### 3.1393 $$\int \frac{1}{\sqrt{b d+2 c d x} (a+b x+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=187 $\frac{40 c \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{3 \sqrt{d} \left (b^2-4 a c\right )^{7/4} \sqrt{a+b x+c x^2}}+\frac{20 c \sqrt{b d+2 c d x}}{3 d \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 \sqrt{b d+2 c d x}}{3 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}$

[Out]

(-2*Sqrt[b*d + 2*c*d*x])/(3*(b^2 - 4*a*c)*d*(a + b*x + c*x^2)^(3/2)) + (20*c*Sqrt[b*d + 2*c*d*x])/(3*(b^2 - 4*
a*c)^2*d*Sqrt[a + b*x + c*x^2]) + (40*c*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])/(3*(b^2 - 4*a*c)^(7/4)*Sqrt[d]*Sqrt[a + b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.140025, antiderivative size = 187, 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 = {687, 691, 689, 221} $\frac{20 c \sqrt{b d+2 c d x}}{3 d \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 \sqrt{b d+2 c d x}}{3 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{40 c \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{3 \sqrt{d} \left (b^2-4 a c\right )^{7/4} \sqrt{a+b x+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^(5/2)),x]

[Out]

(-2*Sqrt[b*d + 2*c*d*x])/(3*(b^2 - 4*a*c)*d*(a + b*x + c*x^2)^(3/2)) + (20*c*Sqrt[b*d + 2*c*d*x])/(3*(b^2 - 4*
a*c)^2*d*Sqrt[a + b*x + c*x^2]) + (40*c*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])/(3*(b^2 - 4*a*c)^(7/4)*Sqrt[d]*Sqrt[a + b*x + c*x^2])

Rule 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), 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] && LtQ[p, -1] &&  !GtQ[m, 1] && 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{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 \sqrt{b d+2 c d x}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}-\frac{(10 c) \int \frac{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=-\frac{2 \sqrt{b d+2 c d x}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac{20 c \sqrt{b d+2 c d x}}{3 \left (b^2-4 a c\right )^2 d \sqrt{a+b x+c x^2}}+\frac{\left (20 c^2\right ) \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx}{3 \left (b^2-4 a c\right )^2}\\ &=-\frac{2 \sqrt{b d+2 c d x}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac{20 c \sqrt{b d+2 c d x}}{3 \left (b^2-4 a c\right )^2 d \sqrt{a+b x+c x^2}}+\frac{\left (20 c^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}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}\\ &=-\frac{2 \sqrt{b d+2 c d x}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac{20 c \sqrt{b d+2 c d x}}{3 \left (b^2-4 a c\right )^2 d \sqrt{a+b x+c x^2}}+\frac{\left (40 c \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 )}{3 \left (b^2-4 a c\right )^2 d \sqrt{a+b x+c x^2}}\\ &=-\frac{2 \sqrt{b d+2 c d x}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac{20 c \sqrt{b d+2 c d x}}{3 \left (b^2-4 a c\right )^2 d \sqrt{a+b x+c x^2}}+\frac{40 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 )}{3 \left (b^2-4 a c\right )^{7/4} \sqrt{d} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.167233, size = 132, normalized size = 0.71 $-\frac{2 \sqrt{d (b+2 c x)} \left (-20 c (a+x (b+c x)) \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )-2 c \left (7 a+5 c x^2\right )+b^2-10 b c x\right )}{3 d \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/(Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^(5/2)),x]

[Out]

(-2*Sqrt[d*(b + 2*c*x)]*(b^2 - 10*b*c*x - 2*c*(7*a + 5*c*x^2) - 20*c*(a + x*(b + c*x))*Sqrt[(c*(a + x*(b + c*x
)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[1/4, 1/2, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(3*(b^2 - 4*a*c)^2*d*(a +
x*(b + c*x))^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.235, size = 491, normalized size = 2.6 \begin{align*}{\frac{2}{3\,d \left ( 2\,cx+b \right ) \left ( 4\,ac-{b}^{2} \right ) ^{2}}\sqrt{d \left ( 2\,cx+b \right ) } \left ( 10\,{\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}{c}^{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}}}}}\sqrt{-4\,ac+{b}^{2}}+10\,{\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 ) xbc\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}}}}}\sqrt{-4\,ac+{b}^{2}}+10\,\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}}ac+20\,{c}^{3}{x}^{3}+30\,b{c}^{2}{x}^{2}+28\,xa{c}^{2}+8\,x{b}^{2}c+14\,abc-{b}^{3} \right ) \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(d*(2*c*x+b))^(1/2)*(10*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*c^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)*(-4*a*c+b^2)^(1/2)+10*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*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)*(-4*a*c+b^
2)^(1/2)+10*((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*c+20*c^3*x^3+30*b*c^2*x^2+28*x*a*c^2+8*x*b^2*c+14*a*b*c-b^3
)/d/(2*c*x+b)/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(3/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{2 \, c d x + b d}{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}}{2 \, c^{4} d x^{7} + 7 \, b c^{3} d x^{6} + 3 \,{\left (3 \, b^{2} c^{2} + 2 \, a c^{3}\right )} d x^{5} + 5 \,{\left (b^{3} c + 3 \, a b c^{2}\right )} d x^{4} + a^{3} b d +{\left (b^{4} + 12 \, a b^{2} c + 6 \, a^{2} c^{2}\right )} d x^{3} + 3 \,{\left (a b^{3} + 3 \, a^{2} b c\right )} d x^{2} +{\left (3 \, a^{2} b^{2} + 2 \, a^{3} c\right )} d x}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(2*c^4*d*x^7 + 7*b*c^3*d*x^6 + 3*(3*b^2*c^2 + 2*a*c^3)*d*x^
5 + 5*(b^3*c + 3*a*b*c^2)*d*x^4 + a^3*b*d + (b^4 + 12*a*b^2*c + 6*a^2*c^2)*d*x^3 + 3*(a*b^3 + 3*a^2*b*c)*d*x^2
+ (3*a^2*b^2 + 2*a^3*c)*d*x), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d \left (b + 2 c x\right )} \left (a + b x + c x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{2 \, c d x + b d}{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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