3.2358 \(\int (a+b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=149 \[ \frac{5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}{512 c^3}-\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}-\frac{5 \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{1024 c^{7/2}}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{12 c} \]

[Out]

(5*(b^2 - 4*a*c)^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(512*c^3) - (5*(b^2 - 4*a*c)*(b + 2*c*x)*(a + b*x + c*x^
2)^(3/2))/(192*c^2) + ((b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(12*c) - (5*(b^2 - 4*a*c)^3*ArcTanh[(b + 2*c*x)/(2
*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(1024*c^(7/2))

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Rubi [A]  time = 0.0503224, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {612, 621, 206} \[ \frac{5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}{512 c^3}-\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}-\frac{5 \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{1024 c^{7/2}}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{12 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2),x]

[Out]

(5*(b^2 - 4*a*c)^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(512*c^3) - (5*(b^2 - 4*a*c)*(b + 2*c*x)*(a + b*x + c*x^
2)^(3/2))/(192*c^2) + ((b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(12*c) - (5*(b^2 - 4*a*c)^3*ArcTanh[(b + 2*c*x)/(2
*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(1024*c^(7/2))

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 \left (a+b x+c x^2\right )^{5/2} \, dx &=\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{12 c}-\frac{\left (5 \left (b^2-4 a c\right )\right ) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{24 c}\\ &=-\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{12 c}+\frac{\left (5 \left (b^2-4 a c\right )^2\right ) \int \sqrt{a+b x+c x^2} \, dx}{128 c^2}\\ &=\frac{5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}{512 c^3}-\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{12 c}-\frac{\left (5 \left (b^2-4 a c\right )^3\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{1024 c^3}\\ &=\frac{5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}{512 c^3}-\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{12 c}-\frac{\left (5 \left (b^2-4 a c\right )^3\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 )}{512 c^3}\\ &=\frac{5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}{512 c^3}-\frac{5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{12 c}-\frac{5 \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{1024 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.580177, size = 162, normalized size = 1.09 \[ \frac{\sqrt{a+x (b+c x)} \left (2 (b+2 c x) \left (16 c^2 \left (33 a^2+26 a c x^2+8 c^2 x^4\right )+8 b^2 c \left (11 c x^2-20 a\right )+32 b c^2 x \left (13 a+8 c x^2\right )-40 b^3 c x+15 b^4\right )+\frac{15 \left (b^2-4 a c\right )^{5/2} \sin ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}\right )}{3072 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2),x]

[Out]

(Sqrt[a + x*(b + c*x)]*(2*(b + 2*c*x)*(15*b^4 - 40*b^3*c*x + 32*b*c^2*x*(13*a + 8*c*x^2) + 8*b^2*c*(-20*a + 11
*c*x^2) + 16*c^2*(33*a^2 + 26*a*c*x^2 + 8*c^2*x^4)) + (15*(b^2 - 4*a*c)^(5/2)*ArcSin[(b + 2*c*x)/Sqrt[b^2 - 4*
a*c]])/Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]))/(3072*c^3)

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Maple [B]  time = 0.044, size = 360, normalized size = 2.4 \begin{align*}{\frac{2\,cx+b}{12\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,ax}{24} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{b}^{2}x}{96\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,ab}{48\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{b}^{3}}{192\,{c}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{a}^{2}x}{16}\sqrt{c{x}^{2}+bx+a}}-{\frac{5\,ax{b}^{2}}{32\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,x{b}^{4}}{256\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,b{a}^{2}}{32\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{5\,a{b}^{3}}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,{b}^{5}}{512\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,{a}^{3}}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{15\,{b}^{2}{a}^{2}}{64}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{15\,{b}^{4}a}{256}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{5\,{b}^{6}}{1024}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*c*x+b)*(c*x^2+b*x+a)^(5/2)/c+5/24*(c*x^2+b*x+a)^(3/2)*x*a-5/96/c*(c*x^2+b*x+a)^(3/2)*x*b^2+5/48/c*(c*x
^2+b*x+a)^(3/2)*b*a-5/192/c^2*(c*x^2+b*x+a)^(3/2)*b^3+5/16*(c*x^2+b*x+a)^(1/2)*x*a^2-5/32/c*(c*x^2+b*x+a)^(1/2
)*x*a*b^2+5/256/c^2*(c*x^2+b*x+a)^(1/2)*x*b^4+5/32/c*(c*x^2+b*x+a)^(1/2)*b*a^2-5/64/c^2*(c*x^2+b*x+a)^(1/2)*b^
3*a+5/512/c^3*(c*x^2+b*x+a)^(1/2)*b^5+5/16/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^3-15/64/c^(3/
2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2*b^2+15/256/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/
2))*b^4*a-5/1024/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^6

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.73123, size = 1000, normalized size = 6.71 \begin{align*} \left [-\frac{15 \,{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right ) - 4 \,{\left (256 \, c^{6} x^{5} + 640 \, b c^{5} x^{4} + 15 \, b^{5} c - 160 \, a b^{3} c^{2} + 528 \, a^{2} b c^{3} + 16 \,{\left (27 \, b^{2} c^{4} + 52 \, a c^{5}\right )} x^{3} + 8 \,{\left (b^{3} c^{3} + 156 \, a b c^{4}\right )} x^{2} - 2 \,{\left (5 \, b^{4} c^{2} - 48 \, a b^{2} c^{3} - 528 \, a^{2} c^{4}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{6144 \, c^{4}}, \frac{15 \,{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \,{\left (256 \, c^{6} x^{5} + 640 \, b c^{5} x^{4} + 15 \, b^{5} c - 160 \, a b^{3} c^{2} + 528 \, a^{2} b c^{3} + 16 \,{\left (27 \, b^{2} c^{4} + 52 \, a c^{5}\right )} x^{3} + 8 \,{\left (b^{3} c^{3} + 156 \, a b c^{4}\right )} x^{2} - 2 \,{\left (5 \, b^{4} c^{2} - 48 \, a b^{2} c^{3} - 528 \, a^{2} c^{4}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{3072 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6144*(15*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*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*(256*c^6*x^5 + 640*b*c^5*x^4 + 15*b^5*c - 160*a*b^3*c^2 + 52
8*a^2*b*c^3 + 16*(27*b^2*c^4 + 52*a*c^5)*x^3 + 8*(b^3*c^3 + 156*a*b*c^4)*x^2 - 2*(5*b^4*c^2 - 48*a*b^2*c^3 - 5
28*a^2*c^4)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/3072*(15*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*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*(256*c^6*x^5 + 640*b*c^5*
x^4 + 15*b^5*c - 160*a*b^3*c^2 + 528*a^2*b*c^3 + 16*(27*b^2*c^4 + 52*a*c^5)*x^3 + 8*(b^3*c^3 + 156*a*b*c^4)*x^
2 - 2*(5*b^4*c^2 - 48*a*b^2*c^3 - 528*a^2*c^4)*x)*sqrt(c*x^2 + b*x + a))/c^4]

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

[Out]

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

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Giac [A]  time = 1.14355, size = 281, normalized size = 1.89 \begin{align*} \frac{1}{1536} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (2 \, c^{2} x + 5 \, b c\right )} x + \frac{27 \, b^{2} c^{5} + 52 \, a c^{6}}{c^{5}}\right )} x + \frac{b^{3} c^{4} + 156 \, a b c^{5}}{c^{5}}\right )} x - \frac{5 \, b^{4} c^{3} - 48 \, a b^{2} c^{4} - 528 \, a^{2} c^{5}}{c^{5}}\right )} x + \frac{15 \, b^{5} c^{2} - 160 \, a b^{3} c^{3} + 528 \, a^{2} b c^{4}}{c^{5}}\right )} + \frac{5 \,{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{1024 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1536*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*c^2*x + 5*b*c)*x + (27*b^2*c^5 + 52*a*c^6)/c^5)*x + (b^3*c^4 + 156
*a*b*c^5)/c^5)*x - (5*b^4*c^3 - 48*a*b^2*c^4 - 528*a^2*c^5)/c^5)*x + (15*b^5*c^2 - 160*a*b^3*c^3 + 528*a^2*b*c
^4)/c^5) + 5/1024*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))*sqrt(c) - b))/c^(7/2)