### 3.1341 $$\int \frac{(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^{9/2}} \, dx$$

Optimal. Leaf size=174 $\frac{\sqrt [4]{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{14 c^3 d^{9/2} \sqrt{a+b x+c x^2}}-\frac{\sqrt{a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}$

[Out]

-Sqrt[a + b*x + c*x^2]/(14*c^2*d^3*(b*d + 2*c*d*x)^(3/2)) - (a + b*x + c*x^2)^(3/2)/(7*c*d*(b*d + 2*c*d*x)^(7/
2)) + ((b^2 - 4*a*c)^(1/4)*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*d^(9/2)*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.14244, antiderivative size = 174, 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 = {684, 691, 689, 221} $\frac{\sqrt [4]{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{14 c^3 d^{9/2} \sqrt{a+b x+c x^2}}-\frac{\sqrt{a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(9/2),x]

[Out]

-Sqrt[a + b*x + c*x^2]/(14*c^2*d^3*(b*d + 2*c*d*x)^(3/2)) - (a + b*x + c*x^2)^(3/2)/(7*c*d*(b*d + 2*c*d*x)^(7/
2)) + ((b^2 - 4*a*c)^(1/4)*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*d^(9/2)*Sqrt[a + b*x + c*x^2])

Rule 684

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 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}}{(b d+2 c d x)^{9/2}} \, dx &=-\frac{\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}+\frac{3 \int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx}{14 c d^2}\\ &=-\frac{\sqrt{a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}+\frac{\int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx}{28 c^2 d^4}\\ &=-\frac{\sqrt{a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}+\frac{\sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \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 d^4 \sqrt{a+b x+c x^2}}\\ &=-\frac{\sqrt{a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}+\frac{\sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \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^5 \sqrt{a+b x+c x^2}}\\ &=-\frac{\sqrt{a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}+\frac{\sqrt [4]{b^2-4 a c} \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 d^{9/2} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0820332, size = 107, normalized size = 0.61 $\frac{\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \sqrt{d (b+2 c x)} \, _2F_1\left (-\frac{7}{4},-\frac{3}{2};-\frac{3}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{56 c^2 d^5 (b+2 c x)^4 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(9/2),x]

[Out]

((b^2 - 4*a*c)*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-7/4, -3/2, -3/4, (b + 2*c*x)^2/(b^
2 - 4*a*c)])/(56*c^2*d^5*(b + 2*c*x)^4*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.22, size = 678, normalized size = 3.9 \begin{align*}{\frac{1}{28\,{d}^{5} \left ( 2\,{c}^{2}{x}^{3}+3\,bc{x}^{2}+2\,acx+{b}^{2}x+ab \right ) \left ( 2\,cx+b \right ) ^{3}{c}^{3}}\sqrt{c{x}^{2}+bx+a}\sqrt{d \left ( 2\,cx+b \right ) } \left ( 8\,\sqrt{-4\,ac+{b}^{2}}\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 ){x}^{3}{c}^{3}+12\,\sqrt{-4\,ac+{b}^{2}}\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 ){x}^{2}b{c}^{2}+6\,\sqrt{-4\,ac+{b}^{2}}\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 ) x{b}^{2}c+\sqrt{-4\,ac+{b}^{2}}\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 ){b}^{3}-12\,{c}^{4}{x}^{4}-24\,b{c}^{3}{x}^{3}-16\,{x}^{2}a{c}^{3}-14\,{x}^{2}{b}^{2}{c}^{2}-16\,ba{c}^{2}x-2\,{b}^{3}cx-4\,{a}^{2}{c}^{2}-2\,ac{b}^{2} \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)^(9/2),x)

[Out]

1/28*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)*(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)*El
lipticF(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))*x^3*c^3+12*(-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))*x^2*b*c^2+6*(-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))*x*b^2*c+(-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))*b^3-12*c^4*x^4-24*b*c^3*x^3-16*x^2*a*c^3-14*x^2*b^2*c^2-16*b*a*c^2*x-2*b^3*c*x-4*a^2*c^2-2*a*c*b^2)/d^5
/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)/(2*c*x+b)^3/c^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac{9}{2}}}\,{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)^(9/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(9/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{2 \, c d x + b d}{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}}{32 \, c^{5} d^{5} x^{5} + 80 \, b c^{4} d^{5} x^{4} + 80 \, b^{2} c^{3} d^{5} x^{3} + 40 \, b^{3} c^{2} d^{5} x^{2} + 10 \, b^{4} c d^{5} x + b^{5} d^{5}}, 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)^(9/2),x, algorithm="fricas")

[Out]

integral(sqrt(2*c*d*x + b*d)*(c*x^2 + b*x + a)^(3/2)/(32*c^5*d^5*x^5 + 80*b*c^4*d^5*x^4 + 80*b^2*c^3*d^5*x^3 +
40*b^3*c^2*d^5*x^2 + 10*b^4*c*d^5*x + b^5*d^5), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac{9}{2}}}\,{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)^(9/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(9/2), x)