\(\int (a+b x+c x^2)^{5/2} \, dx\) [2358]

Optimal result

Integrand size = 14, antiderivative size = 149 \[ \int \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2}}{512 c^3}-\frac {5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{12 c}-\frac {5 \left (b^2-4 a c\right )^3 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{1024 c^{7/2}} \]

[Out]

-5/192*(-4*a*c+b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c^2+1/12*(2*c*x+b)*(c*x^2+b*x+a)^(5/2)/c-5/1024*(-4*a*c+b^2)
^3*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(7/2)+5/512*(-4*a*c+b^2)^2*(2*c*x+b)*(c*x^2+b*x+a)^(1/
2)/c^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {626, 635, 212} \[ \int \left (a+b x+c x^2\right )^{5/2} \, dx=-\frac {5 \left (b^2-4 a c\right )^3 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{1024 c^{7/2}}+\frac {5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2}}{512 c^3}-\frac {5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{12 c} \]

[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 212

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

Rule 626

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 635

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]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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] (verified)

Time = 1.33 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.01 \[ \int \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {\sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)} \left (15 b^4-40 b^3 c x+32 b c^2 x \left (13 a+8 c x^2\right )+8 b^2 c \left (-20 a+11 c x^2\right )+16 c^2 \left (33 a^2+26 a c x^2+8 c^2 x^4\right )\right )-15 \left (b^2-4 a c\right )^3 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{1536 c^{7/2}} \]

[In]

Integrate[(a + b*x + c*x^2)^(5/2),x]

[Out]

(Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b + 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)^3*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] +
Sqrt[a + x*(b + c*x)])])/(1536*c^(7/2))

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{12 c}+\frac {5 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\)	\(143\)
risch	\(\frac {\left (256 c^{5} x^{5}+640 b \,c^{4} x^{4}+832 a \,c^{4} x^{3}+432 b^{2} c^{3} x^{3}+1248 a b \,c^{3} x^{2}+8 b^{3} c^{2} x^{2}+1056 a^{2} c^{3} x +96 a \,b^{2} c^{2} x -10 b^{4} c x +528 a^{2} b \,c^{2}-160 a \,b^{3} c +15 b^{5}\right ) \sqrt {c \,x^{2}+b x +a}}{1536 c^{3}}+\frac {5 \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{1024 c^{\frac {7}{2}}}\)	\(187\)

[In]

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

[Out]

1/12*(2*c*x+b)*(c*x^2+b*x+a)^(5/2)/c+5/24*(4*a*c-b^2)/c*(1/8*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c+3/16*(4*a*c-b^2)/
c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.85 \[ \int \left (a+b x+c x^2\right )^{5/2} \, dx=\left [-\frac {15 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (256 \, c^{6} x^{5} + 640 \, b c^{5} x^{4} + 15 \, b^{5} c - 160 \, a b^{3} c^{2} + 528 \, a^{2} b c^{3} + 16 \, {\left (27 \, b^{2} c^{4} + 52 \, a c^{5}\right )} x^{3} + 8 \, {\left (b^{3} c^{3} + 156 \, a b c^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{4} c^{2} - 48 \, a b^{2} c^{3} - 528 \, a^{2} c^{4}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{6144 \, c^{4}}, \frac {15 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (256 \, c^{6} x^{5} + 640 \, b c^{5} x^{4} + 15 \, b^{5} c - 160 \, a b^{3} c^{2} + 528 \, a^{2} b c^{3} + 16 \, {\left (27 \, b^{2} c^{4} + 52 \, a c^{5}\right )} x^{3} + 8 \, {\left (b^{3} c^{3} + 156 \, a b c^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{4} c^{2} - 48 \, a b^{2} c^{3} - 528 \, a^{2} c^{4}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3072 \, c^{4}}\right ] \]

[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]

Sympy [A] (verification not implemented)

Time = 1.39 (sec) , antiderivative size = 1732, normalized size of antiderivative = 11.62 \[ \int \left (a+b x+c x^2\right )^{5/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

a**2*Piecewise(((a/2 - b**2/(8*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a
- b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + (b/(4*c) + x/2)*sqrt(a +
b*x + c*x**2), Ne(c, 0)), (2*(a + b*x)**(3/2)/(3*b), Ne(b, 0)), (sqrt(a)*x, True)) + 2*a*b*Piecewise(((-a*b/(
12*c) - b*(a/3 - b**2/(8*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a
- b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + sqrt(a + b*x + c*x**2)*
(b*x/(12*c) + x**2/3 + (a/3 - b**2/(8*c))/c), Ne(c, 0)), (2*(-a*(a + b*x)**(3/2)/3 + (a + b*x)**(5/2)/5)/b**2,
Ne(b, 0)), (sqrt(a)*x**2/2, True)) + 2*a*c*Piecewise(((-a*(a/4 - 5*b**2/(48*c))/(2*c) - b*(-a*b/(12*c) - 3*b*
(a/4 - 5*b**2/(48*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a
- b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + sqrt(a + b*x + c*x**2)*
(b*x**2/(24*c) + x**3/4 + x*(a/4 - 5*b**2/(48*c))/(2*c) + (-a*b/(12*c) - 3*b*(a/4 - 5*b**2/(48*c))/(4*c))/c),
Ne(c, 0)), (2*(a**2*(a + b*x)**(3/2)/3 - 2*a*(a + b*x)**(5/2)/5 + (a + b*x)**(7/2)/7)/b**3, Ne(b, 0)), (sqrt(a
)*x**3/3, True)) + b**2*Piecewise(((-a*(a/4 - 5*b**2/(48*c))/(2*c) - b*(-a*b/(12*c) - 3*b*(a/4 - 5*b**2/(48*c)
)/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/(4*c), 0)),
((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + sqrt(a + b*x + c*x**2)*(b*x**2/(24*c) + x**
3/4 + x*(a/4 - 5*b**2/(48*c))/(2*c) + (-a*b/(12*c) - 3*b*(a/4 - 5*b**2/(48*c))/(4*c))/c), Ne(c, 0)), (2*(a**2*
(a + b*x)**(3/2)/3 - 2*a*(a + b*x)**(5/2)/5 + (a + b*x)**(7/2)/7)/b**3, Ne(b, 0)), (sqrt(a)*x**3/3, True)) + 2
*b*c*Piecewise(((-a*(-3*a*b/(40*c) - 5*b*(a/5 - 7*b**2/(80*c))/(6*c))/(2*c) - b*(-2*a*(a/5 - 7*b**2/(80*c))/(3
*c) - 3*b*(-3*a*b/(40*c) - 5*b*(a/5 - 7*b**2/(80*c))/(6*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a
+ b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)
**2), True)) + sqrt(a + b*x + c*x**2)*(b*x**3/(40*c) + x**4/5 + x**2*(a/5 - 7*b**2/(80*c))/(3*c) + x*(-3*a*b/(
40*c) - 5*b*(a/5 - 7*b**2/(80*c))/(6*c))/(2*c) + (-2*a*(a/5 - 7*b**2/(80*c))/(3*c) - 3*b*(-3*a*b/(40*c) - 5*b*
(a/5 - 7*b**2/(80*c))/(6*c))/(4*c))/c), Ne(c, 0)), (2*(-a**3*(a + b*x)**(3/2)/3 + 3*a**2*(a + b*x)**(5/2)/5 -
3*a*(a + b*x)**(7/2)/7 + (a + b*x)**(9/2)/9)/b**4, Ne(b, 0)), (sqrt(a)*x**4/4, True)) + c**2*Piecewise(((-a*(-
3*a*(a/6 - 3*b**2/(40*c))/(4*c) - 5*b*(-a*b/(15*c) - 7*b*(a/6 - 3*b**2/(40*c))/(8*c))/(6*c))/(2*c) - b*(-2*a*(
-a*b/(15*c) - 7*b*(a/6 - 3*b**2/(40*c))/(8*c))/(3*c) - 3*b*(-3*a*(a/6 - 3*b**2/(40*c))/(4*c) - 5*b*(-a*b/(15*c
) - 7*b*(a/6 - 3*b**2/(40*c))/(8*c))/(6*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2)
+ 2*c*x)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + s
qrt(a + b*x + c*x**2)*(b*x**4/(60*c) + x**5/6 + x**3*(a/6 - 3*b**2/(40*c))/(4*c) + x**2*(-a*b/(15*c) - 7*b*(a/
6 - 3*b**2/(40*c))/(8*c))/(3*c) + x*(-3*a*(a/6 - 3*b**2/(40*c))/(4*c) - 5*b*(-a*b/(15*c) - 7*b*(a/6 - 3*b**2/(
40*c))/(8*c))/(6*c))/(2*c) + (-2*a*(-a*b/(15*c) - 7*b*(a/6 - 3*b**2/(40*c))/(8*c))/(3*c) - 3*b*(-3*a*(a/6 - 3*
b**2/(40*c))/(4*c) - 5*b*(-a*b/(15*c) - 7*b*(a/6 - 3*b**2/(40*c))/(8*c))/(6*c))/(4*c))/c), Ne(c, 0)), (2*(a**4
*(a + b*x)**(3/2)/3 - 4*a**3*(a + b*x)**(5/2)/5 + 6*a**2*(a + b*x)**(7/2)/7 - 4*a*(a + b*x)**(9/2)/9 + (a + b*
x)**(11/2)/11)/b**5, Ne(b, 0)), (sqrt(a)*x**5/5, True))

Maxima [F(-2)]

Exception generated. \[ \int \left (a+b x+c x^2\right )^{5/2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.38 \[ \int \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {1}{1536} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, c^{2} x + 5 \, b c\right )} x + \frac {27 \, b^{2} c^{5} + 52 \, a c^{6}}{c^{5}}\right )} x + \frac {b^{3} c^{4} + 156 \, a b c^{5}}{c^{5}}\right )} x - \frac {5 \, b^{4} c^{3} - 48 \, a b^{2} c^{4} - 528 \, a^{2} c^{5}}{c^{5}}\right )} x + \frac {15 \, b^{5} c^{2} - 160 \, a b^{3} c^{3} + 528 \, a^{2} b c^{4}}{c^{5}}\right )} + \frac {5 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{1024 \, c^{\frac {7}{2}}} \]

[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)

Mupad [B] (verification not implemented)

Time = 10.09 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.96 \[ \int \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {\left (\frac {b}{2}+c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{6\,c}+\frac {\left (\frac {\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )\,\left (3\,a\,c-\frac {3\,b^2}{4}\right )}{4\,c}+\frac {\left (\frac {b}{2}+c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}\right )\,\left (5\,a\,c-\frac {5\,b^2}{4}\right )}{6\,c} \]

[In]

int((a + b*x + c*x^2)^(5/2),x)

[Out]

((b/2 + c*x)*(a + b*x + c*x^2)^(5/2))/(6*c) + (((((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c
^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2)))*(3*a*c - (3*b^2)/4))/(4*c) + ((b/2 + c*x)*(a + b
*x + c*x^2)^(3/2))/(4*c))*(5*a*c - (5*b^2)/4))/(6*c)