### 3.2543 $$\int \frac{(d+e x)^2}{(a+b x+c x^2)^{5/4}} \, dx$$

Optimal. Leaf size=594 $\frac{\sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac{1}{2}\right )}{\sqrt{2} c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac{2 (b+2 c x) \sqrt [4]{a+b x+c x^2} \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right )}{c^{3/2} \left (b^2-4 a c\right )^{3/2} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )}-\frac{\sqrt{2} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac{4 e \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{c \left (b^2-4 a c\right )}-\frac{4 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}$

[Out]

(-4*(d + e*x)*(b*d - 2*a*e + (2*c*d - b*e)*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/4)) + (4*e*(2*c*d - b*e)*(a
+ b*x + c*x^2)^(3/4))/(c*(b^2 - 4*a*c)) + (2*(4*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(b*d + 2*a*e))*(b + 2*c*x)*(a + b
*x + c*x^2)^(1/4))/(c^(3/2)*(b^2 - 4*a*c)^(3/2)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])) - (
Sqrt[2]*(4*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(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])*EllipticE[2*
ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(c^(7/4)*(b^2 - 4*a*c)^(1/4)*(b +
2*c*x)) + ((4*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(b*d + 2*a*e))*Sqrt[(b + 2*c*x)^2/((b^2 - 4*a*c)*(1 + (2*Sqrt[c]*Sq
rt[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])*Elliptic
F[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(Sqrt[2]*c^(7/4)*(b^2 - 4*a*c
)^(1/4)*(b + 2*c*x))

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Rubi [A]  time = 0.736935, antiderivative size = 594, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.273, Rules used = {738, 640, 623, 305, 220, 1196} $\frac{2 (b+2 c x) \sqrt [4]{a+b x+c x^2} \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right )}{c^{3/2} \left (b^2-4 a c\right )^{3/2} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )}+\frac{\sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{\sqrt{2} c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}-\frac{\sqrt{2} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac{4 e \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{c \left (b^2-4 a c\right )}-\frac{4 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-4*(d + e*x)*(b*d - 2*a*e + (2*c*d - b*e)*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/4)) + (4*e*(2*c*d - b*e)*(a
+ b*x + c*x^2)^(3/4))/(c*(b^2 - 4*a*c)) + (2*(4*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(b*d + 2*a*e))*(b + 2*c*x)*(a + b
*x + c*x^2)^(1/4))/(c^(3/2)*(b^2 - 4*a*c)^(3/2)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])) - (
Sqrt[2]*(4*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(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])*EllipticE[2*
ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(c^(7/4)*(b^2 - 4*a*c)^(1/4)*(b +
2*c*x)) + ((4*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(b*d + 2*a*e))*Sqrt[(b + 2*c*x)^2/((b^2 - 4*a*c)*(1 + (2*Sqrt[c]*Sq
rt[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])*Elliptic
F[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(Sqrt[2]*c^(7/4)*(b^2 - 4*a*c
)^(1/4)*(b + 2*c*x))

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && LtQ[p, -1] && GtQ[m, 1] && 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 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 305

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

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

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

Mathematica [C]  time = 0.336608, size = 177, normalized size = 0.3 $\frac{\sqrt{2} (b+2 c x) \sqrt [4]{\frac{c (a+x (b+c x))}{4 a c-b^2}} \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )-8 c \left (a b e^2-2 a c e (2 d+e x)+b^2 e^2 x+b c d (d-2 e x)+2 c^2 d^2 x\right )}{2 c^2 \left (b^2-4 a c\right ) \sqrt [4]{a+x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-8*c*(a*b*e^2 + 2*c^2*d^2*x + b^2*e^2*x + b*c*d*(d - 2*e*x) - 2*a*c*e*(2*d + e*x)) + Sqrt[2]*(4*c^2*d^2 + 3*b
^2*e^2 - 4*c*e*(b*d + 2*a*e))*(b + 2*c*x)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/4)*Hypergeometric2F1[1/4,
1/2, 3/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(2*c^2*(b^2 - 4*a*c)*(a + x*(b + c*x))^(1/4))

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Maple [F]  time = 1.151, 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 \frac{{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{4}}}\,{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((e*x + d)^2/(c*x^2 + b*x + a)^(5/4), x)

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

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