3.479 \(\int (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}})^{5/2} \, dx\)

Optimal. Leaf size=370 \[ \frac{f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{3/2}}{12 e^3}+\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{4 e^4}-\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^2 \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{16 e^4 \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}-\frac{5 f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{16 \sqrt{2} e^{9/2}}+\frac{\left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{7/2}}{7 e} \]

[Out]

(f^2*(2*d*e - b*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]])/(4*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])^(3/2))/(12*e^3) + (d + e*x + f*Sqrt[a + b*x + (e^2*
x^2)/f^2])^(7/2)/(7*e) - (f^2*(2*d*e - b*f^2)^2*(4*a*e^2 - b^2*f^2)*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/
f^2]])/(16*e^4*(b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2]))) - (5*f^2*(2*d*e - b*f^2)^(3/2)*(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]])
/(16*Sqrt[2]*e^(9/2))

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Rubi [A]  time = 0.600375, antiderivative size = 370, normalized size of antiderivative = 1., 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 = {2116, 897, 1257, 1810, 208} \[ \frac{f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{3/2}}{12 e^3}+\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{4 e^4}-\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^2 \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{16 e^4 \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}-\frac{5 f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{16 \sqrt{2} e^{9/2}}+\frac{\left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{7/2}}{7 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^2*(2*d*e - b*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]])/(4*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])^(3/2))/(12*e^3) + (d + e*x + f*Sqrt[a + b*x + (e^2*
x^2)/f^2])^(7/2)/(7*e) - (f^2*(2*d*e - b*f^2)^2*(4*a*e^2 - b^2*f^2)*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/
f^2]])/(16*e^4*(b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2]))) - (5*f^2*(2*d*e - b*f^2)^(3/2)*(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]])
/(16*Sqrt[2]*e^(9/2))

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]

Rule 897

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 1257

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[1/(2*e^(2*p +
m/2)*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4
)^p - (-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2)))/(d + e*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] && IGtQ[m/2, 0]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )^{5/2} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^{5/2} \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 )\\ &=4 \operatorname{Subst}\left (\int \frac{x^6 \left (d^2 e-(b d-a e) f^2+\left (-2 d e+b f^2\right ) x^2+e x^4\right )}{\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 (2 d e-b f^2\right )^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}}}}{16 e^4 \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{\operatorname{Subst}\left (\int \frac{-e f^2 \left (2 d e-b f^2\right )^2 \left (4 a e^2-b^2 f^2\right )-4 e^2 f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) x^2-8 e^3 f^2 \left (4 a e^2-b^2 f^2\right ) x^4+16 e^4 \left (2 d e-b f^2\right ) x^6-32 e^5 x^8}{-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 )}{16 e^5}\\ &=-\frac{f^2 \left (2 d e-b f^2\right )^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}}}}{16 e^4 \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{\operatorname{Subst}\left (\int \left (-4 e f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right )-4 e^2 f^2 \left (4 a e^2-b^2 f^2\right ) x^2-16 e^4 x^6-\frac{5 \left (16 a d^2 e^5 f^2-4 b^2 d^2 e^3 f^4-16 a b d e^4 f^4+4 b^3 d e^2 f^6+4 a b^2 e^3 f^6-b^4 e f^8\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 )}{16 e^5}\\ &=\frac{f^2 \left (2 d e-b f^2\right ) \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}}}}{4 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 )^{3/2}}{12 e^3}+\frac{\left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )^{7/2}}{7 e}-\frac{f^2 \left (2 d e-b f^2\right )^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}}}}{16 e^4 \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 (5 f^2 \left (2 d e-b f^2\right )^2 \left (4 a e^2-b^2 f^2\right )\right ) \operatorname{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 )}{16 e^4}\\ &=\frac{f^2 \left (2 d e-b f^2\right ) \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}}}}{4 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 )^{3/2}}{12 e^3}+\frac{\left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )^{7/2}}{7 e}-\frac{f^2 \left (2 d e-b f^2\right )^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}}}}{16 e^4 \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 f^2 \left (2 d e-b f^2\right )^{3/2} \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 )}{16 \sqrt{2} e^{9/2}}\\ \end{align*}

Mathematica [A]  time = 1.10635, size = 357, normalized size = 0.96 \[ \frac{\frac{4}{3} e^2 f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x\right )^{3/2}+4 e f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}-\frac{\left (4 a e^3 f^2-b^2 e f^4\right ) \left (b f^2-2 d e\right )^2 \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{2 e \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+e x\right )+b f^2}-\frac{5 \sqrt{e} f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{\sqrt{2}}+\frac{16}{7} e^4 \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x\right )^{7/2}}{16 e^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*e*f^2*(2*d*e - b*f^2)*(4*a*e^2 - b^2*f^2)*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]] + (4*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)])^(3/2))/3 + (16*e^4*(d + e*x + f*Sqrt[a + x*(b + (e
^2*x)/f^2)])^(7/2))/7 - ((-2*d*e + b*f^2)^2*(4*a*e^3*f^2 - b^2*e*f^4)*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)])) - (5*Sqrt[e]*f^2*(2*d*e - b*f^2)^(3/2)*(4*a*e^2
- b^2*f^2)*ArcTanh[(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]])/Sq
rt[2])/(16*e^5)

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Maple [F]  time = 0.029, size = 0, normalized size = 0. \begin{align*} \int \left ( d+ex+f\sqrt{a+bx+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}} \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

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))^(5/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + \sqrt{b x + \frac{e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac{5}{2}}\,{d x} \end{align*}

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))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.61643, size = 2033, normalized size = 5.49 \begin{align*} \text{result too large to display} \end{align*}

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))^(5/2),x, algorithm="fricas")

[Out]

[1/672*(105*sqrt(1/2)*(b^3*f^6 + 8*a*d*e^3*f^2 - 2*(b^2*d*e + 2*a*b*e^2)*f^4)*sqrt(-(b*f^2 - 2*d*e)/e)*log(-b^
2*f^4 + 4*(b*d*e - a*e^2)*f^2 - 4*(b*e^2*f^2 - 2*d*e^3)*x + 4*(2*sqrt(1/2)*e^2*f*sqrt(-(b*f^2 - 2*d*e)/e)*sqrt
((b*f^2*x + e^2*x^2 + a*f^2)/f^2) - sqrt(1/2)*(b*e*f^2 + 2*e^3*x)*sqrt(-(b*f^2 - 2*d*e)/e))*sqrt(e*x + f*sqrt(
(b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d) + 4*(b*e*f^3 - 2*d*e^2*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2)) + 2*(10
5*b^3*f^6 + 192*e^6*x^3 + 48*d^3*e^3 - 56*(5*b^2*d*e + 6*a*b*e^2)*f^4 + 4*(21*b*d^2*e^2 + 232*a*d*e^3)*f^2 + 1
44*(b*e^4*f^2 + 2*d*e^5)*x^2 + 2*(7*b^2*e^2*f^4 + 156*d^2*e^4 - 4*(3*b*d*e^3 - 32*a*e^4)*f^2)*x - 2*(35*b^2*e*
f^5 - 96*e^5*f*x^2 + 12*d^2*e^3*f - 4*(21*b*d*e^2 + 20*a*e^3)*f^3 - 24*(b*e^3*f^3 + 6*d*e^4*f)*x)*sqrt((b*f^2*
x + e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/e^4, -1/336*(105*sqrt(1/2)
*(b^3*f^6 + 8*a*d*e^3*f^2 - 2*(b^2*d*e + 2*a*b*e^2)*f^4)*sqrt((b*f^2 - 2*d*e)/e)*arctan(2*sqrt(1/2)*sqrt(e*x +
f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d)*e*sqrt((b*f^2 - 2*d*e)/e)/(b*f^2 - 2*d*e)) - (105*b^3*f^6 + 192*
e^6*x^3 + 48*d^3*e^3 - 56*(5*b^2*d*e + 6*a*b*e^2)*f^4 + 4*(21*b*d^2*e^2 + 232*a*d*e^3)*f^2 + 144*(b*e^4*f^2 +
2*d*e^5)*x^2 + 2*(7*b^2*e^2*f^4 + 156*d^2*e^4 - 4*(3*b*d*e^3 - 32*a*e^4)*f^2)*x - 2*(35*b^2*e*f^5 - 96*e^5*f*x
^2 + 12*d^2*e^3*f - 4*(21*b*d*e^2 + 20*a*e^3)*f^3 - 24*(b*e^3*f^3 + 6*d*e^4*f)*x)*sqrt((b*f^2*x + e^2*x^2 + a*
f^2)/f^2))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/e^4]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + \sqrt{b x + \frac{e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac{5}{2}}\,{d x} \end{align*}

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))^(5/2),x, algorithm="giac")

[Out]

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