3.1390 $$\int \frac{(b d+2 c d x)^{11/2}}{(a+b x+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=196 $\frac{40 c d^{11/2} \left (b^2-4 a c\right )^{5/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{\sqrt{a+b x+c x^2}}+80 c^2 d^5 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}-\frac{12 c d^3 (b d+2 c d x)^{5/2}}{\sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}$

[Out]

(-2*d*(b*d + 2*c*d*x)^(9/2))/(3*(a + b*x + c*x^2)^(3/2)) - (12*c*d^3*(b*d + 2*c*d*x)^(5/2))/Sqrt[a + b*x + c*x
^2] + 80*c^2*d^5*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2] + (40*c*(b^2 - 4*a*c)^(5/4)*d^(11/2)*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])/Sqrt[
a + b*x + c*x^2]

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Rubi [A]  time = 0.166264, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.179, Rules used = {686, 692, 691, 689, 221} $\frac{40 c d^{11/2} \left (b^2-4 a c\right )^{5/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{\sqrt{a+b x+c x^2}}+80 c^2 d^5 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}-\frac{12 c d^3 (b d+2 c d x)^{5/2}}{\sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d*(b*d + 2*c*d*x)^(9/2))/(3*(a + b*x + c*x^2)^(3/2)) - (12*c*d^3*(b*d + 2*c*d*x)^(5/2))/Sqrt[a + b*x + c*x
^2] + 80*c^2*d^5*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2] + (40*c*(b^2 - 4*a*c)^(5/4)*d^(11/2)*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])/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 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)^{11/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}+\left (6 c d^2\right ) \int \frac{(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{12 c d^3 (b d+2 c d x)^{5/2}}{\sqrt{a+b x+c x^2}}+\left (60 c^2 d^4\right ) \int \frac{(b d+2 c d x)^{3/2}}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{12 c d^3 (b d+2 c d x)^{5/2}}{\sqrt{a+b x+c x^2}}+80 c^2 d^5 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\left (20 c^2 \left (b^2-4 a c\right ) d^6\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)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{12 c d^3 (b d+2 c d x)^{5/2}}{\sqrt{a+b x+c x^2}}+80 c^2 d^5 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{\left (20 c^2 \left (b^2-4 a c\right ) d^6 \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}{\sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{12 c d^3 (b d+2 c d x)^{5/2}}{\sqrt{a+b x+c x^2}}+80 c^2 d^5 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{\left (40 c \left (b^2-4 a c\right ) d^5 \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 )}{\sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{12 c d^3 (b d+2 c d x)^{5/2}}{\sqrt{a+b x+c x^2}}+80 c^2 d^5 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{40 c \left (b^2-4 a c\right )^{5/4} d^{11/2} \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 )}{\sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.228164, size = 201, normalized size = 1.03 $\frac{2 d^5 \sqrt{d (b+2 c x)} \left (-60 c \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \left (4 a^2 c+a \left (-b^2+4 b c x+4 c^2 x^2\right )-b^2 x (b+c x)\right ) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )+8 c^2 \left (15 a^2+21 a c x^2+4 c^2 x^4\right )+6 b^2 c \left (c x^2-3 a\right )+8 b c^2 x \left (21 a+8 c x^2\right )-26 b^3 c x-b^4\right )}{3 (a+x (b+c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(2*d^5*Sqrt[d*(b + 2*c*x)]*(-b^4 - 26*b^3*c*x + 6*b^2*c*(-3*a + c*x^2) + 8*b*c^2*x*(21*a + 8*c*x^2) + 8*c^2*(1
5*a^2 + 21*a*c*x^2 + 4*c^2*x^4) - 60*c*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*(4*a^2*c - b^2*x*(b + c*x) +
a*(-b^2 + 4*b*c*x + 4*c^2*x^2))*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 = 0.314, size = 958, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(5/2),x)

[Out]

-2/3*(120*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*a*c^3*((-
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)*((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)-30*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^2*c^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)*((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)+1
20*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*a*b*c^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)*((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)-30*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^3*c*((-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)*((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)-64*x^5*c^5+1
20*((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^2*c^2-30*((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-160*x^4*b*c^4-
336*x^3*a*c^4-76*x^3*b^2*c^3-504*x^2*a*b*c^3+46*x^2*b^3*c^2-240*x*a^2*c^3-132*x*a*b^2*c^2+28*x*b^4*c-120*a^2*b
*c^2+18*a*b^3*c+b^5)*d^5*(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{11}{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)^(11/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(11/2)/(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 (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}\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)^(11/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral((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)*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)**(11/2)/(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{11}{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)^(11/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(11/2)/(c*x^2 + b*x + a)^(5/2), x)