3.662 \(\int \frac{x^2}{\sqrt{a+b (d+e x)^3+c (d+e x)^6}} \, dx\)

Optimal. Leaf size=398 \[ \frac{d^2 (d+e x) \sqrt{\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c (d+e x)^3}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{1}{3};\frac{1}{2},\frac{1}{2};\frac{4}{3};-\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}\right )}{e^3 \sqrt{a+b (d+e x)^3+c (d+e x)^6}}-\frac{d (d+e x)^2 \sqrt{\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c (d+e x)^3}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{2}{3};\frac{1}{2},\frac{1}{2};\frac{5}{3};-\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}\right )}{e^3 \sqrt{a+b (d+e x)^3+c (d+e x)^6}}+\frac{\tanh ^{-1}\left (\frac{b+2 c (d+e x)^3}{2 \sqrt{c} \sqrt{a+b (d+e x)^3+c (d+e x)^6}}\right )}{3 \sqrt{c} e^3} \]

[Out]

(d^2*(d + e*x)*Sqrt[1 + (2*c*(d + e*x)^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*(d + e*x)^3)/(b + Sqrt[b^2 -
4*a*c])]*AppellF1[1/3, 1/2, 1/2, 4/3, (-2*c*(d + e*x)^3)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*(d + e*x)^3)/(b + Sqrt
[b^2 - 4*a*c])])/(e^3*Sqrt[a + b*(d + e*x)^3 + c*(d + e*x)^6]) - (d*(d + e*x)^2*Sqrt[1 + (2*c*(d + e*x)^3)/(b
- Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*(d + e*x)^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[2/3, 1/2, 1/2, 5/3, (-2*c*(
d + e*x)^3)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*(d + e*x)^3)/(b + Sqrt[b^2 - 4*a*c])])/(e^3*Sqrt[a + b*(d + e*x)^3
+ c*(d + e*x)^6]) + ArcTanh[(b + 2*c*(d + e*x)^3)/(2*Sqrt[c]*Sqrt[a + b*(d + e*x)^3 + c*(d + e*x)^6])]/(3*Sqrt
[c]*e^3)

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Rubi [A]  time = 0.687049, antiderivative size = 398, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {1389, 1790, 1348, 429, 1385, 510, 1352, 621, 206} \[ \frac{d^2 (d+e x) \sqrt{\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c (d+e x)^3}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{1}{3};\frac{1}{2},\frac{1}{2};\frac{4}{3};-\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}\right )}{e^3 \sqrt{a+b (d+e x)^3+c (d+e x)^6}}-\frac{d (d+e x)^2 \sqrt{\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c (d+e x)^3}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{2}{3};\frac{1}{2},\frac{1}{2};\frac{5}{3};-\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}\right )}{e^3 \sqrt{a+b (d+e x)^3+c (d+e x)^6}}+\frac{\tanh ^{-1}\left (\frac{b+2 c (d+e x)^3}{2 \sqrt{c} \sqrt{a+b (d+e x)^3+c (d+e x)^6}}\right )}{3 \sqrt{c} e^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a + b*(d + e*x)^3 + c*(d + e*x)^6],x]

[Out]

(d^2*(d + e*x)*Sqrt[1 + (2*c*(d + e*x)^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*(d + e*x)^3)/(b + Sqrt[b^2 -
4*a*c])]*AppellF1[1/3, 1/2, 1/2, 4/3, (-2*c*(d + e*x)^3)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*(d + e*x)^3)/(b + Sqrt
[b^2 - 4*a*c])])/(e^3*Sqrt[a + b*(d + e*x)^3 + c*(d + e*x)^6]) - (d*(d + e*x)^2*Sqrt[1 + (2*c*(d + e*x)^3)/(b
- Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*(d + e*x)^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[2/3, 1/2, 1/2, 5/3, (-2*c*(
d + e*x)^3)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*(d + e*x)^3)/(b + Sqrt[b^2 - 4*a*c])])/(e^3*Sqrt[a + b*(d + e*x)^3
+ c*(d + e*x)^6]) + ArcTanh[(b + 2*c*(d + e*x)^3)/(2*Sqrt[c]*Sqrt[a + b*(d + e*x)^3 + c*(d + e*x)^6])]/(3*Sqrt
[c]*e^3)

Rule 1389

Int[((a_.) + (c_.)*(v_)^(n2_.) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/Coefficient[v, x, 1]^(
m + 1), Subst[Int[SimplifyIntegrand[(x - Coefficient[v, x, 0])^m*(a + b*x^n + c*x^(2*n))^p, x], x], x, v], x]
/; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && LinearQ[v, x] && IntegerQ[m] && NeQ[v, x]

Rule 1790

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[x^j*Sum[Coeff[Pq, x, j + k*n]*x^(k*n), {k, 0, (q - j)/n + 1}]*(a + b*x^n + c*x^(2*n))^p, {j, 0, n - 1}], x
]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] &&  !PolyQ[P
q, x^n]

Rule 1348

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n + c*x^(2*n))
^FracPart[p])/((1 + (2*c*x^n)/(b + Rt[b^2 - 4*a*c, 2]))^FracPart[p]*(1 + (2*c*x^n)/(b - Rt[b^2 - 4*a*c, 2]))^F
racPart[p]), Int[(1 + (2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]))^p, x], x] /
; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1385

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a +
b*x^n + c*x^(2*n))^FracPart[p])/((1 + (2*c*x^n)/(b + Rt[b^2 - 4*a*c, 2]))^FracPart[p]*(1 + (2*c*x^n)/(b - Rt[
b^2 - 4*a*c, 2]))^FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^n)/(b - Sqrt
[b^2 - 4*a*c]))^p, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1352

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^2}{\sqrt{a+b (d+e x)^3+c (d+e x)^6}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(-d+x)^2}{\sqrt{a+b x^3+c x^6}} \, dx,x,d+e x\right )}{e^3}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{d^2}{\sqrt{a+b x^3+c x^6}}-\frac{2 d x}{\sqrt{a+b x^3+c x^6}}+\frac{x^2}{\sqrt{a+b x^3+c x^6}}\right ) \, dx,x,d+e x\right )}{e^3}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^3+c x^6}} \, dx,x,d+e x\right )}{e^3}-\frac{(2 d) \operatorname{Subst}\left (\int \frac{x}{\sqrt{a+b x^3+c x^6}} \, dx,x,d+e x\right )}{e^3}+\frac{d^2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^3+c x^6}} \, dx,x,d+e x\right )}{e^3}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,(d+e x)^3\right )}{3 e^3}-\frac{\left (2 d \sqrt{1+\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}}\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+\frac{2 c x^3}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^3}{b+\sqrt{b^2-4 a c}}}} \, dx,x,d+e x\right )}{e^3 \sqrt{a+b (d+e x)^3+c (d+e x)^6}}+\frac{\left (d^2 \sqrt{1+\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{2 c x^3}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^3}{b+\sqrt{b^2-4 a c}}}} \, dx,x,d+e x\right )}{e^3 \sqrt{a+b (d+e x)^3+c (d+e x)^6}}\\ &=\frac{d^2 (d+e x) \sqrt{1+\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}} F_1\left (\frac{1}{3};\frac{1}{2},\frac{1}{2};\frac{4}{3};-\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}\right )}{e^3 \sqrt{a+b (d+e x)^3+c (d+e x)^6}}-\frac{d (d+e x)^2 \sqrt{1+\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}} F_1\left (\frac{2}{3};\frac{1}{2},\frac{1}{2};\frac{5}{3};-\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}\right )}{e^3 \sqrt{a+b (d+e x)^3+c (d+e x)^6}}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c (d+e x)^3}{\sqrt{a+b (d+e x)^3+c (d+e x)^6}}\right )}{3 e^3}\\ &=\frac{d^2 (d+e x) \sqrt{1+\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}} F_1\left (\frac{1}{3};\frac{1}{2},\frac{1}{2};\frac{4}{3};-\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}\right )}{e^3 \sqrt{a+b (d+e x)^3+c (d+e x)^6}}-\frac{d (d+e x)^2 \sqrt{1+\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}} F_1\left (\frac{2}{3};\frac{1}{2},\frac{1}{2};\frac{5}{3};-\frac{2 c (d+e x)^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c (d+e x)^3}{b+\sqrt{b^2-4 a c}}\right )}{e^3 \sqrt{a+b (d+e x)^3+c (d+e x)^6}}+\frac{\tanh ^{-1}\left (\frac{b+2 c (d+e x)^3}{2 \sqrt{c} \sqrt{a+b (d+e x)^3+c (d+e x)^6}}\right )}{3 \sqrt{c} e^3}\\ \end{align*}

Mathematica [F]  time = 0.430273, size = 0, normalized size = 0. \[ \int \frac{x^2}{\sqrt{a+b (d+e x)^3+c (d+e x)^6}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[x^2/Sqrt[a + b*(d + e*x)^3 + c*(d + e*x)^6],x]

[Out]

Integrate[x^2/Sqrt[a + b*(d + e*x)^3 + c*(d + e*x)^6], x]

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Maple [F]  time = 0.042, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2}{\frac{1}{\sqrt{a+b \left ( ex+d \right ) ^{3}+c \left ( ex+d \right ) ^{6}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+b*(e*x+d)^3+c*(e*x+d)^6)^(1/2),x)

[Out]

int(x^2/(a+b*(e*x+d)^3+c*(e*x+d)^6)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{{\left (e x + d\right )}^{6} c +{\left (e x + d\right )}^{3} b + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b*(e*x+d)^3+c*(e*x+d)^6)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2/sqrt((e*x + d)^6*c + (e*x + d)^3*b + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{2}}{\sqrt{c e^{6} x^{6} + 6 \, c d e^{5} x^{5} + 15 \, c d^{2} e^{4} x^{4} + c d^{6} +{\left (20 \, c d^{3} + b\right )} e^{3} x^{3} + 3 \,{\left (5 \, c d^{4} + b d\right )} e^{2} x^{2} + b d^{3} + 3 \,{\left (2 \, c d^{5} + b d^{2}\right )} e x + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b*(e*x+d)^3+c*(e*x+d)^6)^(1/2),x, algorithm="fricas")

[Out]

integral(x^2/sqrt(c*e^6*x^6 + 6*c*d*e^5*x^5 + 15*c*d^2*e^4*x^4 + c*d^6 + (20*c*d^3 + b)*e^3*x^3 + 3*(5*c*d^4 +
b*d)*e^2*x^2 + b*d^3 + 3*(2*c*d^5 + b*d^2)*e*x + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{a + b d^{3} + 3 b d^{2} e x + 3 b d e^{2} x^{2} + b e^{3} x^{3} + c d^{6} + 6 c d^{5} e x + 15 c d^{4} e^{2} x^{2} + 20 c d^{3} e^{3} x^{3} + 15 c d^{2} e^{4} x^{4} + 6 c d e^{5} x^{5} + c e^{6} x^{6}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+b*(e*x+d)**3+c*(e*x+d)**6)**(1/2),x)

[Out]

Integral(x**2/sqrt(a + b*d**3 + 3*b*d**2*e*x + 3*b*d*e**2*x**2 + b*e**3*x**3 + c*d**6 + 6*c*d**5*e*x + 15*c*d*
*4*e**2*x**2 + 20*c*d**3*e**3*x**3 + 15*c*d**2*e**4*x**4 + 6*c*d*e**5*x**5 + c*e**6*x**6), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{{\left (e x + d\right )}^{6} c +{\left (e x + d\right )}^{3} b + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b*(e*x+d)^3+c*(e*x+d)^6)^(1/2),x, algorithm="giac")

[Out]

integrate(x^2/sqrt((e*x + d)^6*c + (e*x + d)^3*b + a), x)