### 3.1338 $$\int (b d+2 c d x)^{3/2} (a+b x+c x^2)^{3/2} \, dx$$

Optimal. Leaf size=227 $\frac{d^{3/2} \left (b^2-4 a c\right )^{13/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{154 c^3 \sqrt{a+b x+c x^2}}+\frac{d \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{77 c^2}-\frac{3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}}{154 c^2 d}+\frac{\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{11 c d}$

[Out]

((b^2 - 4*a*c)^2*d*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(77*c^2) - (3*(b^2 - 4*a*c)*(b*d + 2*c*d*x)^(5/2
)*Sqrt[a + b*x + c*x^2])/(154*c^2*d) + ((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^(3/2))/(11*c*d) + ((b^2 - 4*a*
c)^(13/4)*d^(3/2)*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])/(154*c^3*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.183112, antiderivative size = 227, 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 = {685, 692, 691, 689, 221} $\frac{d^{3/2} \left (b^2-4 a c\right )^{13/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{154 c^3 \sqrt{a+b x+c x^2}}+\frac{d \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{77 c^2}-\frac{3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}}{154 c^2 d}+\frac{\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{11 c d}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^(3/2),x]

[Out]

((b^2 - 4*a*c)^2*d*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(77*c^2) - (3*(b^2 - 4*a*c)*(b*d + 2*c*d*x)^(5/2
)*Sqrt[a + b*x + c*x^2])/(154*c^2*d) + ((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^(3/2))/(11*c*d) + ((b^2 - 4*a*
c)^(13/4)*d^(3/2)*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])/(154*c^3*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 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 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2} \, dx &=\frac{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}-\frac{\left (3 \left (b^2-4 a c\right )\right ) \int (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2} \, dx}{22 c}\\ &=-\frac{3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{154 c^2 d}+\frac{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}+\frac{\left (3 \left (b^2-4 a c\right )^2\right ) \int \frac{(b d+2 c d x)^{3/2}}{\sqrt{a+b x+c x^2}} \, dx}{308 c^2}\\ &=\frac{\left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{77 c^2}-\frac{3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{154 c^2 d}+\frac{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}+\frac{\left (\left (b^2-4 a c\right )^3 d^2\right ) \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx}{308 c^2}\\ &=\frac{\left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{77 c^2}-\frac{3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{154 c^2 d}+\frac{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}+\frac{\left (\left (b^2-4 a c\right )^3 d^2 \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}{308 c^2 \sqrt{a+b x+c x^2}}\\ &=\frac{\left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{77 c^2}-\frac{3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{154 c^2 d}+\frac{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}+\frac{\left (\left (b^2-4 a c\right )^3 d \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 )}{154 c^3 \sqrt{a+b x+c x^2}}\\ &=\frac{\left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{77 c^2}-\frac{3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{154 c^2 d}+\frac{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}}{11 c d}+\frac{\left (b^2-4 a c\right )^{13/4} d^{3/2} \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 )}{154 c^3 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0892676, size = 117, normalized size = 0.52 $\frac{2}{11} d \sqrt{a+x (b+c x)} \sqrt{d (b+2 c x)} \left (2 (a+x (b+c x))^2-\frac{\left (b^2-4 a c\right )^2 \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{16 c^2 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^(3/2),x]

[Out]

(2*d*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*(2*(a + x*(b + c*x))^2 - ((b^2 - 4*a*c)^2*Hypergeometric2F1[-3/
2, 1/4, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(16*c^2*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])))/11

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Maple [B]  time = 0.214, size = 796, 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((2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a)^(3/2),x)

[Out]

-1/308*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d*(-224*x^7*c^7-784*x^6*b*c^6-640*x^5*a*c^6-1016*x^5*b^2*c^5+64
*(-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^3-48*(-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)*E
llipticF(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+12*(-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-(-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^6-1600*x^4*a*b*c^5-580*x^4*b^3*c^4-
544*x^3*a^2*c^5-1328*x^3*a*b^2*c^4-124*x^3*b^4*c^3-816*x^2*a^2*b*c^4-392*x^2*a*b^3*c^3+2*x^2*b^5*c^2-128*x*a^3
*c^4-312*x*a^2*b^2*c^3-20*x*a*b^4*c^2+2*x*b^6*c-64*a^3*b*c^3-20*a^2*b^3*c^2+2*a*b^5*c)/c^3/(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{\left (2 \, c d x + b d\right )}^{\frac{3}{2}}{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(3/2)*(c*x^2 + b*x + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (2 \, c^{2} d x^{3} + 3 \, b c d x^{2} + a b d +{\left (b^{2} + 2 \, a c\right )} d x\right )} \sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

integral((2*c^2*d*x^3 + 3*b*c*d*x^2 + a*b*d + (b^2 + 2*a*c)*d*x)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \left (b + 2 c x\right )\right )^{\frac{3}{2}} \left (a + b x + c x^{2}\right )^{\frac{3}{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**(3/2)*(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((d*(b + 2*c*x))**(3/2)*(a + b*x + c*x**2)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (2 \, c d x + b d\right )}^{\frac{3}{2}}{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(3/2)*(c*x^2 + b*x + a)^(3/2), x)