3.14.40 \(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^{5/2}} \, dx\) [1340]

Optimal. Leaf size=174 \[ \frac {\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{6 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{6 c^3 d^{5/2} \sqrt {a+b x+c x^2}} \]

[Out]

-1/3*(c*x^2+b*x+a)^(3/2)/c/d/(2*c*d*x+b*d)^(3/2)+1/6*(2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^(1/2)/c^2/d^3-1/6*(-4*a
*c+b^2)^(5/4)*EllipticF((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2
)/c^3/d^(5/2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {698, 699, 705, 703, 227} \begin {gather*} -\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\text {ArcSin}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{6 c^3 d^{5/2} \sqrt {a+b x+c x^2}}+\frac {\sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}}{6 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(5/2),x]

[Out]

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

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 698

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 1))), x] - Dist[b*(p/(d*e*(m + 1))), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1
), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] &&
GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0]) && IntegerQ[2*p]

Rule 699

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[d*p*((b^2 - 4*a*c)/(b*e*(m + 2*p + 1))), Int[(d + e*x)^m*(a +
b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] &&
NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

Rule 703

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

Rule 705

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{5/2}} \, dx &=-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}+\frac {\int \frac {\sqrt {a+b x+c x^2}}{\sqrt {b d+2 c d x}} \, dx}{2 c d^2}\\ &=\frac {\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{6 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx}{12 c^2 d^2}\\ &=\frac {\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{6 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}-\frac {\left (\left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {-\frac {a c}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {c^2 x^2}{b^2-4 a c}}} \, dx}{12 c^2 d^2 \sqrt {a+b x+c x^2}}\\ &=\frac {\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{6 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}-\frac {\left (\left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{6 c^3 d^3 \sqrt {a+b x+c x^2}}\\ &=\frac {\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}}{6 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c d (b d+2 c d x)^{3/2}}-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{6 c^3 d^{5/2} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.04, size = 99, normalized size = 0.57 \begin {gather*} \frac {\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \, _2F_1\left (-\frac {3}{2},-\frac {3}{4};\frac {1}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{24 c^2 d (d (b+2 c x))^{3/2} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(5/2),x]

[Out]

((b^2 - 4*a*c)*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-3/2, -3/4, 1/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(24*c^2*
d*(d*(b + 2*c*x))^(3/2)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(625\) vs. \(2(146)=292\).
time = 0.84, size = 626, normalized size = 3.60 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/12*(8*(-4*a*c+b^2)^(1/2)*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1
/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2)
)/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*a*c^2*x-2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)
*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*
((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*(-4*a*c+b^2)^(1/2)*b^2*c*x+4*(-4*a*c+
b^2)^(1/2)*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-
2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(
1/2))^(1/2)*2^(1/2),2^(1/2))*a*b*c-((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c
+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^
2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*(-4*a*c+b^2)^(1/2)*b^3+4*c^4*x^4+8*b*c^3*x^3+6*b^2*c^2*x^
2+2*b^3*c*x-4*a^2*c^2+2*a*c*b^2)/d^3*(d*(2*c*x+b))^(1/2)/(c*x^2+b*x+a)^(1/2)/(2*c*x+b)^2/c^3

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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((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.30, size = 173, normalized size = 0.99 \begin {gather*} -\frac {\sqrt {2} {\left (b^{4} - 4 \, a b^{2} c + 4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 4 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) - 2 \, {\left (2 \, c^{4} x^{2} + 2 \, b c^{3} x + b^{2} c^{2} - 2 \, a c^{3}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{12 \, {\left (4 \, c^{6} d^{3} x^{2} + 4 \, b c^{5} d^{3} x + b^{2} c^{4} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(sqrt(2)*(b^4 - 4*a*b^2*c + 4*(b^2*c^2 - 4*a*c^3)*x^2 + 4*(b^3*c - 4*a*b*c^2)*x)*sqrt(c^2*d)*weierstrass
PInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c) - 2*(2*c^4*x^2 + 2*b*c^3*x + b^2*c^2 - 2*a*c^3)*sqrt(2*c*d*x
+ b*d)*sqrt(c*x^2 + b*x + a))/(4*c^6*d^3*x^2 + 4*b*c^5*d^3*x + b^2*c^4*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(3/2)/(2*c*d*x+b*d)**(5/2),x)

[Out]

Integral((a + b*x + c*x**2)**(3/2)/(d*(b + 2*c*x))**(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((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(5/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(5/2),x)

[Out]

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(5/2), x)

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