### 3.1372 $$\int \frac{1}{(b d+2 c d x)^{7/2} \sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=287 $\frac{6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{5 c d^{7/2} \left (b^2-4 a c\right )^{5/4} \sqrt{a+b x+c x^2}}+\frac{12 \sqrt{a+b x+c x^2}}{5 d^3 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x}}-\frac{6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 c d^{7/2} \left (b^2-4 a c\right )^{5/4} \sqrt{a+b x+c x^2}}+\frac{4 \sqrt{a+b x+c x^2}}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}$

[Out]

(4*Sqrt[a + b*x + c*x^2])/(5*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(5/2)) + (12*Sqrt[a + b*x + c*x^2])/(5*(b^2 - 4*a
*c)^2*d^3*Sqrt[b*d + 2*c*d*x]) - (6*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[b*d + 2
*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(5*c*(b^2 - 4*a*c)^(5/4)*d^(7/2)*Sqrt[a + b*x + c*x^2]) + (6*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])/(5*c*(b^2 - 4*a*c)^(5/4)*d^(7/2)*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.248231, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {693, 691, 690, 307, 221, 1199, 424} $\frac{12 \sqrt{a+b x+c x^2}}{5 d^3 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x}}+\frac{6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 c d^{7/2} \left (b^2-4 a c\right )^{5/4} \sqrt{a+b x+c x^2}}-\frac{6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 c d^{7/2} \left (b^2-4 a c\right )^{5/4} \sqrt{a+b x+c x^2}}+\frac{4 \sqrt{a+b x+c x^2}}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((b*d + 2*c*d*x)^(7/2)*Sqrt[a + b*x + c*x^2]),x]

[Out]

(4*Sqrt[a + b*x + c*x^2])/(5*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(5/2)) + (12*Sqrt[a + b*x + c*x^2])/(5*(b^2 - 4*a
*c)^2*d^3*Sqrt[b*d + 2*c*d*x]) - (6*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[b*d + 2
*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(5*c*(b^2 - 4*a*c)^(5/4)*d^(7/2)*Sqrt[a + b*x + c*x^2]) + (6*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])/(5*c*(b^2 - 4*a*c)^(5/4)*d^(7/2)*Sqrt[a + b*x + c*x^2])

Rule 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

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 690

Int[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[x^2/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 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

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]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rubi steps

\begin{align*} \int \frac{1}{(b d+2 c d x)^{7/2} \sqrt{a+b x+c x^2}} \, dx &=\frac{4 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{3 \int \frac{1}{(b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}} \, dx}{5 \left (b^2-4 a c\right ) d^2}\\ &=\frac{4 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{12 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right )^2 d^3 \sqrt{b d+2 c d x}}-\frac{3 \int \frac{\sqrt{b d+2 c d x}}{\sqrt{a+b x+c x^2}} \, dx}{5 \left (b^2-4 a c\right )^2 d^4}\\ &=\frac{4 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{12 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right )^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (3 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \int \frac{\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}{5 \left (b^2-4 a c\right )^2 d^4 \sqrt{a+b x+c x^2}}\\ &=\frac{4 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{12 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right )^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-\frac{x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{5 c \left (b^2-4 a c\right )^2 d^5 \sqrt{a+b x+c x^2}}\\ &=\frac{4 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{12 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right )^2 d^3 \sqrt{b d+2 c d x}}+\frac{\left (6 \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 )}{5 c \left (b^2-4 a c\right )^{3/2} d^4 \sqrt{a+b x+c x^2}}-\frac{\left (6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1+\frac{x^2}{\sqrt{b^2-4 a c} d}}{\sqrt{1-\frac{x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{5 c \left (b^2-4 a c\right )^{3/2} d^4 \sqrt{a+b x+c x^2}}\\ &=\frac{4 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{12 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right )^2 d^3 \sqrt{b d+2 c d x}}+\frac{6 \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 )}{5 c \left (b^2-4 a c\right )^{5/4} d^{7/2} \sqrt{a+b x+c x^2}}-\frac{\left (6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x^2}{\sqrt{b^2-4 a c} d}}}{\sqrt{1-\frac{x^2}{\sqrt{b^2-4 a c} d}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{5 c \left (b^2-4 a c\right )^{3/2} d^4 \sqrt{a+b x+c x^2}}\\ &=\frac{4 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac{12 \sqrt{a+b x+c x^2}}{5 \left (b^2-4 a c\right )^2 d^3 \sqrt{b d+2 c d x}}-\frac{6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\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 )}{5 c \left (b^2-4 a c\right )^{5/4} d^{7/2} \sqrt{a+b x+c x^2}}+\frac{6 \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 )}{5 c \left (b^2-4 a c\right )^{5/4} d^{7/2} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0613787, size = 91, normalized size = 0.32 $-\frac{2 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (-\frac{5}{4},\frac{1}{2};-\frac{1}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{5 c d \sqrt{a+x (b+c x)} (d (b+2 c x))^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((b*d + 2*c*d*x)^(7/2)*Sqrt[a + b*x + c*x^2]),x]

[Out]

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

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Maple [B]  time = 0.258, size = 874, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(48*EllipticE(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)*(-(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)-12*EllipticE(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)*(-(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)+48*EllipticE(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)*(-(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)-12*Ellipt
icE(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)*(-(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)+12*((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)*EllipticE(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))*a*b^2*c-3*((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)*EllipticE(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^4-48*c^4*x^4-96*b*c^3*x^3-32*x^2*a*c^3-
64*x^2*b^2*c^2-32*b*a*c^2*x-16*b^3*c*x+16*a^2*c^2-16*a*c*b^2)*(d*(2*c*x+b))^(1/2)/d^4/(c*x^2+b*x+a)^(1/2)/(2*c
*x+b)^3/(4*a*c-b^2)^2/c

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

[Out]

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

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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}}{16 \, c^{5} d^{4} x^{6} + 48 \, b c^{4} d^{4} x^{5} + a b^{4} d^{4} + 8 \,{\left (7 \, b^{2} c^{3} + 2 \, a c^{4}\right )} d^{4} x^{4} + 32 \,{\left (b^{3} c^{2} + a b c^{3}\right )} d^{4} x^{3} + 3 \,{\left (3 \, b^{4} c + 8 \, a b^{2} c^{2}\right )} d^{4} x^{2} +{\left (b^{5} + 8 \, a b^{3} c\right )} d^{4} x}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(16*c^5*d^4*x^6 + 48*b*c^4*d^4*x^5 + a*b^4*d^4 + 8*(7*b^2*c
^3 + 2*a*c^4)*d^4*x^4 + 32*(b^3*c^2 + a*b*c^3)*d^4*x^3 + 3*(3*b^4*c + 8*a*b^2*c^2)*d^4*x^2 + (b^5 + 8*a*b^3*c)
*d^4*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \left (b + 2 c x\right )\right )^{\frac{7}{2}} \sqrt{a + b x + c x^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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