3.13.22 \(\int \frac {(a+b x+c x^2)^{5/2}}{b d+2 c d x} \, dx\) [1222]

Optimal. Leaf size=149 \[ \frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^{5/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2} d} \]

[Out]

-1/24*(-4*a*c+b^2)*(c*x^2+b*x+a)^(3/2)/c^2/d+1/10*(c*x^2+b*x+a)^(5/2)/c/d-1/64*(-4*a*c+b^2)^(5/2)*arctan(2*c^(
1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))/c^(7/2)/d+1/32*(-4*a*c+b^2)^2*(c*x^2+b*x+a)^(1/2)/c^3/d

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Rubi [A]
time = 0.08, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {699, 702, 211} \begin {gather*} -\frac {\left (b^2-4 a c\right )^{5/2} \text {ArcTan}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2} d}+\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2])/(32*c^3*d) - ((b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2))/(24*c^2*d) + (a +
b*x + c*x^2)^(5/2)/(10*c*d) - ((b^2 - 4*a*c)^(5/2)*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]
])/(64*c^(7/2)*d)

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 699

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[d*p*((b^2 - 4*a*c)/(b*e*(m + 2*p + 1))), Int[(d + e*x)^m*(a +
b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] &&
NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

Rule 702

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx &=\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right ) \int \frac {\left (a+b x+c x^2\right )^{3/2}}{b d+2 c d x} \, dx}{4 c}\\ &=-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}+\frac {\left (b^2-4 a c\right )^2 \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx}{16 c^2}\\ &=\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^3 \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{64 c^3}\\ &=\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^3 \text {Subst}\left (\int \frac {1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt {a+b x+c x^2}\right )}{16 c^2}\\ &=\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^{5/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2} d}\\ \end {align*}

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Mathematica [A]
time = 0.69, size = 157, normalized size = 1.05 \begin {gather*} \frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (15 b^4-20 b^3 c x+28 b^2 c \left (-5 a+c x^2\right )+16 b c^2 x \left (11 a+6 c x^2\right )+16 c^2 \left (23 a^2+11 a c x^2+3 c^2 x^4\right )\right )+15 \left (b^2-4 a c\right )^{5/2} \tan ^{-1}\left (\frac {b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}}{\sqrt {b^2-4 a c}}\right )}{480 c^{7/2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*Sqrt[a + x*(b + c*x)]*(15*b^4 - 20*b^3*c*x + 28*b^2*c*(-5*a + c*x^2) + 16*b*c^2*x*(11*a + 6*c*x^2) +
16*c^2*(23*a^2 + 11*a*c*x^2 + 3*c^2*x^4)) + 15*(b^2 - 4*a*c)^(5/2)*ArcTan[(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b
+ c*x)])/Sqrt[b^2 - 4*a*c]])/(480*c^(7/2)*d)

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Maple [A]
time = 0.73, size = 245, normalized size = 1.64

method	result	size
default	\(\frac {\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{5}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{3}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{2 c \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 c}\right )}{4 c}}{2 d c}\)	\(245\)
risch	\(\frac {\left (48 c^{4} x^{4}+96 b \,c^{3} x^{3}+176 x^{2} c^{3} a +28 b^{2} c^{2} x^{2}+176 x a b \,c^{2}-20 b^{3} c x +368 a^{2} c^{2}-140 a c \,b^{2}+15 b^{4}\right ) \sqrt {c \,x^{2}+b x +a}}{480 c^{3} d}+\frac {-\frac {\ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right ) a^{3}}{c \sqrt {\frac {4 a c -b^{2}}{c}}}+\frac {3 \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right ) a^{2} b^{2}}{4 c^{2} \sqrt {\frac {4 a c -b^{2}}{c}}}-\frac {3 \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right ) a \,b^{4}}{16 c^{3} \sqrt {\frac {4 a c -b^{2}}{c}}}+\frac {\ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right ) b^{6}}{64 c^{4} \sqrt {\frac {4 a c -b^{2}}{c}}}}{d}\)	\(504\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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Fricas [A]
time = 3.28, size = 372, normalized size = 2.50 \begin {gather*} \left [\frac {15 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-\frac {b^{2} - 4 \, a c}{c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {c x^{2} + b x + a} c \sqrt {-\frac {b^{2} - 4 \, a c}{c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + 4 \, {\left (48 \, c^{4} x^{4} + 96 \, b c^{3} x^{3} + 15 \, b^{4} - 140 \, a b^{2} c + 368 \, a^{2} c^{2} + 4 \, {\left (7 \, b^{2} c^{2} + 44 \, a c^{3}\right )} x^{2} - 4 \, {\left (5 \, b^{3} c - 44 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1920 \, c^{3} d}, \frac {15 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {\frac {b^{2} - 4 \, a c}{c}} \arctan \left (\frac {\sqrt {\frac {b^{2} - 4 \, a c}{c}}}{2 \, \sqrt {c x^{2} + b x + a}}\right ) + 2 \, {\left (48 \, c^{4} x^{4} + 96 \, b c^{3} x^{3} + 15 \, b^{4} - 140 \, a b^{2} c + 368 \, a^{2} c^{2} + 4 \, {\left (7 \, b^{2} c^{2} + 44 \, a c^{3}\right )} x^{2} - 4 \, {\left (5 \, b^{3} c - 44 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{960 \, c^{3} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1920*(15*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-(b^2 - 4*a*c)/c)*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*
sqrt(c*x^2 + b*x + a)*c*sqrt(-(b^2 - 4*a*c)/c))/(4*c^2*x^2 + 4*b*c*x + b^2)) + 4*(48*c^4*x^4 + 96*b*c^3*x^3 +
15*b^4 - 140*a*b^2*c + 368*a^2*c^2 + 4*(7*b^2*c^2 + 44*a*c^3)*x^2 - 4*(5*b^3*c - 44*a*b*c^2)*x)*sqrt(c*x^2 + b
*x + a))/(c^3*d), 1/960*(15*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt((b^2 - 4*a*c)/c)*arctan(1/2*sqrt((b^2 - 4*a*c)
/c)/sqrt(c*x^2 + b*x + a)) + 2*(48*c^4*x^4 + 96*b*c^3*x^3 + 15*b^4 - 140*a*b^2*c + 368*a^2*c^2 + 4*(7*b^2*c^2
+ 44*a*c^3)*x^2 - 4*(5*b^3*c - 44*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c^3*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {b^{2} x^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {c^{2} x^{4} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 a b x \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 a c x^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 b c x^{3} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2*sqrt(a + b*x + c*x**2)/(b + 2*c*x), x) + Integral(b**2*x**2*sqrt(a + b*x + c*x**2)/(b + 2*c*x),
x) + Integral(c**2*x**4*sqrt(a + b*x + c*x**2)/(b + 2*c*x), x) + Integral(2*a*b*x*sqrt(a + b*x + c*x**2)/(b +
2*c*x), x) + Integral(2*a*c*x**2*sqrt(a + b*x + c*x**2)/(b + 2*c*x), x) + Integral(2*b*c*x**3*sqrt(a + b*x +
c*x**2)/(b + 2*c*x), x))/d

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Giac [A]
time = 1.96, size = 237, normalized size = 1.59 \begin {gather*} \frac {1}{480} \, \sqrt {c x^{2} + b x + a} {\left (4 \, {\left ({\left (12 \, {\left (\frac {c x}{d} + \frac {2 \, b}{d}\right )} x + \frac {7 \, b^{2} c^{9} d^{5} + 44 \, a c^{10} d^{5}}{c^{10} d^{6}}\right )} x - \frac {5 \, b^{3} c^{8} d^{5} - 44 \, a b c^{9} d^{5}}{c^{10} d^{6}}\right )} x + \frac {15 \, b^{4} c^{7} d^{5} - 140 \, a b^{2} c^{8} d^{5} + 368 \, a^{2} c^{9} d^{5}}{c^{10} d^{6}}\right )} - \frac {{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \arctan \left (-\frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} c + b \sqrt {c}}{\sqrt {b^{2} c - 4 \, a c^{2}}}\right )}{32 \, \sqrt {b^{2} c - 4 \, a c^{2}} c^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*sqrt(c*x^2 + b*x + a)*(4*((12*(c*x/d + 2*b/d)*x + (7*b^2*c^9*d^5 + 44*a*c^10*d^5)/(c^10*d^6))*x - (5*b^3
*c^8*d^5 - 44*a*b*c^9*d^5)/(c^10*d^6))*x + (15*b^4*c^7*d^5 - 140*a*b^2*c^8*d^5 + 368*a^2*c^9*d^5)/(c^10*d^6))
- 1/32*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*arctan(-(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c + b*s
qrt(c))/sqrt(b^2*c - 4*a*c^2))/(sqrt(b^2*c - 4*a*c^2)*c^3*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{b\,d+2\,c\,d\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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