### 3.1351 $$\int \frac{(a+b x+c x^2)^{5/2}}{\sqrt{b d+2 c d x}} \, dx$$

Optimal. Leaf size=229 $-\frac{5 \left (b^2-4 a c\right )^{13/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{308 c^4 \sqrt{d} \sqrt{a+b x+c x^2}}+\frac{5 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{308 c^3 d}-\frac{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} \sqrt{b d+2 c d x}}{154 c^2 d}+\frac{\left (a+b x+c x^2\right )^{5/2} \sqrt{b d+2 c d x}}{11 c d}$

[Out]

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

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Rubi [A]  time = 0.187023, antiderivative size = 229, 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 = {685, 691, 689, 221} $\frac{5 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{308 c^3 d}-\frac{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} \sqrt{b d+2 c d x}}{154 c^2 d}-\frac{5 \left (b^2-4 a c\right )^{13/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{308 c^4 \sqrt{d} \sqrt{a+b x+c x^2}}+\frac{\left (a+b x+c x^2\right )^{5/2} \sqrt{b d+2 c d x}}{11 c d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 685

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 + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), Int[(d + e*x)^m*(a +
b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && 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}}{\sqrt{b d+2 c d x}} \, dx &=\frac{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{5/2}}{11 c d}-\frac{\left (5 \left (b^2-4 a c\right )\right ) \int \frac{\left (a+b x+c x^2\right )^{3/2}}{\sqrt{b d+2 c d x}} \, dx}{22 c}\\ &=-\frac{5 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d}+\frac{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{5/2}}{11 c d}+\frac{\left (15 \left (b^2-4 a c\right )^2\right ) \int \frac{\sqrt{a+b x+c x^2}}{\sqrt{b d+2 c d x}} \, dx}{308 c^2}\\ &=\frac{5 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{308 c^3 d}-\frac{5 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d}+\frac{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{5/2}}{11 c d}-\frac{\left (5 \left (b^2-4 a c\right )^3\right ) \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx}{616 c^3}\\ &=\frac{5 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{308 c^3 d}-\frac{5 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d}+\frac{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{5/2}}{11 c d}-\frac{\left (5 \left (b^2-4 a c\right )^3 \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 \sqrt{a+b x+c x^2}}\\ &=\frac{5 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{308 c^3 d}-\frac{5 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d}+\frac{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{5/2}}{11 c d}-\frac{\left (5 \left (b^2-4 a c\right )^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 )}{308 c^4 d \sqrt{a+b x+c x^2}}\\ &=\frac{5 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{308 c^3 d}-\frac{5 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{154 c^2 d}+\frac{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{5/2}}{11 c d}-\frac{5 \left (b^2-4 a c\right )^{13/4} \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 d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{308 c^4 \sqrt{d} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0801335, size = 101, normalized size = 0.44 $\frac{\left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)} \sqrt{d (b+2 c x)} \, _2F_1\left (-\frac{5}{2},\frac{1}{4};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{32 c^3 d \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-5/2, 1/4, 5/4, (b + 2*c*x)^2/(b^
2 - 4*a*c)])/(32*c^3*d*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.243, size = 798, normalized size = 3.5 \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)^(1/2),x)

[Out]

1/616*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)/d*(112*x^7*c^7+392*x^6*b*c^6+496*x^5*a*c^6+464*x^5*b^2*c^5+320*(
-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))*a^3*c^3-240*(-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))*a^2*b^2*c^2+60*(-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))*a*b^4*c-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^6+1240*x^4*a*b*c^5+180*x^4*b^3*c^4
+976*x^3*a^2*c^5+752*x^3*a*b^2*c^4-4*x^3*b^4*c^3+1464*x^2*a^2*b*c^4-112*x^2*a*b^3*c^3+10*x^2*b^5*c^2+592*x*a^3
*c^4+288*x*a^2*b^2*c^3-100*x*a*b^4*c^2+10*x*b^6*c+296*a^3*b*c^3-100*a^2*b^3*c^2+10*a*b^5*c)/c^4/(2*c^2*x^3+3*b
*c*x^2+2*a*c*x+b^2*x+a*b)

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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}}}{\sqrt{2 \, c d x + b d}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/sqrt(2*c*d*x + b*d), 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{c x^{2} + b x + a}}{\sqrt{2 \, c d x + b d}}, 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)^(1/2),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(c*x^2 + b*x + a)/sqrt(2*c*d*x + b*d),
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)**(1/2),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}}}{\sqrt{2 \, c d x + b d}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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