### 3.1354 $$\int \frac{(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{13/2}} \, dx$$

Optimal. Leaf size=211 $\frac{5 \sqrt [4]{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{308 c^4 d^{13/2} \sqrt{a+b x+c x^2}}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac{5 \sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}$

[Out]

(-5*Sqrt[a + b*x + c*x^2])/(308*c^3*d^5*(b*d + 2*c*d*x)^(3/2)) - (5*(a + b*x + c*x^2)^(3/2))/(154*c^2*d^3*(b*d
+ 2*c*d*x)^(7/2)) - (a + b*x + c*x^2)^(5/2)/(11*c*d*(b*d + 2*c*d*x)^(11/2)) + (5*(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])
/(308*c^4*d^(13/2)*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.178951, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {684, 691, 689, 221} $\frac{5 \sqrt [4]{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{308 c^4 d^{13/2} \sqrt{a+b x+c x^2}}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac{5 \sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(13/2),x]

[Out]

(-5*Sqrt[a + b*x + c*x^2])/(308*c^3*d^5*(b*d + 2*c*d*x)^(3/2)) - (5*(a + b*x + c*x^2)^(3/2))/(154*c^2*d^3*(b*d
+ 2*c*d*x)^(7/2)) - (a + b*x + c*x^2)^(5/2)/(11*c*d*(b*d + 2*c*d*x)^(11/2)) + (5*(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])
/(308*c^4*d^(13/2)*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 )^{5/2}}{(b d+2 c d x)^{13/2}} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac{5 \int \frac{\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{9/2}} \, dx}{22 c d^2}\\ &=-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac{15 \int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx}{308 c^2 d^4}\\ &=-\frac{5 \sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac{5 \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx}{616 c^3 d^6}\\ &=-\frac{5 \sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac{\left (5 \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}{616 c^3 d^6 \sqrt{a+b x+c x^2}}\\ &=-\frac{5 \sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac{\left (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 )}{308 c^4 d^7 \sqrt{a+b x+c x^2}}\\ &=-\frac{5 \sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{3/2}}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d^3 (b d+2 c d x)^{7/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{11 c d (b d+2 c d x)^{11/2}}+\frac{5 \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 )}{308 c^4 d^{13/2} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0913246, size = 109, normalized size = 0.52 $-\frac{\left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)} \sqrt{d (b+2 c x)} \, _2F_1\left (-\frac{11}{4},-\frac{5}{2};-\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{352 c^3 d^7 (b+2 c x)^6 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(13/2),x]

[Out]

-((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-11/4, -5/2, -7/4, (b + 2*c*x)^2
/(b^2 - 4*a*c)])/(352*c^3*d^7*(b + 2*c*x)^6*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.222, size = 1035, 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((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x)

[Out]

1/616*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)*(160*(-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^5*c^5+400*(-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^4*b*c^4+400*(-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^3*b^2*c^3+200*(-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^3*c^2-296*x^6*c^6+50*(-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*b^4*c-888*x^5*b*c^5+5*(
-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^5-488*x^4*a*c^5-988*x^4*b^2*c^4-976*x^3*a*b*c^4-496*x^3*b^3*c^3-248*x^2*
a^2*c^4-608*x^2*a*b^2*c^3-110*x^2*b^4*c^2-248*x*a^2*b*c^3-120*x*a*b^3*c^2-10*x*b^5*c-56*a^3*c^3-20*a^2*b^2*c^2
-10*a*b^4*c)/d^7/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)/(2*c*x+b)^5/c^4

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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{5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac{13}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(13/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}}{128 \, c^{7} d^{7} x^{7} + 448 \, b c^{6} d^{7} x^{6} + 672 \, b^{2} c^{5} d^{7} x^{5} + 560 \, b^{3} c^{4} d^{7} x^{4} + 280 \, b^{4} c^{3} d^{7} x^{3} + 84 \, b^{5} c^{2} d^{7} x^{2} + 14 \, b^{6} c d^{7} x + b^{7} d^{7}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="fricas")

[Out]

integral((c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(
128*c^7*d^7*x^7 + 448*b*c^6*d^7*x^6 + 672*b^2*c^5*d^7*x^5 + 560*b^3*c^4*d^7*x^4 + 280*b^4*c^3*d^7*x^3 + 84*b^5
*c^2*d^7*x^2 + 14*b^6*c*d^7*x + b^7*d^7), 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)**(5/2)/(2*c*d*x+b*d)**(13/2),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{5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac{13}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(13/2), x)