### 3.1363 $$\int \frac{(b d+2 c d x)^{7/2}}{\sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=174 $\frac{10 d^{7/2} \left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{21 c \sqrt{a+b x+c x^2}}+\frac{20}{21} d^3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}+\frac{4}{7} d \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}$

[Out]

(20*(b^2 - 4*a*c)*d^3*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/21 + (4*d*(b*d + 2*c*d*x)^(5/2)*Sqrt[a + b*x
+ c*x^2])/7 + (10*(b^2 - 4*a*c)^(9/4)*d^(7/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sq
rt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(21*c*Sqrt[a + b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.144182, 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 = {692, 691, 689, 221} $\frac{20}{21} d^3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}+\frac{10 d^{7/2} \left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{21 c \sqrt{a+b x+c x^2}}+\frac{4}{7} d \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^(7/2)/Sqrt[a + b*x + c*x^2],x]

[Out]

(20*(b^2 - 4*a*c)*d^3*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/21 + (4*d*(b*d + 2*c*d*x)^(5/2)*Sqrt[a + b*x
+ c*x^2])/7 + (10*(b^2 - 4*a*c)^(9/4)*d^(7/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sq
rt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(21*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 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{(b d+2 c d x)^{7/2}}{\sqrt{a+b x+c x^2}} \, dx &=\frac{4}{7} d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{1}{7} \left (5 \left (b^2-4 a c\right ) d^2\right ) \int \frac{(b d+2 c d x)^{3/2}}{\sqrt{a+b x+c x^2}} \, dx\\ &=\frac{20}{21} \left (b^2-4 a c\right ) d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{4}{7} d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{1}{21} \left (5 \left (b^2-4 a c\right )^2 d^4\right ) \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx\\ &=\frac{20}{21} \left (b^2-4 a c\right ) d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{4}{7} d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{\left (5 \left (b^2-4 a c\right )^2 d^4 \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}{21 \sqrt{a+b x+c x^2}}\\ &=\frac{20}{21} \left (b^2-4 a c\right ) d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{4}{7} d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{\left (10 \left (b^2-4 a c\right )^2 d^3 \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 )}{21 c \sqrt{a+b x+c x^2}}\\ &=\frac{20}{21} \left (b^2-4 a c\right ) d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{4}{7} d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{10 \left (b^2-4 a c\right )^{9/4} d^{7/2} \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 )}{21 c \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.181379, size = 166, normalized size = 0.95 $\frac{2 d^3 \sqrt{d (b+2 c x)} \left (8 c \left (-5 a^2 c+2 a \left (b^2-b c x-c^2 x^2\right )+x \left (5 b^2 c x+2 b^3+6 b c^2 x^2+3 c^3 x^3\right )\right )+5 \left (b^2-4 a c\right )^2 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )\right )}{21 c \sqrt{a+x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^(7/2)/Sqrt[a + b*x + c*x^2],x]

[Out]

(2*d^3*Sqrt[d*(b + 2*c*x)]*(8*c*(-5*a^2*c + 2*a*(b^2 - b*c*x - c^2*x^2) + x*(2*b^3 + 5*b^2*c*x + 6*b*c^2*x^2 +
3*c^3*x^3)) + 5*(b^2 - 4*a*c)^2*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)]))/(21*c*Sqrt[a + x*(b + c*x)])

________________________________________________________________________________________

Maple [B]  time = 0.232, size = 567, normalized size = 3.3 \begin{align*}{\frac{{d}^{3}}{21\,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 ( 96\,{x}^{5}{c}^{5}+80\,\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}-40\,\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+5\,\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}}{b}^{4}+240\,{x}^{4}b{c}^{4}-64\,{x}^{3}a{c}^{4}+256\,{x}^{3}{b}^{2}{c}^{3}-96\,{x}^{2}ab{c}^{3}+144\,{x}^{2}{b}^{3}{c}^{2}-160\,x{a}^{2}{c}^{3}+32\,xa{b}^{2}{c}^{2}+32\,x{b}^{4}c-80\,{a}^{2}b{c}^{2}+32\,a{b}^{3}c \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(1/2),x)

[Out]

1/21*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d^3*(96*x^5*c^5+80*((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)*Ellip
ticF(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^2-4
0*((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+5*((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+240*x^4*b*c^4-64*x^3
*a*c^4+256*x^3*b^2*c^3-96*x^2*a*b*c^3+144*x^2*b^3*c^2-160*x*a^2*c^3+32*x*a*b^2*c^2+32*x*b^4*c-80*a^2*b*c^2+32*
a*b^3*c)/c/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c d x + b d\right )}^{\frac{7}{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)^(7/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(7/2)/sqrt(c*x^2 + b*x + a), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (8 \, c^{3} d^{3} x^{3} + 12 \, b c^{2} d^{3} x^{2} + 6 \, b^{2} c d^{3} x + b^{3} d^{3}\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)^(7/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral((8*c^3*d^3*x^3 + 12*b*c^2*d^3*x^2 + 6*b^2*c*d^3*x + b^3*d^3)*sqrt(2*c*d*x + b*d)/sqrt(c*x^2 + b*x + a
), x)

________________________________________________________________________________________

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((2*c*d*x+b*d)**(7/2)/(c*x**2+b*x+a)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c d x + b d\right )}^{\frac{7}{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)^(7/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(7/2)/sqrt(c*x^2 + b*x + a), x)