### 3.2522 $$\int (d+e x)^3 (a+b x+c x^2)^{5/4} \, dx$$

Optimal. Leaf size=448 $\frac{5 \left (b^2-4 a c\right )^{9/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) (2 c d-b e) \left (-4 c e (6 a e+11 b d)+17 b^2 e^2+44 c^2 d^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac{1}{2}\right )}{14784 \sqrt{2} c^{21/4} (b+2 c x)}-\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2} (2 c d-b e) \left (-4 c e (6 a e+11 b d)+17 b^2 e^2+44 c^2 d^2\right )}{7392 c^5}+\frac{e \left (a+b x+c x^2\right )^{9/4} \left (-2 c e (88 a e+507 b d)+221 b^2 e^2+306 c e x (2 c d-b e)+1320 c^2 d^2\right )}{2574 c^3}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/4} (2 c d-b e) \left (-4 c e (6 a e+11 b d)+17 b^2 e^2+44 c^2 d^2\right )}{616 c^4}+\frac{2 e (d+e x)^2 \left (a+b x+c x^2\right )^{9/4}}{13 c}$

[Out]

(-5*(b^2 - 4*a*c)*(2*c*d - b*e)*(44*c^2*d^2 + 17*b^2*e^2 - 4*c*e*(11*b*d + 6*a*e))*(b + 2*c*x)*(a + b*x + c*x^
2)^(1/4))/(7392*c^5) + ((2*c*d - b*e)*(44*c^2*d^2 + 17*b^2*e^2 - 4*c*e*(11*b*d + 6*a*e))*(b + 2*c*x)*(a + b*x
+ c*x^2)^(5/4))/(616*c^4) + (2*e*(d + e*x)^2*(a + b*x + c*x^2)^(9/4))/(13*c) + (e*(1320*c^2*d^2 + 221*b^2*e^2
- 2*c*e*(507*b*d + 88*a*e) + 306*c*e*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(9/4))/(2574*c^3) + (5*(b^2 - 4*a*c)^(
9/4)*(2*c*d - b*e)*(44*c^2*d^2 + 17*b^2*e^2 - 4*c*e*(11*b*d + 6*a*e))*Sqrt[(b + 2*c*x)^2/((b^2 - 4*a*c)*(1 + (
2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*
c])*EllipticF[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(14784*Sqrt[2]*c^
(21/4)*(b + 2*c*x))

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Rubi [A]  time = 0.559006, antiderivative size = 448, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.227, Rules used = {742, 779, 612, 623, 220} $-\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2} (2 c d-b e) \left (-4 c e (6 a e+11 b d)+17 b^2 e^2+44 c^2 d^2\right )}{7392 c^5}+\frac{e \left (a+b x+c x^2\right )^{9/4} \left (-2 c e (88 a e+507 b d)+221 b^2 e^2+306 c e x (2 c d-b e)+1320 c^2 d^2\right )}{2574 c^3}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/4} (2 c d-b e) \left (-4 c e (6 a e+11 b d)+17 b^2 e^2+44 c^2 d^2\right )}{616 c^4}+\frac{5 \left (b^2-4 a c\right )^{9/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) (2 c d-b e) \left (-4 c e (6 a e+11 b d)+17 b^2 e^2+44 c^2 d^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{14784 \sqrt{2} c^{21/4} (b+2 c x)}+\frac{2 e (d+e x)^2 \left (a+b x+c x^2\right )^{9/4}}{13 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^3*(a + b*x + c*x^2)^(5/4),x]

[Out]

(-5*(b^2 - 4*a*c)*(2*c*d - b*e)*(44*c^2*d^2 + 17*b^2*e^2 - 4*c*e*(11*b*d + 6*a*e))*(b + 2*c*x)*(a + b*x + c*x^
2)^(1/4))/(7392*c^5) + ((2*c*d - b*e)*(44*c^2*d^2 + 17*b^2*e^2 - 4*c*e*(11*b*d + 6*a*e))*(b + 2*c*x)*(a + b*x
+ c*x^2)^(5/4))/(616*c^4) + (2*e*(d + e*x)^2*(a + b*x + c*x^2)^(9/4))/(13*c) + (e*(1320*c^2*d^2 + 221*b^2*e^2
- 2*c*e*(507*b*d + 88*a*e) + 306*c*e*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(9/4))/(2574*c^3) + (5*(b^2 - 4*a*c)^(
9/4)*(2*c*d - b*e)*(44*c^2*d^2 + 17*b^2*e^2 - 4*c*e*(11*b*d + 6*a*e))*Sqrt[(b + 2*c*x)^2/((b^2 - 4*a*c)*(1 + (
2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*
c])*EllipticF[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(14784*Sqrt[2]*c^
(21/4)*(b + 2*c*x))

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 623

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[(d*Sqrt[(b + 2*c*x)
^2])/(b + 2*c*x), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rule 220

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

Rubi steps

\begin{align*} \int (d+e x)^3 \left (a+b x+c x^2\right )^{5/4} \, dx &=\frac{2 e (d+e x)^2 \left (a+b x+c x^2\right )^{9/4}}{13 c}+\frac{2 \int (d+e x) \left (\frac{1}{4} \left (26 c d^2-9 b d e-8 a e^2\right )+\frac{17}{4} e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{5/4} \, dx}{13 c}\\ &=\frac{2 e (d+e x)^2 \left (a+b x+c x^2\right )^{9/4}}{13 c}+\frac{e \left (1320 c^2 d^2+221 b^2 e^2-2 c e (507 b d+88 a e)+306 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{9/4}}{2574 c^3}+\frac{\left ((2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right )\right ) \int \left (a+b x+c x^2\right )^{5/4} \, dx}{88 c^3}\\ &=\frac{(2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{616 c^4}+\frac{2 e (d+e x)^2 \left (a+b x+c x^2\right )^{9/4}}{13 c}+\frac{e \left (1320 c^2 d^2+221 b^2 e^2-2 c e (507 b d+88 a e)+306 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{9/4}}{2574 c^3}-\frac{\left (5 \left (b^2-4 a c\right ) (2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right )\right ) \int \sqrt [4]{a+b x+c x^2} \, dx}{2464 c^4}\\ &=-\frac{5 \left (b^2-4 a c\right ) (2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{7392 c^5}+\frac{(2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{616 c^4}+\frac{2 e (d+e x)^2 \left (a+b x+c x^2\right )^{9/4}}{13 c}+\frac{e \left (1320 c^2 d^2+221 b^2 e^2-2 c e (507 b d+88 a e)+306 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{9/4}}{2574 c^3}+\frac{\left (5 \left (b^2-4 a c\right )^2 (2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right )\right ) \int \frac{1}{\left (a+b x+c x^2\right )^{3/4}} \, dx}{29568 c^5}\\ &=-\frac{5 \left (b^2-4 a c\right ) (2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{7392 c^5}+\frac{(2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{616 c^4}+\frac{2 e (d+e x)^2 \left (a+b x+c x^2\right )^{9/4}}{13 c}+\frac{e \left (1320 c^2 d^2+221 b^2 e^2-2 c e (507 b d+88 a e)+306 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{9/4}}{2574 c^3}+\frac{\left (5 \left (b^2-4 a c\right )^2 (2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right ) \sqrt{(b+2 c x)^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{7392 c^5 (b+2 c x)}\\ &=-\frac{5 \left (b^2-4 a c\right ) (2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{7392 c^5}+\frac{(2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{616 c^4}+\frac{2 e (d+e x)^2 \left (a+b x+c x^2\right )^{9/4}}{13 c}+\frac{e \left (1320 c^2 d^2+221 b^2 e^2-2 c e (507 b d+88 a e)+306 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{9/4}}{2574 c^3}+\frac{5 \left (b^2-4 a c\right )^{9/4} (2 c d-b e) \left (44 c^2 d^2+17 b^2 e^2-4 c e (11 b d+6 a e)\right ) \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )^2}} \left (1+\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{14784 \sqrt{2} c^{21/4} (b+2 c x)}\\ \end{align*}

Mathematica [A]  time = 1.17966, size = 286, normalized size = 0.64 $\frac{39 (2 c d-b e) \left (-4 c e (6 a e+11 b d)+17 b^2 e^2+44 c^2 d^2\right ) \left (5 \sqrt{2} \left (b^2-4 a c\right )^{5/2} \left (\frac{c (a+x (b+c x))}{4 a c-b^2}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ),2\right )+2 c (b+2 c x) \left (32 a^2 c+a \left (-5 b^2+44 b c x+44 c^2 x^2\right )+x \left (7 b^2 c x-5 b^3+24 b c^2 x^2+12 c^3 x^3\right )\right )\right )+224 c^3 e (a+x (b+c x))^3 \left (-2 c e (88 a e+507 b d+153 b e x)+221 b^2 e^2+12 c^2 d (110 d+51 e x)\right )+88704 c^5 e (d+e x)^2 (a+x (b+c x))^3}{576576 c^6 (a+x (b+c x))^{3/4}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^3*(a + b*x + c*x^2)^(5/4),x]

[Out]

(88704*c^5*e*(d + e*x)^2*(a + x*(b + c*x))^3 + 224*c^3*e*(a + x*(b + c*x))^3*(221*b^2*e^2 + 12*c^2*d*(110*d +
51*e*x) - 2*c*e*(507*b*d + 88*a*e + 153*b*e*x)) + 39*(2*c*d - b*e)*(44*c^2*d^2 + 17*b^2*e^2 - 4*c*e*(11*b*d +
6*a*e))*(2*c*(b + 2*c*x)*(32*a^2*c + a*(-5*b^2 + 44*b*c*x + 44*c^2*x^2) + x*(-5*b^3 + 7*b^2*c*x + 24*b*c^2*x^2
+ 12*c^3*x^3)) + 5*Sqrt[2]*(b^2 - 4*a*c)^(5/2)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(3/4)*EllipticF[ArcSin[
(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]/2, 2]))/(576576*c^6*(a + x*(b + c*x))^(3/4))

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Maple [F]  time = 1.085, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{3} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{5}{4}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(c*x^2+b*x+a)^(5/4),x)

[Out]

int((e*x+d)^3*(c*x^2+b*x+a)^(5/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{5}{4}}{\left (e x + d\right )}^{3}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(c*x^2+b*x+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/4)*(e*x + d)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c e^{3} x^{5} +{\left (3 \, c d e^{2} + b e^{3}\right )} x^{4} + a d^{3} +{\left (3 \, c d^{2} e + 3 \, b d e^{2} + a e^{3}\right )} x^{3} +{\left (c d^{3} + 3 \, b d^{2} e + 3 \, a d e^{2}\right )} x^{2} +{\left (b d^{3} + 3 \, a d^{2} e\right )} x\right )}{\left (c x^{2} + b x + a\right )}^{\frac{1}{4}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(c*x^2+b*x+a)^(5/4),x, algorithm="fricas")

[Out]

integral((c*e^3*x^5 + (3*c*d*e^2 + b*e^3)*x^4 + a*d^3 + (3*c*d^2*e + 3*b*d*e^2 + a*e^3)*x^3 + (c*d^3 + 3*b*d^2
*e + 3*a*d*e^2)*x^2 + (b*d^3 + 3*a*d^2*e)*x)*(c*x^2 + b*x + a)^(1/4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d + e x\right )^{3} \left (a + b x + c x^{2}\right )^{\frac{5}{4}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(c*x**2+b*x+a)**(5/4),x)

[Out]

Integral((d + e*x)**3*(a + b*x + c*x**2)**(5/4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{5}{4}}{\left (e x + d\right )}^{3}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(c*x^2+b*x+a)^(5/4),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(5/4)*(e*x + d)^3, x)