3.11.36 \(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^3} \, dx\)

Optimal. Leaf size=147 \[ \frac {5 \left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{128 c^{7/2} d^3}-\frac {5 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}{64 c^3 d^3}+\frac {5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2} \]

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Rubi [A]  time = 0.11, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {684, 685, 688, 205} \begin {gather*} -\frac {5 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}{64 c^3 d^3}+\frac {5 \left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{128 c^{7/2} d^3}+\frac {5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^3,x]

[Out]

(-5*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])/(64*c^3*d^3) + (5*(a + b*x + c*x^2)^(3/2))/(48*c^2*d^3) - (a + b*x +
c*x^2)^(5/2)/(4*c*d^3*(b + 2*c*x)^2) + (5*(b^2 - 4*a*c)^(3/2)*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^
2 - 4*a*c]])/(128*c^(7/2)*d^3)

Rule 205

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

Rule 684

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 + 1)), x] - Dist[(b*p)/(d*e*(m + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1
), x], x] /; FreeQ[{a, b, c, d, e}, 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] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0]) && IntegerQ[2*p]

Rule 685

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 688

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)^3} \, dx &=-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}+\frac {5 \int \frac {\left (a+b x+c x^2\right )^{3/2}}{b d+2 c d x} \, dx}{8 c d^2}\\ &=\frac {5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}-\frac {\left (5 \left (b^2-4 a c\right )\right ) \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx}{32 c^2 d^2}\\ &=-\frac {5 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}{64 c^3 d^3}+\frac {5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}+\frac {\left (5 \left (b^2-4 a c\right )^2\right ) \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{128 c^3 d^2}\\ &=-\frac {5 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}{64 c^3 d^3}+\frac {5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}+\frac {\left (5 \left (b^2-4 a c\right )^2\right ) \operatorname {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 )}{32 c^2 d^2}\\ &=-\frac {5 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}{64 c^3 d^3}+\frac {5 \left (a+b x+c x^2\right )^{3/2}}{48 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 c d^3 (b+2 c x)^2}+\frac {5 \left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{128 c^{7/2} d^3}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 62, normalized size = 0.42 \begin {gather*} \frac {2 (a+x (b+c x))^{7/2} \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};\frac {4 c (a+x (b+c x))}{4 a c-b^2}\right )}{7 d^3 \left (b^2-4 a c\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^3,x]

[Out]

(2*(a + x*(b + c*x))^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(7*(b^2 - 4
*a*c)^2*d^3)

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IntegrateAlgebraic [A]  time = 1.48, size = 195, normalized size = 1.33 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (-48 a^2 c^2+80 a b^2 c+224 a b c^2 x+224 a c^3 x^2-15 b^4-40 b^3 c x-8 b^2 c^2 x^2+64 b c^3 x^3+32 c^4 x^4\right )}{192 c^3 d^3 (b+2 c x)^2}-\frac {5 \left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (-\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}+\frac {b}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2} d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^3,x]

[Out]

(Sqrt[a + b*x + c*x^2]*(-15*b^4 + 80*a*b^2*c - 48*a^2*c^2 - 40*b^3*c*x + 224*a*b*c^2*x - 8*b^2*c^2*x^2 + 224*a
*c^3*x^2 + 64*b*c^3*x^3 + 32*c^4*x^4))/(192*c^3*d^3*(b + 2*c*x)^2) - (5*(b^2 - 4*a*c)^(3/2)*ArcTan[b/Sqrt[b^2
- 4*a*c] + (2*c*x)/Sqrt[b^2 - 4*a*c] - (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/(64*c^(7/2)*d^3)

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fricas [A]  time = 0.71, size = 480, normalized size = 3.27 \begin {gather*} \left [-\frac {15 \, {\left (b^{4} - 4 \, a b^{2} c + 4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 4 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x\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 (32 \, c^{4} x^{4} + 64 \, b c^{3} x^{3} - 15 \, b^{4} + 80 \, a b^{2} c - 48 \, a^{2} c^{2} - 8 \, {\left (b^{2} c^{2} - 28 \, a c^{3}\right )} x^{2} - 8 \, {\left (5 \, b^{3} c - 28 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, {\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\right )}}, -\frac {15 \, {\left (b^{4} - 4 \, a b^{2} c + 4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 4 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x\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 (32 \, c^{4} x^{4} + 64 \, b c^{3} x^{3} - 15 \, b^{4} + 80 \, a b^{2} c - 48 \, a^{2} c^{2} - 8 \, {\left (b^{2} c^{2} - 28 \, a c^{3}\right )} x^{2} - 8 \, {\left (5 \, b^{3} c - 28 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, {\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\right )}}\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)^3,x, algorithm="fricas")

[Out]

[-1/768*(15*(b^4 - 4*a*b^2*c + 4*(b^2*c^2 - 4*a*c^3)*x^2 + 4*(b^3*c - 4*a*b*c^2)*x)*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*(32*c^4*x^4 + 64*b*c^3*x^3 - 15*b^4 + 80*a*b^2*c - 48*a^2*c^2 - 8*(b^2*c^2 - 28*a*c^3)*x^2 - 8*(5*
b^3*c - 28*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(4*c^5*d^3*x^2 + 4*b*c^4*d^3*x + b^2*c^3*d^3), -1/384*(15*(b^4 -
4*a*b^2*c + 4*(b^2*c^2 - 4*a*c^3)*x^2 + 4*(b^3*c - 4*a*b*c^2)*x)*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*(32*c^4*x^4 + 64*b*c^3*x^3 - 15*b^4 + 80*a*b^2*c - 48*a^2*c^2 - 8*(b^2*c
^2 - 28*a*c^3)*x^2 - 8*(5*b^3*c - 28*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(4*c^5*d^3*x^2 + 4*b*c^4*d^3*x + b^2*c
^3*d^3)]

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giac [B]  time = 0.36, size = 519, normalized size = 3.53 \begin {gather*} \frac {1}{48} \, \sqrt {c x^{2} + b x + a} {\left (2 \, x {\left (\frac {x}{c d^{3}} + \frac {b}{c^{2} d^{3}}\right )} - \frac {3 \, b^{2} c^{5} d^{9} - 14 \, a c^{6} d^{9}}{c^{8} d^{12}}\right )} + \frac {5 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\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 )}{64 \, \sqrt {b^{2} c - 4 \, a c^{2}} c^{3} d^{3}} + \frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} b^{4} c^{\frac {3}{2}} - 16 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} a b^{2} c^{\frac {5}{2}} + 32 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} a^{2} c^{\frac {7}{2}} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} b^{5} c - 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} a b^{3} c^{2} + 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} a^{2} b c^{3} + {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b^{6} \sqrt {c} - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} a b^{4} c^{\frac {3}{2}} + 32 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} a^{3} c^{\frac {7}{2}} + a b^{5} c - 8 \, a^{2} b^{3} c^{2} + 16 \, a^{3} b c^{3}}{64 \, {\left (2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} c^{\frac {3}{2}} + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b c + b^{2} \sqrt {c} - 2 \, a c^{\frac {3}{2}}\right )}^{2} c^{\frac {5}{2}} d^{3}} \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)^3,x, algorithm="giac")

[Out]

1/48*sqrt(c*x^2 + b*x + a)*(2*x*(x/(c*d^3) + b/(c^2*d^3)) - (3*b^2*c^5*d^9 - 14*a*c^6*d^9)/(c^8*d^12)) + 5/64*
(b^4 - 8*a*b^2*c + 16*a^2*c^2)*arctan(-(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c + b*sqrt(c))/sqrt(b^2*c - 4*a*
c^2))/(sqrt(b^2*c - 4*a*c^2)*c^3*d^3) + 1/64*(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^4*c^(3/2) - 16*(sqrt(c
)*x - sqrt(c*x^2 + b*x + a))^3*a*b^2*c^(5/2) + 32*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^2*c^(7/2) + 3*(sqrt(
c)*x - sqrt(c*x^2 + b*x + a))^2*b^5*c - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b^3*c^2 + 48*(sqrt(c)*x - s
qrt(c*x^2 + b*x + a))^2*a^2*b*c^3 + (sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^6*sqrt(c) - 6*(sqrt(c)*x - sqrt(c*x^
2 + b*x + a))*a*b^4*c^(3/2) + 32*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^3*c^(7/2) + a*b^5*c - 8*a^2*b^3*c^2 + 1
6*a^3*b*c^3)/((2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c^(3/2) + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b*c + b
^2*sqrt(c) - 2*a*c^(3/2))^2*c^(5/2)*d^3)

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maple [B]  time = 0.06, size = 840, normalized size = 5.71 \begin {gather*} -\frac {5 a^{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 )}{2 \left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}\, c \,d^{3}}+\frac {15 a^{2} b^{2} \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 )}{8 \left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{2} d^{3}}-\frac {15 a \,b^{4} \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 )}{32 \left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{3} d^{3}}+\frac {5 b^{6} \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 )}{128 \left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{4} d^{3}}+\frac {5 \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, a^{2}}{8 \left (4 a c -b^{2}\right ) c \,d^{3}}-\frac {5 \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, a \,b^{2}}{16 \left (4 a c -b^{2}\right ) c^{2} d^{3}}+\frac {5 \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, b^{4}}{128 \left (4 a c -b^{2}\right ) c^{3} d^{3}}+\frac {5 \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}} a}{12 \left (4 a c -b^{2}\right ) c \,d^{3}}-\frac {5 \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}} b^{2}}{48 \left (4 a c -b^{2}\right ) c^{2} d^{3}}+\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{4 \left (4 a c -b^{2}\right ) c \,d^{3}}-\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {7}{2}}}{4 \left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2} c^{2} d^{3}} \end {gather*}

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)^3,x)

[Out]

-1/4/d^3/c^2/(4*a*c-b^2)/(x+1/2*b/c)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)+1/4/d^3/c/(4*a*c-b^2)*((x+1/2
*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+5/12/d^3/c/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*a-5/48/d^3
/c^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b^2+5/8/d^3/c/(4*a*c-b^2)*(4*(x+1/2*b/c)^2*c+(4*a*c
-b^2)/c)^(1/2)*a^2-5/16/d^3/c^2/(4*a*c-b^2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a*b^2+5/128/d^3/c^3/(4*a*c
-b^2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*b^4-5/2/d^3/c/(4*a*c-b^2)/((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))*a^3+15/8/d^3/c^2/(4*a*c-
b^2)/((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))*a^2*b^2-15/32/d^3/c^3/(4*a*c-b^2)/((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))*a*b^4+5/128/d^3/c^4/(4*a*c-b^2)/((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
))*b^6

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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)^3,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 zero or nonzero?

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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}}{{\left (b\,d+2\,c\,d\,x\right )}^3} \,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)^3,x)

[Out]

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^3, x)

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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^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {b^{2} x^{2} \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {c^{2} x^{4} \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {2 a b x \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {2 a c x^{2} \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {2 b c x^{3} \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx}{d^{3}} \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)**3,x)

[Out]

(Integral(a**2*sqrt(a + b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(b**2*x
**2*sqrt(a + b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(c**2*x**4*sqrt(a
+ b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(2*a*b*x*sqrt(a + b*x + c*x**
2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(2*a*c*x**2*sqrt(a + b*x + c*x**2)/(b**3 +
6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(2*b*c*x**3*sqrt(a + b*x + c*x**2)/(b**3 + 6*b**2*c*
x + 12*b*c**2*x**2 + 8*c**3*x**3), x))/d**3

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