3.1369 $$\int \frac{(b d+2 c d x)^{5/2}}{\sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=231 $-\frac{6 d^{5/2} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{5 c \sqrt{a+b x+c x^2}}+\frac{6 d^{5/2} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 c \sqrt{a+b x+c x^2}}+\frac{4}{5} d \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}$

[Out]

(4*d*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/5 + (6*(b^2 - 4*a*c)^(7/4)*d^(5/2)*Sqrt[-((c*(a + b*x + c*x^
2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(5*c*Sqrt[a + b*
x + c*x^2]) - (6*(b^2 - 4*a*c)^(7/4)*d^(5/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sqr
t[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(5*c*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.215393, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {692, 691, 690, 307, 221, 1199, 424} $-\frac{6 d^{5/2} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 c \sqrt{a+b x+c x^2}}+\frac{6 d^{5/2} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 c \sqrt{a+b x+c x^2}}+\frac{4}{5} d \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*d*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/5 + (6*(b^2 - 4*a*c)^(7/4)*d^(5/2)*Sqrt[-((c*(a + b*x + c*x^
2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(5*c*Sqrt[a + b*
x + c*x^2]) - (6*(b^2 - 4*a*c)^(7/4)*d^(5/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sqr
t[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(5*c*Sqrt[a + b*x + c*x^2])

Rule 692

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

Rule 691

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]

Rule 690

Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(4*Sqrt[-(c/(b^2 - 4*a*
c))])/e, Subst[Int[x^2/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 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 221

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 1199

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

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rubi steps

\begin{align*} \int \frac{(b d+2 c d x)^{5/2}}{\sqrt{a+b x+c x^2}} \, dx &=\frac{4}{5} d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}+\frac{1}{5} \left (3 \left (b^2-4 a c\right ) d^2\right ) \int \frac{\sqrt{b d+2 c d x}}{\sqrt{a+b x+c x^2}} \, dx\\ &=\frac{4}{5} d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}+\frac{\left (3 \left (b^2-4 a c\right ) d^2 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \int \frac{\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}{5 \sqrt{a+b x+c x^2}}\\ &=\frac{4}{5} d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}+\frac{\left (6 \left (b^2-4 a c\right ) d \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-\frac{x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{5 c \sqrt{a+b x+c x^2}}\\ &=\frac{4}{5} d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}-\frac{\left (6 \left (b^2-4 a c\right )^{3/2} d^2 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{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 )}{5 c \sqrt{a+b x+c x^2}}+\frac{\left (6 \left (b^2-4 a c\right )^{3/2} d^2 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1+\frac{x^2}{\sqrt{b^2-4 a c} d}}{\sqrt{1-\frac{x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{5 c \sqrt{a+b x+c x^2}}\\ &=\frac{4}{5} d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}-\frac{6 \left (b^2-4 a c\right )^{7/4} d^{5/2} \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 )}{5 c \sqrt{a+b x+c x^2}}+\frac{\left (6 \left (b^2-4 a c\right )^{3/2} d^2 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x^2}{\sqrt{b^2-4 a c} d}}}{\sqrt{1-\frac{x^2}{\sqrt{b^2-4 a c} d}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{5 c \sqrt{a+b x+c x^2}}\\ &=\frac{4}{5} d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}+\frac{6 \left (b^2-4 a c\right )^{7/4} d^{5/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\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 )}{5 c \sqrt{a+b x+c x^2}}-\frac{6 \left (b^2-4 a c\right )^{7/4} d^{5/2} \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 )}{5 c \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.126672, size = 111, normalized size = 0.48 $\frac{2 d (d (b+2 c x))^{3/2} \left (\left (b^2-4 a c\right ) \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )+2 c (a+x (b+c x))\right )}{5 c \sqrt{a+x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*d*(d*(b + 2*c*x))^(3/2)*(2*c*(a + x*(b + c*x)) + (b^2 - 4*a*c)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*H
ypergeometric2F1[1/2, 3/4, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(5*c*Sqrt[a + x*(b + c*x)])

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Maple [B]  time = 0.208, size = 496, normalized size = 2.2 \begin{align*} -{\frac{{d}^{2}}{5\,c \left ( 2\,{c}^{2}{x}^{3}+3\,bc{x}^{2}+2\,acx+{b}^{2}x+ab \right ) }\sqrt{d \left ( 2\,cx+b \right ) }\sqrt{c{x}^{2}+bx+a} \left ( 48\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{-{\frac{2\,cx+b}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{{\frac{-b-2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}{\it EllipticE} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ){a}^{2}{c}^{2}-24\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{-{\frac{2\,cx+b}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{{\frac{-b-2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}{\it EllipticE} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ) a{b}^{2}c+3\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{-{\frac{2\,cx+b}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{{\frac{-b-2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}{\it EllipticE} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ){b}^{4}-16\,{c}^{4}{x}^{4}-32\,b{c}^{3}{x}^{3}-16\,{x}^{2}a{c}^{3}-20\,{x}^{2}{b}^{2}{c}^{2}-16\,ba{c}^{2}x-4\,{b}^{3}cx-4\,ac{b}^{2} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d^2*(48*((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)*EllipticE(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^2*c^2-24*((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)*EllipticE(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^2*c+3*
((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)*EllipticE(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))*b^4-16*c^4*x^4-32*b*c^3*x^3-16*x^2*a*c^3-20*x^2*b^2*c^2-16*b*a*c^2*x-4*b^3*c*x-4*a*c*b^2)/c/
(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral((4*c^2*d^2*x^2 + 4*b*c*d^2*x + b^2*d^2)*sqrt(2*c*d*x + b*d)/sqrt(c*x^2 + b*x + a), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**(5/2)/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((d*(b + 2*c*x))**(5/2)/sqrt(a + b*x + c*x**2), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

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