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

Optimal. Leaf size=197 \[ \frac {\sqrt {a x^2+b x+c} \left (256 a^5 x^5+640 a^4 b x^4+832 a^4 c x^3+432 a^3 b^2 x^3+1248 a^3 b c x^2+1056 a^3 c^2 x+8 a^2 b^3 x^2+96 a^2 b^2 c x+528 a^2 b c^2-10 a b^4 x-160 a b^3 c+15 b^5\right )}{1536 a^3}-\frac {5 \left (64 a^3 c^3-48 a^2 b^2 c^2+12 a b^4 c-b^6\right ) \log \left (-2 \sqrt {a} \sqrt {a x^2+b x+c}+2 a x+b\right )}{1024 a^{7/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 149, normalized size of antiderivative = 0.76, 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} \begin {gather*} -\frac {5 \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac {2 a x+b}{2 \sqrt {a} \sqrt {a x^2+b x+c}}\right )}{1024 a^{7/2}}+\frac {5 \left (b^2-4 a c\right )^2 (2 a x+b) \sqrt {a x^2+b x+c}}{512 a^3}-\frac {5 \left (b^2-4 a c\right ) (2 a x+b) \left (a x^2+b x+c\right )^{3/2}}{192 a^2}+\frac {(2 a x+b) \left (a x^2+b x+c\right )^{5/2}}{12 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + b*x + a*x^2)^(5/2),x]

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.64, size = 162, normalized size = 0.82 \begin {gather*} \frac {\sqrt {x (a x+b)+c} \left (2 (2 a x+b) \left (32 a^2 b x \left (8 a x^2+13 c\right )+16 a^2 \left (8 a^2 x^4+26 a c x^2+33 c^2\right )-40 a b^3 x+8 a b^2 \left (11 a x^2-20 c\right )+15 b^4\right )+\frac {15 \left (b^2-4 a c\right )^{5/2} \sin ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{\sqrt {\frac {a (x (a x+b)+c)}{4 a c-b^2}}}\right )}{3072 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + b*x + a*x^2)^(5/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.66, size = 197, normalized size = 1.00 \begin {gather*} \frac {\sqrt {c+b x+a x^2} \left (15 b^5-160 a b^3 c+528 a^2 b c^2-10 a b^4 x+96 a^2 b^2 c x+1056 a^3 c^2 x+8 a^2 b^3 x^2+1248 a^3 b c x^2+432 a^3 b^2 x^3+832 a^4 c x^3+640 a^4 b x^4+256 a^5 x^5\right )}{1536 a^3}-\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 (b+2 a x-2 \sqrt {a} \sqrt {c+b x+a x^2}\right )}{1024 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + b*x + a*x^2)^(5/2),x]

[Out]

(Sqrt[c + b*x + a*x^2]*(15*b^5 - 160*a*b^3*c + 528*a^2*b*c^2 - 10*a*b^4*x + 96*a^2*b^2*c*x + 1056*a^3*c^2*x +
8*a^2*b^3*x^2 + 1248*a^3*b*c*x^2 + 432*a^3*b^2*x^3 + 832*a^4*c*x^3 + 640*a^4*b*x^4 + 256*a^5*x^5))/(1536*a^3)
- (5*(-b^6 + 12*a*b^4*c - 48*a^2*b^2*c^2 + 64*a^3*c^3)*Log[b + 2*a*x - 2*Sqrt[a]*Sqrt[c + b*x + a*x^2]])/(1024
*a^(7/2))

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fricas [A]  time = 0.53, size = 425, normalized size = 2.16 \begin {gather*} \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 {a} \log \left (-8 \, a^{2} x^{2} - 8 \, a b x - 4 \, \sqrt {a x^{2} + b x + c} {\left (2 \, a x + b\right )} \sqrt {a} - b^{2} - 4 \, a c\right ) - 4 \, {\left (256 \, a^{6} x^{5} + 640 \, a^{5} b x^{4} + 15 \, a b^{5} - 160 \, a^{2} b^{3} c + 528 \, a^{3} b c^{2} + 16 \, {\left (27 \, a^{4} b^{2} + 52 \, a^{5} c\right )} x^{3} + 8 \, {\left (a^{3} b^{3} + 156 \, a^{4} b c\right )} x^{2} - 2 \, {\left (5 \, a^{2} b^{4} - 48 \, a^{3} b^{2} c - 528 \, a^{4} c^{2}\right )} x\right )} \sqrt {a x^{2} + b x + c}}{6144 \, a^{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 {-a} \arctan \left (\frac {\sqrt {a x^{2} + b x + c} {\left (2 \, a x + b\right )} \sqrt {-a}}{2 \, {\left (a^{2} x^{2} + a b x + a c\right )}}\right ) + 2 \, {\left (256 \, a^{6} x^{5} + 640 \, a^{5} b x^{4} + 15 \, a b^{5} - 160 \, a^{2} b^{3} c + 528 \, a^{3} b c^{2} + 16 \, {\left (27 \, a^{4} b^{2} + 52 \, a^{5} c\right )} x^{3} + 8 \, {\left (a^{3} b^{3} + 156 \, a^{4} b c\right )} x^{2} - 2 \, {\left (5 \, a^{2} b^{4} - 48 \, a^{3} b^{2} c - 528 \, a^{4} c^{2}\right )} x\right )} \sqrt {a x^{2} + b x + c}}{3072 \, a^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.49, size = 208, normalized size = 1.06 \begin {gather*} \frac {1}{1536} \, \sqrt {a x^{2} + b x + c} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, a^{2} x + 5 \, a b\right )} x + \frac {27 \, a^{5} b^{2} + 52 \, a^{6} c}{a^{5}}\right )} x + \frac {a^{4} b^{3} + 156 \, a^{5} b c}{a^{5}}\right )} x - \frac {5 \, a^{3} b^{4} - 48 \, a^{4} b^{2} c - 528 \, a^{5} c^{2}}{a^{5}}\right )} x + \frac {15 \, a^{2} b^{5} - 160 \, a^{3} b^{3} c + 528 \, a^{4} b c^{2}}{a^{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 {a} x - \sqrt {a x^{2} + b x + c}\right )} \sqrt {a} - b \right |}\right )}{1024 \, a^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.30, size = 143, normalized size = 0.73

method	result	size
default	\(\frac {\left (2 a x +b \right ) \left (a \,x^{2}+b x +c \right )^{\frac {5}{2}}}{12 a}+\frac {5 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 a x +b \right ) \left (a \,x^{2}+b x +c \right )^{\frac {3}{2}}}{8 a}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 a x +b \right ) \sqrt {a \,x^{2}+b x +c}}{4 a}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x +c}\right )}{8 a^{\frac {3}{2}}}\right )}{16 a}\right )}{24 a}\)	\(143\)
risch	\(\frac {\sqrt {a \,x^{2}+b x +c}\, \left (256 a^{5} x^{5}+640 a^{4} b \,x^{4}+832 a^{4} c \,x^{3}+432 a^{3} b^{2} x^{3}+1248 a^{3} b c \,x^{2}+8 a^{2} b^{3} x^{2}+1056 a^{3} c^{2} x +96 a^{2} b^{2} c x -10 a \,b^{4} x +528 a^{2} b \,c^{2}-160 a \,b^{3} c +15 b^{5}\right )}{1536 a^{3}}+\frac {5 \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x +c}\right ) c^{3}}{16 \sqrt {a}}-\frac {15 \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x +c}\right ) b^{2} c^{2}}{64 a^{\frac {3}{2}}}+\frac {15 \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x +c}\right ) b^{4} c}{256 a^{\frac {5}{2}}}-\frac {5 \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x +c}\right ) b^{6}}{1024 a^{\frac {7}{2}}}\)	\(261\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^2+b*x+c)^(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 details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 1.76, size = 143, normalized size = 0.73 \begin {gather*} \frac {\left (\frac {b}{2}+a\,x\right )\,{\left (a\,x^2+b\,x+c\right )}^{5/2}}{6\,a}+\frac {\left (5\,a\,c-\frac {5\,b^2}{4}\right )\,\left (\frac {\left (\left (\frac {x}{2}+\frac {b}{4\,a}\right )\,\sqrt {a\,x^2+b\,x+c}+\frac {\ln \left (\frac {\frac {b}{2}+a\,x}{\sqrt {a}}+\sqrt {a\,x^2+b\,x+c}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,a^{3/2}}\right )\,\left (3\,a\,c-\frac {3\,b^2}{4}\right )}{4\,a}+\frac {\left (\frac {b}{2}+a\,x\right )\,{\left (a\,x^2+b\,x+c\right )}^{3/2}}{4\,a}\right )}{6\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + b*x + a*x^2)^(5/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a x^{2} + b x + c\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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