### 3.2357 $$\int (d+e x) (a+b x+c x^2)^{5/2} \, dx$$

Optimal. Leaf size=207 $-\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{384 c^3}+\frac{5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2} (2 c d-b e)}{1024 c^4}-\frac{5 \left (b^2-4 a c\right )^3 (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2048 c^{9/2}}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/2} (2 c d-b e)}{24 c^2}+\frac{e \left (a+b x+c x^2\right )^{7/2}}{7 c}$

[Out]

(5*(b^2 - 4*a*c)^2*(2*c*d - b*e)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(1024*c^4) - (5*(b^2 - 4*a*c)*(2*c*d - b*e
)*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(384*c^3) + ((2*c*d - b*e)*(b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(24*c^2
) + (e*(a + b*x + c*x^2)^(7/2))/(7*c) - (5*(b^2 - 4*a*c)^3*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a
+ b*x + c*x^2])])/(2048*c^(9/2))

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Rubi [A]  time = 0.0885637, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {640, 612, 621, 206} $-\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{384 c^3}+\frac{5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2} (2 c d-b e)}{1024 c^4}-\frac{5 \left (b^2-4 a c\right )^3 (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2048 c^{9/2}}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/2} (2 c d-b e)}{24 c^2}+\frac{e \left (a+b x+c x^2\right )^{7/2}}{7 c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(5*(b^2 - 4*a*c)^2*(2*c*d - b*e)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(1024*c^4) - (5*(b^2 - 4*a*c)*(2*c*d - b*e
)*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(384*c^3) + ((2*c*d - b*e)*(b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(24*c^2
) + (e*(a + b*x + c*x^2)^(7/2))/(7*c) - (5*(b^2 - 4*a*c)^3*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a
+ b*x + c*x^2])])/(2048*c^(9/2))

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 621

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int (d+e x) \left (a+b x+c x^2\right )^{5/2} \, dx &=\frac{e \left (a+b x+c x^2\right )^{7/2}}{7 c}+\frac{(2 c d-b e) \int \left (a+b x+c x^2\right )^{5/2} \, dx}{2 c}\\ &=\frac{(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{24 c^2}+\frac{e \left (a+b x+c x^2\right )^{7/2}}{7 c}-\frac{\left (5 \left (b^2-4 a c\right ) (2 c d-b e)\right ) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{48 c^2}\\ &=-\frac{5 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{384 c^3}+\frac{(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{24 c^2}+\frac{e \left (a+b x+c x^2\right )^{7/2}}{7 c}+\frac{\left (5 \left (b^2-4 a c\right )^2 (2 c d-b e)\right ) \int \sqrt{a+b x+c x^2} \, dx}{256 c^3}\\ &=\frac{5 \left (b^2-4 a c\right )^2 (2 c d-b e) (b+2 c x) \sqrt{a+b x+c x^2}}{1024 c^4}-\frac{5 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{384 c^3}+\frac{(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{24 c^2}+\frac{e \left (a+b x+c x^2\right )^{7/2}}{7 c}-\frac{\left (5 \left (b^2-4 a c\right )^3 (2 c d-b e)\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{2048 c^4}\\ &=\frac{5 \left (b^2-4 a c\right )^2 (2 c d-b e) (b+2 c x) \sqrt{a+b x+c x^2}}{1024 c^4}-\frac{5 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{384 c^3}+\frac{(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{24 c^2}+\frac{e \left (a+b x+c x^2\right )^{7/2}}{7 c}-\frac{\left (5 \left (b^2-4 a c\right )^3 (2 c d-b e)\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{1024 c^4}\\ &=\frac{5 \left (b^2-4 a c\right )^2 (2 c d-b e) (b+2 c x) \sqrt{a+b x+c x^2}}{1024 c^4}-\frac{5 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{384 c^3}+\frac{(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{24 c^2}+\frac{e \left (a+b x+c x^2\right )^{7/2}}{7 c}-\frac{5 \left (b^2-4 a c\right )^3 (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2048 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.278456, size = 180, normalized size = 0.87 $\frac{(2 c d-b e) \left (256 c^{5/2} (b+2 c x) (a+x (b+c x))^{5/2}-5 \left (b^2-4 a c\right ) \left (16 c^{3/2} (b+2 c x) (a+x (b+c x))^{3/2}-3 \left (b^2-4 a c\right ) \left (2 \sqrt{c} (b+2 c x) \sqrt{a+x (b+c x)}-\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right )\right )\right )}{6144 c^{9/2}}+\frac{e (a+x (b+c x))^{7/2}}{7 c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(e*(a + x*(b + c*x))^(7/2))/(7*c) + ((2*c*d - b*e)*(256*c^(5/2)*(b + 2*c*x)*(a + x*(b + c*x))^(5/2) - 5*(b^2 -
4*a*c)*(16*c^(3/2)*(b + 2*c*x)*(a + x*(b + c*x))^(3/2) - 3*(b^2 - 4*a*c)*(2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b
+ c*x)] - (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))))/(6144*c^(9/2))

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Maple [B]  time = 0.047, size = 807, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*e*(c*x^2+b*x+a)^(7/2)/c+5/64*e*b^3/c^2*(c*x^2+b*x+a)^(1/2)*x*a+5/384*e*b^4/c^3*(c*x^2+b*x+a)^(3/2)+5/512*d
/c^3*(c*x^2+b*x+a)^(1/2)*b^5+5/16*d/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^3-5/1024*d/c^(7/2)*l
n((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^6+5/24*d*(c*x^2+b*x+a)^(3/2)*x*a-5/192*d/c^2*(c*x^2+b*x+a)^(3/2)*
b^3+1/12*d/c*(c*x^2+b*x+a)^(5/2)*b-5/1024*e*b^6/c^4*(c*x^2+b*x+a)^(1/2)+5/2048*e*b^7/c^(9/2)*ln((1/2*b+c*x)/c^
(1/2)+(c*x^2+b*x+a)^(1/2))-1/24*e*b^2/c^2*(c*x^2+b*x+a)^(5/2)+5/16*d*(c*x^2+b*x+a)^(1/2)*x*a^2-5/32*d/c*(c*x^2
+b*x+a)^(1/2)*x*a*b^2-5/48*e*b/c*(c*x^2+b*x+a)^(3/2)*x*a-5/32*e*b/c*(c*x^2+b*x+a)^(1/2)*x*a^2+1/6*d*x*(c*x^2+b
*x+a)^(5/2)+5/32*d/c*(c*x^2+b*x+a)^(1/2)*b*a^2-5/64*d/c^2*(c*x^2+b*x+a)^(1/2)*b^3*a-5/32*e*b/c^(3/2)*ln((1/2*b
+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^3-1/12*e*b/c*x*(c*x^2+b*x+a)^(5/2)+15/128*e*b^3/c^(5/2)*ln((1/2*b+c*x)/c^
(1/2)+(c*x^2+b*x+a)^(1/2))*a^2-15/512*e*b^5/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+5/128*e*b^4/
c^3*(c*x^2+b*x+a)^(1/2)*a-5/64*e*b^2/c^2*(c*x^2+b*x+a)^(1/2)*a^2-5/96*e*b^2/c^2*(c*x^2+b*x+a)^(3/2)*a-5/512*e*
b^5/c^3*(c*x^2+b*x+a)^(1/2)*x+5/192*e*b^3/c^2*(c*x^2+b*x+a)^(3/2)*x+5/256*d/c^2*(c*x^2+b*x+a)^(1/2)*x*b^4-5/96
*d/c*(c*x^2+b*x+a)^(3/2)*x*b^2-15/64*d/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^2*a^2+15/256*d/c^
(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^4*a+5/48*d/c*(c*x^2+b*x+a)^(3/2)*b*a

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.28142, size = 1974, normalized size = 9.54 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/86016*(105*(2*(b^6*c - 12*a*b^4*c^2 + 48*a^2*b^2*c^3 - 64*a^3*c^4)*d - (b^7 - 12*a*b^5*c + 48*a^2*b^3*c^2 -
64*a^3*b*c^3)*e)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c
) + 4*(3072*c^7*e*x^6 + 256*(14*c^7*d + 29*b*c^6*e)*x^5 + 128*(70*b*c^6*d + (37*b^2*c^5 + 72*a*c^6)*e)*x^4 + 1
6*(14*(27*b^2*c^5 + 52*a*c^6)*d + (3*b^3*c^4 + 788*a*b*c^5)*e)*x^3 + 8*(14*(b^3*c^4 + 156*a*b*c^5)*d - (7*b^4*
c^3 - 60*a*b^2*c^4 - 1152*a^2*c^5)*e)*x^2 + 14*(15*b^5*c^2 - 160*a*b^3*c^3 + 528*a^2*b*c^4)*d - (105*b^6*c - 1
120*a*b^4*c^2 + 3696*a^2*b^2*c^3 - 3072*a^3*c^4)*e - 2*(14*(5*b^4*c^3 - 48*a*b^2*c^4 - 528*a^2*c^5)*d - (35*b^
5*c^2 - 336*a*b^3*c^3 + 912*a^2*b*c^4)*e)*x)*sqrt(c*x^2 + b*x + a))/c^5, 1/43008*(105*(2*(b^6*c - 12*a*b^4*c^2
+ 48*a^2*b^2*c^3 - 64*a^3*c^4)*d - (b^7 - 12*a*b^5*c + 48*a^2*b^3*c^2 - 64*a^3*b*c^3)*e)*sqrt(-c)*arctan(1/2*
sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(3072*c^7*e*x^6 + 256*(14*c^7*d + 29*b
*c^6*e)*x^5 + 128*(70*b*c^6*d + (37*b^2*c^5 + 72*a*c^6)*e)*x^4 + 16*(14*(27*b^2*c^5 + 52*a*c^6)*d + (3*b^3*c^4
+ 788*a*b*c^5)*e)*x^3 + 8*(14*(b^3*c^4 + 156*a*b*c^5)*d - (7*b^4*c^3 - 60*a*b^2*c^4 - 1152*a^2*c^5)*e)*x^2 +
14*(15*b^5*c^2 - 160*a*b^3*c^3 + 528*a^2*b*c^4)*d - (105*b^6*c - 1120*a*b^4*c^2 + 3696*a^2*b^2*c^3 - 3072*a^3*
c^4)*e - 2*(14*(5*b^4*c^3 - 48*a*b^2*c^4 - 528*a^2*c^5)*d - (35*b^5*c^2 - 336*a*b^3*c^3 + 912*a^2*b*c^4)*e)*x)
*sqrt(c*x^2 + b*x + a))/c^5]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.15808, size = 601, normalized size = 2.9 \begin{align*} \frac{1}{21504} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (2 \,{\left (12 \, c^{2} x e + \frac{14 \, c^{8} d + 29 \, b c^{7} e}{c^{6}}\right )} x + \frac{70 \, b c^{7} d + 37 \, b^{2} c^{6} e + 72 \, a c^{7} e}{c^{6}}\right )} x + \frac{378 \, b^{2} c^{6} d + 728 \, a c^{7} d + 3 \, b^{3} c^{5} e + 788 \, a b c^{6} e}{c^{6}}\right )} x + \frac{14 \, b^{3} c^{5} d + 2184 \, a b c^{6} d - 7 \, b^{4} c^{4} e + 60 \, a b^{2} c^{5} e + 1152 \, a^{2} c^{6} e}{c^{6}}\right )} x - \frac{70 \, b^{4} c^{4} d - 672 \, a b^{2} c^{5} d - 7392 \, a^{2} c^{6} d - 35 \, b^{5} c^{3} e + 336 \, a b^{3} c^{4} e - 912 \, a^{2} b c^{5} e}{c^{6}}\right )} x + \frac{210 \, b^{5} c^{3} d - 2240 \, a b^{3} c^{4} d + 7392 \, a^{2} b c^{5} d - 105 \, b^{6} c^{2} e + 1120 \, a b^{4} c^{3} e - 3696 \, a^{2} b^{2} c^{4} e + 3072 \, a^{3} c^{5} e}{c^{6}}\right )} + \frac{5 \,{\left (2 \, b^{6} c d - 24 \, a b^{4} c^{2} d + 96 \, a^{2} b^{2} c^{3} d - 128 \, a^{3} c^{4} d - b^{7} e + 12 \, a b^{5} c e - 48 \, a^{2} b^{3} c^{2} e + 64 \, a^{3} b c^{3} e\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{2048 \, c^{\frac{9}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21504*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*(12*c^2*x*e + (14*c^8*d + 29*b*c^7*e)/c^6)*x + (70*b*c^7*d + 37*b
^2*c^6*e + 72*a*c^7*e)/c^6)*x + (378*b^2*c^6*d + 728*a*c^7*d + 3*b^3*c^5*e + 788*a*b*c^6*e)/c^6)*x + (14*b^3*c
^5*d + 2184*a*b*c^6*d - 7*b^4*c^4*e + 60*a*b^2*c^5*e + 1152*a^2*c^6*e)/c^6)*x - (70*b^4*c^4*d - 672*a*b^2*c^5*
d - 7392*a^2*c^6*d - 35*b^5*c^3*e + 336*a*b^3*c^4*e - 912*a^2*b*c^5*e)/c^6)*x + (210*b^5*c^3*d - 2240*a*b^3*c^
4*d + 7392*a^2*b*c^5*d - 105*b^6*c^2*e + 1120*a*b^4*c^3*e - 3696*a^2*b^2*c^4*e + 3072*a^3*c^5*e)/c^6) + 5/2048
*(2*b^6*c*d - 24*a*b^4*c^2*d + 96*a^2*b^2*c^3*d - 128*a^3*c^4*d - b^7*e + 12*a*b^5*c*e - 48*a^2*b^3*c^2*e + 64
*a^3*b*c^3*e)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(9/2)