3.2523 \(\int (d+e x)^2 (a+b x+c x^2)^{5/4} \, dx\)

Optimal. Leaf size=384 \[ \frac{5 \left (b^2-4 a c\right )^{9/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+11 b d)+13 b^2 e^2+44 c^2 d^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac{1}{2}\right )}{7392 \sqrt{2} c^{17/4} (b+2 c x)}-\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2} \left (-4 c e (2 a e+11 b d)+13 b^2 e^2+44 c^2 d^2\right )}{3696 c^4}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/4} \left (-4 c e (2 a e+11 b d)+13 b^2 e^2+44 c^2 d^2\right )}{308 c^3}+\frac{13 e \left (a+b x+c x^2\right )^{9/4} (2 c d-b e)}{99 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{9/4}}{11 c} \]

[Out]

(-5*(b^2 - 4*a*c)*(44*c^2*d^2 + 13*b^2*e^2 - 4*c*e*(11*b*d + 2*a*e))*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(369
6*c^4) + ((44*c^2*d^2 + 13*b^2*e^2 - 4*c*e*(11*b*d + 2*a*e))*(b + 2*c*x)*(a + b*x + c*x^2)^(5/4))/(308*c^3) +
(13*e*(2*c*d - b*e)*(a + b*x + c*x^2)^(9/4))/(99*c^2) + (2*e*(d + e*x)*(a + b*x + c*x^2)^(9/4))/(11*c) + (5*(b
^2 - 4*a*c)^(9/4)*(44*c^2*d^2 + 13*b^2*e^2 - 4*c*e*(11*b*d + 2*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])/(7392*Sqrt[2]*c^(1
7/4)*(b + 2*c*x))

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Rubi [A]  time = 0.541275, antiderivative size = 384, 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, 640, 612, 623, 220} \[ -\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2} \left (-4 c e (2 a e+11 b d)+13 b^2 e^2+44 c^2 d^2\right )}{3696 c^4}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/4} \left (-4 c e (2 a e+11 b d)+13 b^2 e^2+44 c^2 d^2\right )}{308 c^3}+\frac{5 \left (b^2-4 a c\right )^{9/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+11 b d)+13 b^2 e^2+44 c^2 d^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{7392 \sqrt{2} c^{17/4} (b+2 c x)}+\frac{13 e \left (a+b x+c x^2\right )^{9/4} (2 c d-b e)}{99 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{9/4}}{11 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(b^2 - 4*a*c)*(44*c^2*d^2 + 13*b^2*e^2 - 4*c*e*(11*b*d + 2*a*e))*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(369
6*c^4) + ((44*c^2*d^2 + 13*b^2*e^2 - 4*c*e*(11*b*d + 2*a*e))*(b + 2*c*x)*(a + b*x + c*x^2)^(5/4))/(308*c^3) +
(13*e*(2*c*d - b*e)*(a + b*x + c*x^2)^(9/4))/(99*c^2) + (2*e*(d + e*x)*(a + b*x + c*x^2)^(9/4))/(11*c) + (5*(b
^2 - 4*a*c)^(9/4)*(44*c^2*d^2 + 13*b^2*e^2 - 4*c*e*(11*b*d + 2*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])/(7392*Sqrt[2]*c^(1
7/4)*(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 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[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)^2 \left (a+b x+c x^2\right )^{5/4} \, dx &=\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{9/4}}{11 c}+\frac{2 \int \left (\frac{1}{4} \left (22 c d^2-4 e \left (\frac{9 b d}{4}+a e\right )\right )+\frac{13}{4} e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{5/4} \, dx}{11 c}\\ &=\frac{13 e (2 c d-b e) \left (a+b x+c x^2\right )^{9/4}}{99 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{9/4}}{11 c}+\frac{\left (-\frac{13}{4} b e (2 c d-b e)+\frac{1}{2} c \left (22 c d^2-4 e \left (\frac{9 b d}{4}+a e\right )\right )\right ) \int \left (a+b x+c x^2\right )^{5/4} \, dx}{11 c^2}\\ &=\frac{\left (44 c^2 d^2+13 b^2 e^2-4 c e (11 b d+2 a e)\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{308 c^3}+\frac{13 e (2 c d-b e) \left (a+b x+c x^2\right )^{9/4}}{99 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{9/4}}{11 c}-\frac{\left (5 \left (b^2-4 a c\right ) \left (44 c^2 d^2+13 b^2 e^2-4 c e (11 b d+2 a e)\right )\right ) \int \sqrt [4]{a+b x+c x^2} \, dx}{1232 c^3}\\ &=-\frac{5 \left (b^2-4 a c\right ) \left (44 c^2 d^2+13 b^2 e^2-4 c e (11 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{3696 c^4}+\frac{\left (44 c^2 d^2+13 b^2 e^2-4 c e (11 b d+2 a e)\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{308 c^3}+\frac{13 e (2 c d-b e) \left (a+b x+c x^2\right )^{9/4}}{99 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{9/4}}{11 c}+\frac{\left (5 \left (b^2-4 a c\right )^2 \left (44 c^2 d^2+13 b^2 e^2-4 c e (11 b d+2 a e)\right )\right ) \int \frac{1}{\left (a+b x+c x^2\right )^{3/4}} \, dx}{14784 c^4}\\ &=-\frac{5 \left (b^2-4 a c\right ) \left (44 c^2 d^2+13 b^2 e^2-4 c e (11 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{3696 c^4}+\frac{\left (44 c^2 d^2+13 b^2 e^2-4 c e (11 b d+2 a e)\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{308 c^3}+\frac{13 e (2 c d-b e) \left (a+b x+c x^2\right )^{9/4}}{99 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{9/4}}{11 c}+\frac{\left (5 \left (b^2-4 a c\right )^2 \left (44 c^2 d^2+13 b^2 e^2-4 c e (11 b d+2 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 )}{3696 c^4 (b+2 c x)}\\ &=-\frac{5 \left (b^2-4 a c\right ) \left (44 c^2 d^2+13 b^2 e^2-4 c e (11 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{3696 c^4}+\frac{\left (44 c^2 d^2+13 b^2 e^2-4 c e (11 b d+2 a e)\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{308 c^3}+\frac{13 e (2 c d-b e) \left (a+b x+c x^2\right )^{9/4}}{99 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{9/4}}{11 c}+\frac{5 \left (b^2-4 a c\right )^{9/4} \left (44 c^2 d^2+13 b^2 e^2-4 c e (11 b d+2 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 )}{7392 \sqrt{2} c^{17/4} (b+2 c x)}\\ \end{align*}

Mathematica [A]  time = 0.604829, size = 233, normalized size = 0.61 \[ \frac{2 \left (-\frac{\left (c e (2 a e+11 b d)-\frac{13 b^2 e^2}{4}-11 c^2 d^2\right ) \left (24 c^2 (b+2 c x) (a+x (b+c x))^2-5 \left (b^2-4 a c\right ) \left (2 c (b+2 c x) (a+x (b+c x))-\sqrt{2} \left (b^2-4 a c\right )^{3/2} \left (\frac{c (a+x (b+c x))}{4 a c-b^2}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ),2\right )\right )\right )}{336 c^4 (a+x (b+c x))^{3/4}}+\frac{13 e (a+x (b+c x))^{9/4} (2 c d-b e)}{18 c}+e (d+e x) (a+x (b+c x))^{9/4}\right )}{11 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((13*e*(2*c*d - b*e)*(a + x*(b + c*x))^(9/4))/(18*c) + e*(d + e*x)*(a + x*(b + c*x))^(9/4) - ((-11*c^2*d^2
- (13*b^2*e^2)/4 + c*e*(11*b*d + 2*a*e))*(24*c^2*(b + 2*c*x)*(a + x*(b + c*x))^2 - 5*(b^2 - 4*a*c)*(2*c*(b + 2
*c*x)*(a + x*(b + c*x)) - Sqrt[2]*(b^2 - 4*a*c)^(3/2)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(3/4)*EllipticF[A
rcSin[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]/2, 2])))/(336*c^4*(a + x*(b + c*x))^(3/4))))/(11*c)

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Maple [F]  time = 1.01, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{2} \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)^2*(c*x^2+b*x+a)^(5/4),x)

[Out]

int((e*x+d)^2*(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 )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Integral((d + e*x)**2*(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 )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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