### 3.1391 $$\int \frac{(b d+2 c d x)^{7/2}}{(a+b x+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=165 $\frac{40 c d^{7/2} \sqrt [4]{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{3 \sqrt{a+b x+c x^2}}-\frac{20 c d^3 \sqrt{b d+2 c d x}}{3 \sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}$

[Out]

(-2*d*(b*d + 2*c*d*x)^(5/2))/(3*(a + b*x + c*x^2)^(3/2)) - (20*c*d^3*Sqrt[b*d + 2*c*d*x])/(3*Sqrt[a + b*x + c*
x^2]) + (40*c*(b^2 - 4*a*c)^(1/4)*d^(7/2)*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*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.136209, antiderivative size = 165, 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 = {686, 691, 689, 221} $\frac{40 c d^{7/2} \sqrt [4]{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{3 \sqrt{a+b x+c x^2}}-\frac{20 c d^3 \sqrt{b d+2 c d x}}{3 \sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d*(b*d + 2*c*d*x)^(5/2))/(3*(a + b*x + c*x^2)^(3/2)) - (20*c*d^3*Sqrt[b*d + 2*c*d*x])/(3*Sqrt[a + b*x + c*
x^2]) + (40*c*(b^2 - 4*a*c)^(1/4)*d^(7/2)*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*Sqrt[a + b*x + c*x^2])

Rule 686

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

Mathematica [C]  time = 0.132878, size = 122, normalized size = 0.74 $-\frac{2 d^3 \sqrt{d (b+2 c x)} \left (-20 c (a+x (b+c x)) \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )+2 c \left (5 a+7 c x^2\right )+b^2+14 b c x\right )}{3 (a+x (b+c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d^3*Sqrt[d*(b + 2*c*x)]*(b^2 + 14*b*c*x + 2*c*(5*a + 7*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*(a + x*(b + c*x))^(
3/2))

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Maple [B]  time = 0.286, size = 479, normalized size = 2.9 \begin{align*}{\frac{2\,{d}^{3}}{6\,cx+3\,b} \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-28\,{c}^{3}{x}^{3}-42\,b{c}^{2}{x}^{2}-20\,xa{c}^{2}-16\,x{b}^{2}c-10\,abc-{b}^{3} \right ) \sqrt{d \left ( 2\,cx+b \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((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(5/2),x)

[Out]

2/3*(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-28*c^3*x^3-42*b*c^2*x^2-20*x*a*c^2-16*x*b^2*c-10*a*b*c-b^3)*d^3*(d*(2*c*x+b))
^(1/2)/(2*c*x+b)/(c*x^2+b*x+a)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c d x + b d\right )}^{\frac{7}{2}}}{{\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((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(7/2)/(c*x^2 + b*x + a)^(5/2), x)

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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}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \,{\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x +{\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \,{\left (a b^{2} + a^{2} c\right )} x^{2}}, 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)^(5/2),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
)/(c^3*x^6 + 3*b*c^2*x^5 + 3*(b^2*c + a*c^2)*x^4 + 3*a^2*b*x + (b^3 + 6*a*b*c)*x^3 + a^3 + 3*(a*b^2 + a^2*c)*x
^2), 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((2*c*d*x+b*d)**(7/2)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c d x + b d\right )}^{\frac{7}{2}}}{{\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((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(7/2)/(c*x^2 + b*x + a)^(5/2), x)