### 3.1334 $$\int \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2} \, dx$$

Optimal. Leaf size=236 $\frac{\sqrt{d} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{5 c^2 \sqrt{a+b x+c x^2}}-\frac{\sqrt{d} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 c^2 \sqrt{a+b x+c x^2}}+\frac{\sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{5 c d}$

[Out]

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

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Rubi [A]  time = 0.215207, antiderivative size = 236, 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 = {685, 691, 690, 307, 221, 1199, 424} $\frac{\sqrt{d} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 c^2 \sqrt{a+b x+c x^2}}-\frac{\sqrt{d} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 c^2 \sqrt{a+b x+c x^2}}+\frac{\sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{5 c d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2],x]

[Out]

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

Rule 685

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 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 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2} \, dx &=\frac{(b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{5 c d}-\frac{\left (b^2-4 a c\right ) \int \frac{\sqrt{b d+2 c d x}}{\sqrt{a+b x+c x^2}} \, dx}{10 c}\\ &=\frac{(b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{5 c d}-\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{\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}{10 c \sqrt{a+b x+c x^2}}\\ &=\frac{(b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{5 c d}-\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 ) \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^2 d \sqrt{a+b x+c x^2}}\\ &=\frac{(b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{5 c d}+\frac{\left (\left (b^2-4 a c\right )^{3/2} \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^2 \sqrt{a+b x+c x^2}}-\frac{\left (\left (b^2-4 a c\right )^{3/2} \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^2 \sqrt{a+b x+c x^2}}\\ &=\frac{(b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{5 c d}+\frac{\left (b^2-4 a c\right )^{7/4} \sqrt{d} \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^2 \sqrt{a+b x+c x^2}}-\frac{\left (\left (b^2-4 a c\right )^{3/2} \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^2 \sqrt{a+b x+c x^2}}\\ &=\frac{(b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{5 c d}-\frac{\left (b^2-4 a c\right )^{7/4} \sqrt{d} \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^2 \sqrt{a+b x+c x^2}}+\frac{\left (b^2-4 a c\right )^{7/4} \sqrt{d} \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^2 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0502355, size = 91, normalized size = 0.39 $\frac{\sqrt{a+x (b+c x)} (d (b+2 c x))^{3/2} \, _2F_1\left (-\frac{1}{2},\frac{3}{4};\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{6 c d \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2],x]

[Out]

((d*(b + 2*c*x))^(3/2)*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-1/2, 3/4, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(
6*c*d*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.207, size = 492, normalized size = 2.1 \begin{align*}{\frac{1}{ \left ( 20\,{c}^{2}{x}^{3}+30\,bc{x}^{2}+20\,acx+10\,{b}^{2}x+10\,ab \right ){c}^{2}}\sqrt{d \left ( 2\,cx+b \right ) }\sqrt{c{x}^{2}+bx+a} \left ( 16\,\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}-8\,\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+\sqrt{{ \left ( b+2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}}}\sqrt{-{(2\,cx+b){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}}}\sqrt{{ \left ( -b-2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}}}{\it EllipticE} \left ({\frac{\sqrt{2}}{2}\sqrt{{ \left ( b+2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}}}},\sqrt{2} \right ){b}^{4}+8\,{c}^{4}{x}^{4}+16\,b{c}^{3}{x}^{3}+8\,{x}^{2}a{c}^{3}+10\,{x}^{2}{b}^{2}{c}^{2}+8\,ba{c}^{2}x+2\,{b}^{3}cx+2\,ac{b}^{2} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*(16*((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^2-8*((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+((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+8*c^4*x^4+16*b*c^3*x^3+8*x^2*a*c^3+10*x^2*b^2*c^2+8*b*a*c^2*x+2*b^3*c*x+2*a*c*b^2)/(2*c^2*x^3+3
*b*c*x^2+2*a*c*x+b^2*x+a*b)/c^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{2 \, c d x + b d} \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)^(1/2)*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(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 \sqrt{d \left (b + 2 c x\right )} \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)**(1/2)*(c*x**2+b*x+a)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{2 \, c d x + b d} \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)^(1/2)*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

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