3.5.84 \(\int \frac {1}{(d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}})^{5/2}} \, dx\) [484]

Optimal. Leaf size=335 \[ -\frac {4 \left (d^2 e-b d f^2+a e f^2\right )}{3 \left (2 d e-b f^2\right )^2 \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}}-\frac {4 f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^3 \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}-\frac {2 e f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\left (2 d e-b f^2\right )^3 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac {5 \sqrt {2} \sqrt {e} f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\sqrt {2 d e-b f^2}}\right )}{\left (2 d e-b f^2\right )^{7/2}} \]

[Out]

5*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
))*2^(1/2)*e^(1/2)/(-b*f^2+2*d*e)^(7/2)-4/3*(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))^(3/2)-4*f^2*(-b^2*f^2+4*a*e^2)/(-b*f^2+2*d*e)^3/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2)-2*e*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)^3/(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 = 0.37, antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2141, 911, 1273, 1275, 214} \begin {gather*} -\frac {4 f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^3 \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}-\frac {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}}{\left (2 d e-b f^2\right )^3 \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 {5 \sqrt {2} \sqrt {e} 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 )}{\left (2 d e-b f^2\right )^{7/2}}-\frac {4 \left (a e f^2-b d f^2+d^2 e\right )}{3 \left (2 d e-b f^2\right )^2 \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(-5/2),x]

[Out]

(-4*(d^2*e - b*d*f^2 + a*e*f^2))/(3*(2*d*e - b*f^2)^2*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(3/2)) - (4*
f^2*(4*a*e^2 - b^2*f^2))/((2*d*e - b*f^2)^3*Sqrt[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]])/((2*d*e - b*f^2)^3*(b*f^2 + 2*e*(e*x + f*Sqrt[a
+ (x*(b*f^2 + e^2*x))/f^2]))) + (5*Sqrt[2]*Sqrt[e]*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]])/(2*d*e - b*f^2)^(7/2)

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 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 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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 )^{5/2}} \, 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^{5/2} \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^4 \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 {2 e 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 )^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 {\text {Subst}\left (\int \frac {8 e^2 \left (2 d e-b f^2\right )^2 \left (d^2 e-b d f^2+a e f^2\right )-8 e^2 \left (2 d e-b f^2\right ) \left (2 d^2 e^2-2 b d e f^2-2 a e^2 f^2+b^2 f^4\right ) x^2+4 e^3 f^2 \left (4 a e^2-b^2 f^2\right ) x^4}{x^4 \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 )^3}\\ &=-\frac {2 e 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 )^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 {\text {Subst}\left (\int \left (-\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 )}{x^4}-\frac {8 \left (4 a e^4 f^2-b^2 e^2 f^4\right )}{x^2}-\frac {20 \left (4 a e^5 f^2-b^2 e^3 f^4\right )}{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 )^3}\\ &=-\frac {4 \left (d^2 e-b d f^2+a e f^2\right )}{3 \left (2 d e-b f^2\right )^2 \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}}-\frac {4 f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^3 \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}-\frac {2 e 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 )^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 {\left (10 e 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 )^3}\\ &=-\frac {4 \left (d^2 e-b d f^2+a e f^2\right )}{3 \left (2 d e-b f^2\right )^2 \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}}-\frac {4 f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^3 \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}-\frac {2 e 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 )^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 {5 \sqrt {2} \sqrt {e} 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 )}{\left (2 d e-b f^2\right )^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 3.64, size = 557, normalized size = 1.66 \begin {gather*} \frac {2 b^3 f^6 \left (4 d+21 e x+6 f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+2 b^2 e f^4 \left (9 d^2+17 a f^2+14 d \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+30 e x \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )-8 b e^2 f^2 \left (d^3+7 a d f^2-3 d^2 \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+5 a f^2 \left (4 e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )-8 e^3 \left (15 a^2 f^4+2 d^3 \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+a f^2 \left (3 d^2+20 d \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+30 e x \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )\right )}{3 \left (2 d e-b f^2\right )^3 \left (d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )^{3/2} \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 {20 \sqrt {2} a e^{5/2} 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 )^{7/2}}-\frac {5 \sqrt {2} b^2 \sqrt {e} 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 )}{\left (-2 d e+b f^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(-5/2),x]

[Out]

(2*b^3*f^6*(4*d + 21*e*x + 6*f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + 2*b^2*e*f^4*(9*d^2 + 17*a*f^2 + 14*d*(e*x + f*
Sqrt[a + x*(b + (e^2*x)/f^2)]) + 30*e*x*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])) - 8*b*e^2*f^2*(d^3 + 7*a*d*f^
2 - 3*d^2*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + 5*a*f^2*(4*e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])) - 8*e^3
*(15*a^2*f^4 + 2*d^3*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + a*f^2*(3*d^2 + 20*d*(e*x + f*Sqrt[a + x*(b + (e
^2*x)/f^2)]) + 30*e*x*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]))))/(3*(2*d*e - b*f^2)^3*(d + e*x + f*Sqrt[a + x*
(b + (e^2*x)/f^2)])^(3/2)*(b*f^2 + 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]))) + (20*Sqrt[2]*a*e^(5/2)*f^2*A
rcTan[(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
)^(7/2) - (5*Sqrt[2]*b^2*Sqrt[e]*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]])/(-2*d*e + b*f^2)^(7/2)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (d +e x +f \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {5}{2}}}\, dx\]

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))^(5/2),x)

[Out]

int(1/(d+e*x+f*(a+b*x+1/f^2*e^2*x^2)^(1/2))^(5/2),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))^(5/2),x, algorithm="maxima")

[Out]

integrate((x*e + sqrt(b*x + a + x^2*e^2/f^2)*f + d)^(-5/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1227 vs. \(2 (302) = 604\).
time = 1.49, size = 2523, normalized size = 7.53 \begin {gather*} \text {Too large to display} \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))^(5/2),x, algorithm="fricas")

[Out]

[1/6*(15*sqrt(2)*(b^4*f^8*x^2 + a^2*b^2*f^8 - 2*a*b^2*d^2*f^6 + b^2*d^4*f^4 - 16*a*d^2*f^2*x^2*e^4 + 2*(a*b^3*
f^8 - b^3*d^2*f^6)*x + 16*(a*b*d*f^4*x^2 + (a^2*d*f^4 - a*d^3*f^2)*x)*e^3 - 4*(a^3*f^6 - 2*a^2*d^2*f^4 + a*d^4
*f^2 + (a*b^2*f^6 - b^2*d^2*f^4)*x^2 + 2*(a^2*b*f^6 - a*b*d^2*f^4)*x)*e^2 - 4*(b^3*d*f^6*x^2 + (a*b^2*d*f^6 -
b^2*d^3*f^4)*x)*e)*sqrt(-e/(b*f^2 - 2*d*e))*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^3*e - 2*d*f*e^2)*sqrt(-e/(b*f^2 - 2*d*e))*sqrt((b*f^2*x + a*f^2 + x^2*e^2)/f^2) - sqrt(2)*(b
^2*f^4 + 2*b*f^2*x*e^2 - 2*b*d*f^2*e - 4*d*x*e^3)*sqrt(-e/(b*f^2 - 2*d*e)))*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*b^3*d*f^6*x + 8*
a*b^2*d*f^6 - 4*b^2*d^3*f^4 - 24*d^2*x^3*e^5 + 8*(3*b*d*f^2*x^3 + 2*(5*a*d*f^2 - d^3)*x^2)*e^4 - 2*(3*b^2*f^4*
x^3 + 2*(10*a*b*f^4 + 3*b*d^2*f^2)*x^2 + (15*a^2*f^4 - 46*a*d^2*f^2 - d^4)*x)*e^3 + 4*(b^2*d*f^4*x^2 - 5*a^2*d
*f^4 + 8*a*d^3*f^2 + d^5 - (3*a*b*d*f^4 + 5*b*d^3*f^2)*x)*e^2 + (3*b^3*f^6*x^2 - 5*a^2*b*f^6 - 2*a*b*d^2*f^4 -
9*b*d^4*f^2 - 2*(a*b^2*f^6 + 7*b^2*d^2*f^4)*x)*e - 2*(3*b^3*f^7*x + 3*a*b^2*f^7 - b^2*d^2*f^5 - 12*d^2*f*x^2*
e^4 + 4*(3*b*d*f^3*x^2 + 2*(5*a*d*f^3 - d^3*f)*x)*e^3 - (3*b^2*f^5*x^2 + 20*a*b*f^5*x + 15*a^2*f^5 - 22*a*d^2*
f^3 - d^4*f)*e^2 - 2*(2*b^2*d*f^5*x + a*b*d*f^5 + 3*b*d^3*f^3)*e)*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))/(b^5*f^10*x^2 + a^2*b^3*f^10 - 2*a*b^3*d^2*f^8 + b^3*d^4*f
^6 - 32*d^5*x^2*e^5 + 2*(a*b^4*f^10 - b^4*d^2*f^8)*x + 16*(5*b*d^4*f^2*x^2 + 2*(a*d^4*f^2 - d^6)*x)*e^4 - 8*(1
0*b^2*d^3*f^4*x^2 + a^2*d^3*f^4 - 2*a*d^5*f^2 + d^7 + 8*(a*b*d^3*f^4 - b*d^5*f^2)*x)*e^3 + 4*(10*b^3*d^2*f^6*x
^2 + 3*a^2*b*d^2*f^6 - 6*a*b*d^4*f^4 + 3*b*d^6*f^2 + 12*(a*b^2*d^2*f^6 - b^2*d^4*f^4)*x)*e^2 - 2*(5*b^4*d*f^8*
x^2 + 3*a^2*b^2*d*f^8 - 6*a*b^2*d^3*f^6 + 3*b^2*d^5*f^4 + 8*(a*b^3*d*f^8 - b^3*d^3*f^6)*x)*e), 1/3*(15*sqrt(2)
*(b^4*f^8*x^2 + a^2*b^2*f^8 - 2*a*b^2*d^2*f^6 + b^2*d^4*f^4 - 16*a*d^2*f^2*x^2*e^4 + 2*(a*b^3*f^8 - b^3*d^2*f^
6)*x + 16*(a*b*d*f^4*x^2 + (a^2*d*f^4 - a*d^3*f^2)*x)*e^3 - 4*(a^3*f^6 - 2*a^2*d^2*f^4 + a*d^4*f^2 + (a*b^2*f^
6 - b^2*d^2*f^4)*x^2 + 2*(a^2*b*f^6 - a*b*d^2*f^4)*x)*e^2 - 4*(b^3*d*f^6*x^2 + (a*b^2*d*f^6 - b^2*d^3*f^4)*x)*
e)*arctan(-1/2*(sqrt(2)*(b*f^3 - 2*d*f*e)*sqrt((b*f^2*x + a*f^2 + x^2*e^2)/f^2)*e^(1/2)/sqrt(b*f^2 - 2*d*e) -
sqrt(2)*(b*d*f^2 - 2*d*x*e^2 + (b*f^2*x - 2*d^2)*e)*e^(1/2)/sqrt(b*f^2 - 2*d*e))*sqrt(x*e + f*sqrt((b*f^2*x +
a*f^2 + x^2*e^2)/f^2) + d)/(2*d*x*e^2 - (b*f^2*x + a*f^2 - d^2)*e))*e^(1/2)/sqrt(b*f^2 - 2*d*e) + 2*(8*b^3*d*f
^6*x + 8*a*b^2*d*f^6 - 4*b^2*d^3*f^4 - 24*d^2*x^3*e^5 + 8*(3*b*d*f^2*x^3 + 2*(5*a*d*f^2 - d^3)*x^2)*e^4 - 2*(3
*b^2*f^4*x^3 + 2*(10*a*b*f^4 + 3*b*d^2*f^2)*x^2 + (15*a^2*f^4 - 46*a*d^2*f^2 - d^4)*x)*e^3 + 4*(b^2*d*f^4*x^2
- 5*a^2*d*f^4 + 8*a*d^3*f^2 + d^5 - (3*a*b*d*f^4 + 5*b*d^3*f^2)*x)*e^2 + (3*b^3*f^6*x^2 - 5*a^2*b*f^6 - 2*a*b*
d^2*f^4 - 9*b*d^4*f^2 - 2*(a*b^2*f^6 + 7*b^2*d^2*f^4)*x)*e - 2*(3*b^3*f^7*x + 3*a*b^2*f^7 - b^2*d^2*f^5 - 12*d
^2*f*x^2*e^4 + 4*(3*b*d*f^3*x^2 + 2*(5*a*d*f^3 - d^3*f)*x)*e^3 - (3*b^2*f^5*x^2 + 20*a*b*f^5*x + 15*a^2*f^5 -
22*a*d^2*f^3 - d^4*f)*e^2 - 2*(2*b^2*d*f^5*x + a*b*d*f^5 + 3*b*d^3*f^3)*e)*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))/(b^5*f^10*x^2 + a^2*b^3*f^10 - 2*a*b^3*d^2*f^8 +
b^3*d^4*f^6 - 32*d^5*x^2*e^5 + 2*(a*b^4*f^10 - b^4*d^2*f^8)*x + 16*(5*b*d^4*f^2*x^2 + 2*(a*d^4*f^2 - d^6)*x)*e
^4 - 8*(10*b^2*d^3*f^4*x^2 + a^2*d^3*f^4 - 2*a*d^5*f^2 + d^7 + 8*(a*b*d^3*f^4 - b*d^5*f^2)*x)*e^3 + 4*(10*b^3*
d^2*f^6*x^2 + 3*a^2*b*d^2*f^6 - 6*a*b*d^4*f^4 + 3*b*d^6*f^2 + 12*(a*b^2*d^2*f^6 - b^2*d^4*f^4)*x)*e^2 - 2*(5*b
^4*d*f^8*x^2 + 3*a^2*b^2*d*f^8 - 6*a*b^2*d^3*f^6 + 3*b^2*d^5*f^4 + 8*(a*b^3*d*f^8 - b^3*d^3*f^6)*x)*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 {5}{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))**(5/2),x)

[Out]

Integral((d + e*x + f*sqrt(a + b*x + e**2*x**2/f**2))**(-5/2), 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))^(5/2),x, algorithm="giac")

[Out]

integrate((x*e + sqrt(b*x + a + x^2*e^2/f^2)*f + d)^(-5/2), 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 )}^{5/2}} \,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))^(5/2),x)

[Out]

int(1/(d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^(5/2), x)

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