*b*
f^4*x + a^2*f^4 - a*d^2*f^2)*e^2)*sqrt(2*b*f^2*e - 4*d*e^2)*arctan(-sqrt(2*b*f^2*e - 4*d*e^2)*sqrt(x*e + f*sqr
t((b*f^2*x + a*f^2 + x^2*e^2)/f^2) + d)/(b*f^2 - 2*d*e)) - 2*(8*d^2*x^2*e^5 - 4*(2*b*d*f^2*x^2 + (3*a*d*f^2 +
d^3)*x)*e^4 + 2*(b^2*f^4*x^2 - 4*a*d^2*f^2 - 4*d^4 + (3*a*b*f^4 + 7*b*d^2*f^2)*x)*e^3 - 2*(4*b^2*d*f^4*x - a*b
*d*f^4 - 7*b*d^3*f^2)*e^2 + (b^3*f^6*x + a*b^2*f^6 - 5*b^2*d^2*f^4)*e + 2*(2*b^2*d*f^5*e - 4*d^2*f*x*e^4 + 2*(
2*b*d*f^3*x + 3*a*d*f^3 + d^3*f)*e^3 - (b^2*f^5*x + 3*a*b*f^5 + 5*b*d^2*f^3)*e^2)*sqrt((b*f^2*x + a*f^2 + x^2*
e^2)/f^2))*sqrt(x*e + f*sqrt((b*f^2*x + a*f^2 + x^2*e^2)/f^2) + d))/(16*d^4*x*e^5 - 8*(4*b*d^3*f^2*x + a*d^3*f
^2 - d^5)*e^4 + 12*(2*b^2*d^2*f^4*x + a*b*d^2*f^4 - b*d^4*f^2)*e^3 - 2*(4*b^3*d*f^6*x + 3*a*b^2*d*f^6 - 3*b^2*
d^3*f^4)*e^2 + (b^4*f^8*x + a*b^3*f^8 - b^3*d^2*f^6)*e)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d + e x + f \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x*e + sqrt(b*x + a + x^2*e^2/f^2)*f + d)^(-3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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