3.1339 \(\int \frac{(a+b x+c x^2)^{3/2}}{\sqrt{b d+2 c d x}} \, dx\)

Optimal. Leaf size=182 \[ \frac{\left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{14 c^3 \sqrt{d} \sqrt{a+b x+c x^2}}-\frac{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{14 c^2 d}+\frac{\left (a+b x+c x^2\right )^{3/2} \sqrt{b d+2 c d x}}{7 c d} \]

[Out]

-((b^2 - 4*a*c)*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(14*c^2*d) + (Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)
^(3/2))/(7*c*d) + ((b^2 - 4*a*c)^(9/4)*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])/(14*c^3*Sqrt[d]*Sqrt[a + b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.149949, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {685, 691, 689, 221} \[ -\frac{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{14 c^2 d}+\frac{\left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{14 c^3 \sqrt{d} \sqrt{a+b x+c x^2}}+\frac{\left (a+b x+c x^2\right )^{3/2} \sqrt{b d+2 c d x}}{7 c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(14*c^2*d) + (Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)
^(3/2))/(7*c*d) + ((b^2 - 4*a*c)^(9/4)*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])/(14*c^3*Sqrt[d]*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 689

Int[1/(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[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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{3/2}}{\sqrt{b d+2 c d x}} \, dx &=\frac{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{7 c d}-\frac{\left (3 \left (b^2-4 a c\right )\right ) \int \frac{\sqrt{a+b x+c x^2}}{\sqrt{b d+2 c d x}} \, dx}{14 c}\\ &=-\frac{\left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{14 c^2 d}+\frac{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{7 c d}+\frac{\left (b^2-4 a c\right )^2 \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx}{28 c^2}\\ &=-\frac{\left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{14 c^2 d}+\frac{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{7 c d}+\frac{\left (\left (b^2-4 a c\right )^2 \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}{28 c^2 \sqrt{a+b x+c x^2}}\\ &=-\frac{\left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{14 c^2 d}+\frac{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{7 c d}+\frac{\left (\left (b^2-4 a c\right )^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 )}{14 c^3 d \sqrt{a+b x+c x^2}}\\ &=-\frac{\left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{14 c^2 d}+\frac{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{7 c d}+\frac{\left (b^2-4 a c\right )^{9/4} \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 )}{14 c^3 \sqrt{d} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0705426, size = 99, normalized size = 0.54 \[ -\frac{\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \sqrt{d (b+2 c x)} \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{8 c^2 d \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-3/2, 1/4, 5/4, (b + 2*c*x)^2/(b^2
- 4*a*c)])/(8*c^2*d*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

________________________________________________________________________________________

Maple [B]  time = 0.207, size = 566, normalized size = 3.1 \begin{align*}{\frac{1}{28\,d \left ( 2\,{c}^{2}{x}^{3}+3\,bc{x}^{2}+2\,acx+{b}^{2}x+ab \right ){c}^{3}}\sqrt{c{x}^{2}+bx+a}\sqrt{d \left ( 2\,cx+b \right ) } \left ( 8\,{x}^{5}{c}^{5}+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 EllipticF} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ) \sqrt{-4\,ac+{b}^{2}}{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 EllipticF} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ) \sqrt{-4\,ac+{b}^{2}}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 EllipticF} \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 ) \sqrt{-4\,ac+{b}^{2}}{b}^{4}+20\,{x}^{4}b{c}^{4}+32\,{x}^{3}a{c}^{4}+12\,{x}^{3}{b}^{2}{c}^{3}+48\,{x}^{2}ab{c}^{3}-2\,{x}^{2}{b}^{3}{c}^{2}+24\,x{a}^{2}{c}^{3}+12\,xa{b}^{2}{c}^{2}-2\,x{b}^{4}c+12\,{a}^{2}b{c}^{2}-2\,a{b}^{3}c \right ) } \end{align*}

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)^(1/2),x)

[Out]

1/28*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)/d*(8*x^5*c^5+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)*Elliptic
F(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)*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)*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)*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)*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^4+20*x^4*b*c^4+32*x^3*a*c^4+
12*x^3*b^2*c^3+48*x^2*a*b*c^3-2*x^2*b^3*c^2+24*x*a^2*c^3+12*x*a*b^2*c^2-2*x*b^4*c+12*a^2*b*c^2-2*a*b^3*c)/(2*c
^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)/c^3

________________________________________________________________________________________

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

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)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

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)^(1/2),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^(3/2)/sqrt(2*c*d*x + b*d), x)

________________________________________________________________________________________

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

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)**(1/2),x)

[Out]

Integral((a + b*x + c*x**2)**(3/2)/sqrt(d*(b + 2*c*x)), x)

________________________________________________________________________________________

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

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)^(1/2),x, algorithm="giac")

[Out]

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