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

Optimal. Leaf size=147 \[ -\frac {15 \sqrt {b^2-4 a c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{512 c^{7/2} d^5}+\frac {15 \sqrt {a+b x+c x^2}}{256 c^3 d^5}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4} \]

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Rubi [A]  time = 0.10, 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 {15 \sqrt {b^2-4 a c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{512 c^{7/2} d^5}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}+\frac {15 \sqrt {a+b x+c x^2}}{256 c^3 d^5}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Sqrt[a + b*x + c*x^2])/(256*c^3*d^5) - (5*(a + b*x + c*x^2)^(3/2))/(64*c^2*d^5*(b + 2*c*x)^2) - (a + b*x +
c*x^2)^(5/2)/(8*c*d^5*(b + 2*c*x)^4) - (15*Sqrt[b^2 - 4*a*c]*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^
2 - 4*a*c]])/(512*c^(7/2)*d^5)

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)^5} \, dx &=-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4}+\frac {5 \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^3} \, dx}{16 c d^2}\\ &=-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4}+\frac {15 \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx}{128 c^2 d^4}\\ &=\frac {15 \sqrt {a+b x+c x^2}}{256 c^3 d^5}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4}-\frac {\left (15 \left (b^2-4 a c\right )\right ) \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{512 c^3 d^4}\\ &=\frac {15 \sqrt {a+b x+c x^2}}{256 c^3 d^5}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4}-\frac {\left (15 \left (b^2-4 a c\right )\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 )}{128 c^2 d^4}\\ &=\frac {15 \sqrt {a+b x+c x^2}}{256 c^3 d^5}-\frac {5 \left (a+b x+c x^2\right )^{3/2}}{64 c^2 d^5 (b+2 c x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{8 c d^5 (b+2 c x)^4}-\frac {15 \sqrt {b^2-4 a c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{512 c^{7/2} d^5}\\ \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 (3,\frac {7}{2};\frac {9}{2};\frac {4 c (a+x (b+c x))}{4 a c-b^2}\right )}{7 d^5 \left (b^2-4 a c\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + x*(b + c*x))^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(7*(b^2 - 4
*a*c)^3*d^5)

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IntegrateAlgebraic [A]  time = 2.48, size = 195, normalized size = 1.33 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (-32 a^2 c^2-20 a b^2 c-144 a b c^2 x-144 a c^3 x^2+15 b^4+100 b^3 c x+228 b^2 c^2 x^2+256 b c^3 x^3+128 c^4 x^4\right )}{256 c^3 d^5 (b+2 c x)^4}+\frac {15 \sqrt {b^2-4 a c} \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 )}{256 c^{7/2} d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(15*b^4 - 20*a*b^2*c - 32*a^2*c^2 + 100*b^3*c*x - 144*a*b*c^2*x + 228*b^2*c^2*x^2 - 144
*a*c^3*x^2 + 256*b*c^3*x^3 + 128*c^4*x^4))/(256*c^3*d^5*(b + 2*c*x)^4) + (15*Sqrt[b^2 - 4*a*c]*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]])/(256*c^(7/2)*d
^5)

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fricas [A]  time = 1.53, size = 524, normalized size = 3.56 \begin {gather*} \left [\frac {15 \, {\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\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 (128 \, c^{4} x^{4} + 256 \, b c^{3} x^{3} + 15 \, b^{4} - 20 \, a b^{2} c - 32 \, a^{2} c^{2} + 12 \, {\left (19 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{2} + 4 \, {\left (25 \, b^{3} c - 36 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1024 \, {\left (16 \, c^{7} d^{5} x^{4} + 32 \, b c^{6} d^{5} x^{3} + 24 \, b^{2} c^{5} d^{5} x^{2} + 8 \, b^{3} c^{4} d^{5} x + b^{4} c^{3} d^{5}\right )}}, \frac {15 \, {\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\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 (128 \, c^{4} x^{4} + 256 \, b c^{3} x^{3} + 15 \, b^{4} - 20 \, a b^{2} c - 32 \, a^{2} c^{2} + 12 \, {\left (19 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{2} + 4 \, {\left (25 \, b^{3} c - 36 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{512 \, {\left (16 \, c^{7} d^{5} x^{4} + 32 \, b c^{6} d^{5} x^{3} + 24 \, b^{2} c^{5} d^{5} x^{2} + 8 \, b^{3} c^{4} d^{5} x + b^{4} c^{3} d^{5}\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)^5,x, algorithm="fricas")

[Out]

[1/1024*(15*(16*c^4*x^4 + 32*b*c^3*x^3 + 24*b^2*c^2*x^2 + 8*b^3*c*x + b^4)*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*(128*c^4*x^4 + 256*b*c^3*x^3 + 15*b^4 - 20*a*b^2*c - 32*a^2*c^2 + 12*(19*b^2*c^2 - 12*a*c^3)*x^2 + 4*(25*b^
3*c - 36*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/(16*c^7*d^5*x^4 + 32*b*c^6*d^5*x^3 + 24*b^2*c^5*d^5*x^2 + 8*b^3*c^
4*d^5*x + b^4*c^3*d^5), 1/512*(15*(16*c^4*x^4 + 32*b*c^3*x^3 + 24*b^2*c^2*x^2 + 8*b^3*c*x + b^4)*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*(128*c^4*x^4 + 256*b*c^3*x^3 + 15*b^4 - 2
0*a*b^2*c - 32*a^2*c^2 + 12*(19*b^2*c^2 - 12*a*c^3)*x^2 + 4*(25*b^3*c - 36*a*b*c^2)*x)*sqrt(c*x^2 + b*x + a))/
(16*c^7*d^5*x^4 + 32*b*c^6*d^5*x^3 + 24*b^2*c^5*d^5*x^2 + 8*b^3*c^4*d^5*x + b^4*c^3*d^5)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)^5,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.06, size = 900, normalized size = 6.12 \begin {gather*} -\frac {15 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 )}{8 \left (4 a c -b^{2}\right )^{2} \sqrt {\frac {4 a c -b^{2}}{c}}\, c \,d^{5}}+\frac {45 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 )}{32 \left (4 a c -b^{2}\right )^{2} \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{2} d^{5}}-\frac {45 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 )}{128 \left (4 a c -b^{2}\right )^{2} \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{3} d^{5}}+\frac {15 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 )}{512 \left (4 a c -b^{2}\right )^{2} \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{4} d^{5}}+\frac {15 \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, a^{2}}{32 \left (4 a c -b^{2}\right )^{2} c \,d^{5}}-\frac {15 \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, a \,b^{2}}{64 \left (4 a c -b^{2}\right )^{2} c^{2} d^{5}}+\frac {15 \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, b^{4}}{512 \left (4 a c -b^{2}\right )^{2} c^{3} d^{5}}+\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}{16 \left (4 a c -b^{2}\right )^{2} c \,d^{5}}-\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}}{64 \left (4 a c -b^{2}\right )^{2} c^{2} d^{5}}+\frac {3 \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{16 \left (4 a c -b^{2}\right )^{2} c \,d^{5}}-\frac {3 \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {7}{2}}}{16 \left (4 a c -b^{2}\right )^{2} \left (x +\frac {b}{2 c}\right )^{2} c^{2} d^{5}}-\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {7}{2}}}{32 \left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{4} c^{4} d^{5}} \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)^5,x)

[Out]

-1/32/d^5/c^4/(4*a*c-b^2)/(x+1/2*b/c)^4*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)-3/16/d^5/c^2/(4*a*c-b^2)^2/(
x+1/2*b/c)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)+3/16/d^5/c/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^
2)/c)^(5/2)+5/16/d^5/c/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*a-5/64/d^5/c^2/(4*a*c-b^2)^2*((
x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b^2+15/32/d^5/c/(4*a*c-b^2)^2*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*
a^2-15/64/d^5/c^2/(4*a*c-b^2)^2*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a*b^2+15/512/d^5/c^3/(4*a*c-b^2)^2*(4*
(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*b^4-15/8/d^5/c/(4*a*c-b^2)^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+45/32/d^5/c^2/(4*a*c-b^2)^
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^2-45/128/d^5/c^3/(4*a*c-b^2)^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+15/512/d^5/c^4/(4*a*c-b^2)^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)^5,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 )}^5} \,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)^5,x)

[Out]

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^5, 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^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {b^{2} x^{2} \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {c^{2} x^{4} \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {2 a b x \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {2 a c x^{2} \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx + \int \frac {2 b c x^{3} \sqrt {a + b x + c x^{2}}}{b^{5} + 10 b^{4} c x + 40 b^{3} c^{2} x^{2} + 80 b^{2} c^{3} x^{3} + 80 b c^{4} x^{4} + 32 c^{5} x^{5}}\, dx}{d^{5}} \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)**5,x)

[Out]

(Integral(a**2*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c**3*x**3 + 80*b*c**4*
x**4 + 32*c**5*x**5), x) + Integral(b**2*x**2*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 +
80*b**2*c**3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5), x) + Integral(c**2*x**4*sqrt(a + b*x + c*x**2)/(b**5 + 10
*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c**3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5), x) + Integral(2*a*b*x*sqrt
(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c**3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5
), x) + Integral(2*a*c*x**2*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b**3*c**2*x**2 + 80*b**2*c**3*x**3
+ 80*b*c**4*x**4 + 32*c**5*x**5), x) + Integral(2*b*c*x**3*sqrt(a + b*x + c*x**2)/(b**5 + 10*b**4*c*x + 40*b*
*3*c**2*x**2 + 80*b**2*c**3*x**3 + 80*b*c**4*x**4 + 32*c**5*x**5), x))/d**5

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