3.7.62 \(\int \frac {x^2}{\sqrt {a+b (d+e x)^3+c (d+e x)^6}} \, dx\) [662]

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

[Out]

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

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Rubi [A]
time = 0.47, antiderivative size = 398, normalized size of antiderivative = 1.00, 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 = {1403, 1804, 1362, 440, 1399, 524, 1366, 635, 212} \begin {gather*} \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} \end {gather*}

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 212

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

Rule 440

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 635

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 1362

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])))^Fr
acPart[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 1366

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 1399

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 1403

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 1804

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]

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt {a+b (d+e x)^3+c (d+e x)^6}} \, dx &=\frac {\text {Subst}\left (\int \frac {(-d+x)^2}{\sqrt {a+b x^3+c x^6}} \, dx,x,d+e x\right )}{e^3}\\ &=\frac {\text {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 {\text {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) \text {Subst}\left (\int \frac {x}{\sqrt {a+b x^3+c x^6}} \, dx,x,d+e x\right )}{e^3}+\frac {d^2 \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^3+c x^6}} \, dx,x,d+e x\right )}{e^3}\\ &=\frac {\text {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 ) \text {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 ) \text {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 \text {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*}

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Mathematica [F]
time = 10.33, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2}{\sqrt {a+b (d+e x)^3+c (d+e x)^6}} \, dx \end {gather*}

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 = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{\sqrt {a +b \left (e x +d \right )^{3}+c \left (e x +d \right )^{6}}}\, dx\]

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 = 0.00, 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(x^2/(a+b*(e*x+d)^3+c*(e*x+d)^6)^(1/2),x, algorithm="maxima")

[Out]

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

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, 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(x^2/(a+b*(e*x+d)^3+c*(e*x+d)^6)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2}{\sqrt {a+b\,{\left (d+e\,x\right )}^3+c\,{\left (d+e\,x\right )}^6}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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