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

Optimal. Leaf size=252 \[ -\frac {5 \left (b^2-4 a c\right )^3 \left (-4 a B c-16 A b c+9 b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32768 c^{11/2}}+\frac {5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2} \left (-4 a B c-16 A b c+9 b^2 B\right )}{16384 c^5}-\frac {5 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2} \left (-4 a B c-16 A b c+9 b^2 B\right )}{6144 c^4}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/2} \left (-4 a B c-16 A b c+9 b^2 B\right )}{384 c^3}-\frac {\left (a+b x+c x^2\right )^{7/2} (-16 A c+9 b B-14 B c x)}{112 c^2} \]

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Rubi [A]  time = 0.12, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {779, 612, 621, 206} \begin {gather*} \frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/2} \left (-4 a B c-16 A b c+9 b^2 B\right )}{384 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} \left (-4 a B c-16 A b c+9 b^2 B\right )}{6144 c^4}+\frac {5 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2} \left (-4 a B c-16 A b c+9 b^2 B\right )}{16384 c^5}-\frac {5 \left (b^2-4 a c\right )^3 \left (-4 a B c-16 A b c+9 b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32768 c^{11/2}}-\frac {\left (a+b x+c x^2\right )^{7/2} (-16 A c+9 b B-14 B c x)}{112 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(b^2 - 4*a*c)^2*(9*b^2*B - 16*A*b*c - 4*a*B*c)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(16384*c^5) - (5*(b^2 - 4
*a*c)*(9*b^2*B - 16*A*b*c - 4*a*B*c)*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(6144*c^4) + ((9*b^2*B - 16*A*b*c -
4*a*B*c)*(b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(384*c^3) - ((9*b*B - 16*A*c - 14*B*c*x)*(a + b*x + c*x^2)^(7/2)
)/(112*c^2) - (5*(b^2 - 4*a*c)^3*(9*b^2*B - 16*A*b*c - 4*a*B*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x +
c*x^2])])/(32768*c^(11/2))

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]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int x (A+B x) \left (a+b x+c x^2\right )^{5/2} \, dx &=-\frac {(9 b B-16 A c-14 B c x) \left (a+b x+c x^2\right )^{7/2}}{112 c^2}+\frac {\left (9 b^2 B-16 A b c-4 a B c\right ) \int \left (a+b x+c x^2\right )^{5/2} \, dx}{32 c^2}\\ &=\frac {\left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{384 c^3}-\frac {(9 b B-16 A c-14 B c x) \left (a+b x+c x^2\right )^{7/2}}{112 c^2}-\frac {\left (5 \left (b^2-4 a c\right ) \left (9 b^2 B-16 A b c-4 a B c\right )\right ) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{768 c^3}\\ &=-\frac {5 \left (b^2-4 a c\right ) \left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{6144 c^4}+\frac {\left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{384 c^3}-\frac {(9 b B-16 A c-14 B c x) \left (a+b x+c x^2\right )^{7/2}}{112 c^2}+\frac {\left (5 \left (b^2-4 a c\right )^2 \left (9 b^2 B-16 A b c-4 a B c\right )\right ) \int \sqrt {a+b x+c x^2} \, dx}{4096 c^4}\\ &=\frac {5 \left (b^2-4 a c\right )^2 \left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{16384 c^5}-\frac {5 \left (b^2-4 a c\right ) \left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{6144 c^4}+\frac {\left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{384 c^3}-\frac {(9 b B-16 A c-14 B c x) \left (a+b x+c x^2\right )^{7/2}}{112 c^2}-\frac {\left (5 \left (b^2-4 a c\right )^3 \left (9 b^2 B-16 A b c-4 a B c\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{32768 c^5}\\ &=\frac {5 \left (b^2-4 a c\right )^2 \left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{16384 c^5}-\frac {5 \left (b^2-4 a c\right ) \left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{6144 c^4}+\frac {\left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{384 c^3}-\frac {(9 b B-16 A c-14 B c x) \left (a+b x+c x^2\right )^{7/2}}{112 c^2}-\frac {\left (5 \left (b^2-4 a c\right )^3 \left (9 b^2 B-16 A b c-4 a B c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{16384 c^5}\\ &=\frac {5 \left (b^2-4 a c\right )^2 \left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{16384 c^5}-\frac {5 \left (b^2-4 a c\right ) \left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{6144 c^4}+\frac {\left (9 b^2 B-16 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{384 c^3}-\frac {(9 b B-16 A c-14 B c x) \left (a+b x+c x^2\right )^{7/2}}{112 c^2}-\frac {5 \left (b^2-4 a c\right )^3 \left (9 b^2 B-16 A b c-4 a B c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32768 c^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 205, normalized size = 0.81 \begin {gather*} \frac {\frac {7 \left (-2 a B c-8 A b c+\frac {9 b^2 B}{2}\right ) \left (2 (b+2 c x) (a+x (b+c x))^{5/2}-5 \left (b^2-4 a c\right ) \left (\frac {3 \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)}\right )}{128 c^{5/2}}+\frac {(b+2 c x) (a+x (b+c x))^{3/2}}{8 c}\right )\right )}{24 c}+(a+x (b+c x))^{7/2} (2 c (8 A+7 B x)-9 b B)}{112 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-9*b*B + 2*c*(8*A + 7*B*x))*(a + x*(b + c*x))^(7/2) + (7*((9*b^2*B)/2 - 8*A*b*c - 2*a*B*c)*(2*(b + 2*c*x)*(a
+ x*(b + c*x))^(5/2) - 5*(b^2 - 4*a*c)*(((b + 2*c*x)*(a + x*(b + c*x))^(3/2))/(8*c) + (3*(b^2 - 4*a*c)*(-2*Sq
rt[c]*(b + 2*c*x)*Sqrt[a + x*(b + c*x)] + (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]
))/(128*c^(5/2)))))/(24*c))/(112*c^2)

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IntegrateAlgebraic [B]  time = 2.45, size = 535, normalized size = 2.12 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (49152 a^3 A c^4-42432 a^3 b B c^3+13440 a^3 B c^4 x-59136 a^2 A b^2 c^3+29184 a^2 A b c^4 x+147456 a^2 A c^5 x^2+37744 a^2 b^3 B c^2-19104 a^2 b^2 B c^3 x+11136 a^2 b B c^4 x^2+105728 a^2 B c^5 x^3+17920 a A b^4 c^2-10752 a A b^3 c^3 x+7680 a A b^2 c^4 x^2+201728 a A b c^5 x^3+147456 a A c^6 x^4-10500 a b^5 B c+6328 a b^4 B c^2 x-4544 a b^3 B c^3 x^2+3456 a b^2 B c^4 x^3+157184 a b B c^5 x^4+121856 a B c^6 x^5-1680 A b^6 c+1120 A b^5 c^2 x-896 A b^4 c^3 x^2+768 A b^3 c^4 x^3+75776 A b^2 c^5 x^4+118784 A b c^6 x^5+49152 A c^7 x^6+945 b^7 B-630 b^6 B c x+504 b^5 B c^2 x^2-432 b^4 B c^3 x^3+384 b^3 B c^4 x^4+62208 b^2 B c^5 x^5+101376 b B c^6 x^6+43008 B c^7 x^7\right )}{344064 c^5}+\frac {5 \left (256 a^4 B c^4+1024 a^3 A b c^4-768 a^3 b^2 B c^3-768 a^2 A b^3 c^3+480 a^2 b^4 B c^2+192 a A b^5 c^2-112 a b^6 B c-16 A b^7 c+9 b^8 B\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{32768 c^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(945*b^7*B - 1680*A*b^6*c - 10500*a*b^5*B*c + 17920*a*A*b^4*c^2 + 37744*a^2*b^3*B*c^2 -
59136*a^2*A*b^2*c^3 - 42432*a^3*b*B*c^3 + 49152*a^3*A*c^4 - 630*b^6*B*c*x + 1120*A*b^5*c^2*x + 6328*a*b^4*B*c
^2*x - 10752*a*A*b^3*c^3*x - 19104*a^2*b^2*B*c^3*x + 29184*a^2*A*b*c^4*x + 13440*a^3*B*c^4*x + 504*b^5*B*c^2*x
^2 - 896*A*b^4*c^3*x^2 - 4544*a*b^3*B*c^3*x^2 + 7680*a*A*b^2*c^4*x^2 + 11136*a^2*b*B*c^4*x^2 + 147456*a^2*A*c^
5*x^2 - 432*b^4*B*c^3*x^3 + 768*A*b^3*c^4*x^3 + 3456*a*b^2*B*c^4*x^3 + 201728*a*A*b*c^5*x^3 + 105728*a^2*B*c^5
*x^3 + 384*b^3*B*c^4*x^4 + 75776*A*b^2*c^5*x^4 + 157184*a*b*B*c^5*x^4 + 147456*a*A*c^6*x^4 + 62208*b^2*B*c^5*x
^5 + 118784*A*b*c^6*x^5 + 121856*a*B*c^6*x^5 + 101376*b*B*c^6*x^6 + 49152*A*c^7*x^6 + 43008*B*c^7*x^7))/(34406
4*c^5) + (5*(9*b^8*B - 16*A*b^7*c - 112*a*b^6*B*c + 192*a*A*b^5*c^2 + 480*a^2*b^4*B*c^2 - 768*a^2*A*b^3*c^3 -
768*a^3*b^2*B*c^3 + 1024*a^3*A*b*c^4 + 256*a^4*B*c^4)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(32768
*c^(11/2))

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fricas [B]  time = 0.59, size = 1039, normalized size = 4.12

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1376256*(105*(9*B*b^8 + 256*(B*a^4 + 4*A*a^3*b)*c^4 - 768*(B*a^3*b^2 + A*a^2*b^3)*c^3 + 96*(5*B*a^2*b^4 + 2
*A*a*b^5)*c^2 - 16*(7*B*a*b^6 + A*b^7)*c)*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*(43008*B*c^8*x^7 + 945*B*b^7*c + 49152*A*a^3*c^5 + 3072*(33*B*b*c^7 + 16*A*c^8)*
x^6 + 256*(243*B*b^2*c^6 + 4*(119*B*a + 116*A*b)*c^7)*x^5 - 192*(221*B*a^3*b + 308*A*a^2*b^2)*c^4 + 128*(3*B*b
^3*c^5 + 1152*A*a*c^7 + 4*(307*B*a*b + 148*A*b^2)*c^6)*x^4 + 112*(337*B*a^2*b^3 + 160*A*a*b^4)*c^3 - 16*(27*B*
b^4*c^4 - 16*(413*B*a^2 + 788*A*a*b)*c^6 - 24*(9*B*a*b^2 + 2*A*b^3)*c^5)*x^3 - 420*(25*B*a*b^5 + 4*A*b^6)*c^2
+ 8*(63*B*b^5*c^3 + 18432*A*a^2*c^6 + 48*(29*B*a^2*b + 20*A*a*b^2)*c^5 - 8*(71*B*a*b^3 + 14*A*b^4)*c^4)*x^2 -
2*(315*B*b^6*c^2 - 192*(35*B*a^3 + 76*A*a^2*b)*c^5 + 48*(199*B*a^2*b^2 + 112*A*a*b^3)*c^4 - 28*(113*B*a*b^4 +
20*A*b^5)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^6, 1/688128*(105*(9*B*b^8 + 256*(B*a^4 + 4*A*a^3*b)*c^4 - 768*(B*a^
3*b^2 + A*a^2*b^3)*c^3 + 96*(5*B*a^2*b^4 + 2*A*a*b^5)*c^2 - 16*(7*B*a*b^6 + A*b^7)*c)*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*(43008*B*c^8*x^7 + 945*B*b^7*c + 49152*A*a
^3*c^5 + 3072*(33*B*b*c^7 + 16*A*c^8)*x^6 + 256*(243*B*b^2*c^6 + 4*(119*B*a + 116*A*b)*c^7)*x^5 - 192*(221*B*a
^3*b + 308*A*a^2*b^2)*c^4 + 128*(3*B*b^3*c^5 + 1152*A*a*c^7 + 4*(307*B*a*b + 148*A*b^2)*c^6)*x^4 + 112*(337*B*
a^2*b^3 + 160*A*a*b^4)*c^3 - 16*(27*B*b^4*c^4 - 16*(413*B*a^2 + 788*A*a*b)*c^6 - 24*(9*B*a*b^2 + 2*A*b^3)*c^5)
*x^3 - 420*(25*B*a*b^5 + 4*A*b^6)*c^2 + 8*(63*B*b^5*c^3 + 18432*A*a^2*c^6 + 48*(29*B*a^2*b + 20*A*a*b^2)*c^5 -
8*(71*B*a*b^3 + 14*A*b^4)*c^4)*x^2 - 2*(315*B*b^6*c^2 - 192*(35*B*a^3 + 76*A*a^2*b)*c^5 + 48*(199*B*a^2*b^2 +
112*A*a*b^3)*c^4 - 28*(113*B*a*b^4 + 20*A*b^5)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^6]

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giac [B]  time = 0.31, size = 528, normalized size = 2.10 \begin {gather*} \frac {1}{344064} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, {\left (12 \, {\left (14 \, B c^{2} x + \frac {33 \, B b c^{8} + 16 \, A c^{9}}{c^{7}}\right )} x + \frac {243 \, B b^{2} c^{7} + 476 \, B a c^{8} + 464 \, A b c^{8}}{c^{7}}\right )} x + \frac {3 \, B b^{3} c^{6} + 1228 \, B a b c^{7} + 592 \, A b^{2} c^{7} + 1152 \, A a c^{8}}{c^{7}}\right )} x - \frac {27 \, B b^{4} c^{5} - 216 \, B a b^{2} c^{6} - 48 \, A b^{3} c^{6} - 6608 \, B a^{2} c^{7} - 12608 \, A a b c^{7}}{c^{7}}\right )} x + \frac {63 \, B b^{5} c^{4} - 568 \, B a b^{3} c^{5} - 112 \, A b^{4} c^{5} + 1392 \, B a^{2} b c^{6} + 960 \, A a b^{2} c^{6} + 18432 \, A a^{2} c^{7}}{c^{7}}\right )} x - \frac {315 \, B b^{6} c^{3} - 3164 \, B a b^{4} c^{4} - 560 \, A b^{5} c^{4} + 9552 \, B a^{2} b^{2} c^{5} + 5376 \, A a b^{3} c^{5} - 6720 \, B a^{3} c^{6} - 14592 \, A a^{2} b c^{6}}{c^{7}}\right )} x + \frac {945 \, B b^{7} c^{2} - 10500 \, B a b^{5} c^{3} - 1680 \, A b^{6} c^{3} + 37744 \, B a^{2} b^{3} c^{4} + 17920 \, A a b^{4} c^{4} - 42432 \, B a^{3} b c^{5} - 59136 \, A a^{2} b^{2} c^{5} + 49152 \, A a^{3} c^{6}}{c^{7}}\right )} + \frac {5 \, {\left (9 \, B b^{8} - 112 \, B a b^{6} c - 16 \, A b^{7} c + 480 \, B a^{2} b^{4} c^{2} + 192 \, A a b^{5} c^{2} - 768 \, B a^{3} b^{2} c^{3} - 768 \, A a^{2} b^{3} c^{3} + 256 \, B a^{4} c^{4} + 1024 \, A a^{3} b c^{4}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{32768 \, c^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/344064*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*(12*(14*B*c^2*x + (33*B*b*c^8 + 16*A*c^9)/c^7)*x + (243*B*b^2*c^
7 + 476*B*a*c^8 + 464*A*b*c^8)/c^7)*x + (3*B*b^3*c^6 + 1228*B*a*b*c^7 + 592*A*b^2*c^7 + 1152*A*a*c^8)/c^7)*x -
(27*B*b^4*c^5 - 216*B*a*b^2*c^6 - 48*A*b^3*c^6 - 6608*B*a^2*c^7 - 12608*A*a*b*c^7)/c^7)*x + (63*B*b^5*c^4 - 5
68*B*a*b^3*c^5 - 112*A*b^4*c^5 + 1392*B*a^2*b*c^6 + 960*A*a*b^2*c^6 + 18432*A*a^2*c^7)/c^7)*x - (315*B*b^6*c^3
- 3164*B*a*b^4*c^4 - 560*A*b^5*c^4 + 9552*B*a^2*b^2*c^5 + 5376*A*a*b^3*c^5 - 6720*B*a^3*c^6 - 14592*A*a^2*b*c
^6)/c^7)*x + (945*B*b^7*c^2 - 10500*B*a*b^5*c^3 - 1680*A*b^6*c^3 + 37744*B*a^2*b^3*c^4 + 17920*A*a*b^4*c^4 - 4
2432*B*a^3*b*c^5 - 59136*A*a^2*b^2*c^5 + 49152*A*a^3*c^6)/c^7) + 5/32768*(9*B*b^8 - 112*B*a*b^6*c - 16*A*b^7*c
+ 480*B*a^2*b^4*c^2 + 192*A*a*b^5*c^2 - 768*B*a^3*b^2*c^3 - 768*A*a^2*b^3*c^3 + 256*B*a^4*c^4 + 1024*A*a^3*b*
c^4)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(11/2)

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maple [B]  time = 0.06, size = 1034, normalized size = 4.10

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-95/2048*B*b^4/c^3*(c*x^2+b*x+a)^(1/2)*x*a-5/32*A*b/c*(c*x^2+b*x+a)^(1/2)*x*a^2+5/64*A*b^3/c^2*(c*x^2+b*x+a)^(
1/2)*x*a-5/48*A*b/c*(c*x^2+b*x+a)^(3/2)*x*a+25/384*B*b^2/c^2*(c*x^2+b*x+a)^(3/2)*x*a+55/512*B*b^2/c^2*(c*x^2+b
*x+a)^(1/2)*x*a^2+15/128*B*b^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^3-5/256*B*a^3/c^2*(c*x^2+
b*x+a)^(1/2)*b+55/1024*B*b^3/c^3*(c*x^2+b*x+a)^(1/2)*a^2-15/1024*B*b^4/c^3*(c*x^2+b*x+a)^(3/2)*x-1/48*B*a/c*x*
(c*x^2+b*x+a)^(5/2)+5/192*A*b^3/c^2*(c*x^2+b*x+a)^(3/2)*x-5/128*B*a^3/c*(c*x^2+b*x+a)^(1/2)*x+3/128*B*b^3/c^3*
(c*x^2+b*x+a)^(5/2)-9/112*B*b/c^2*(c*x^2+b*x+a)^(7/2)+25/768*B*b^3/c^3*(c*x^2+b*x+a)^(3/2)*a-5/192*B*a^2/c*(c*
x^2+b*x+a)^(3/2)*x+45/8192*B*b^6/c^4*(c*x^2+b*x+a)^(1/2)*x+3/64*B*b^2/c^2*x*(c*x^2+b*x+a)^(5/2)-45/32768*B*b^8
/c^(11/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-5/128*B*a^4/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)
^(1/2))+1/8*B*x*(c*x^2+b*x+a)^(7/2)/c-1/12*A*b/c*x*(c*x^2+b*x+a)^(5/2)+45/16384*B*b^7/c^5*(c*x^2+b*x+a)^(1/2)+
1/7*A*(c*x^2+b*x+a)^(7/2)/c-5/96*A*b^2/c^2*(c*x^2+b*x+a)^(3/2)*a-75/1024*B*b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+
(c*x^2+b*x+a)^(1/2))*a^2-15/512*A*b^5/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+35/2048*B*b^6/c^(9
/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+15/128*A*b^3/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1
/2))*a^2-5/384*B*a^2/c^2*(c*x^2+b*x+a)^(3/2)*b-1/96*B*a/c^2*(c*x^2+b*x+a)^(5/2)*b-95/4096*B*b^5/c^4*(c*x^2+b*x
+a)^(1/2)*a-5/512*A*b^5/c^3*(c*x^2+b*x+a)^(1/2)*x-5/64*A*b^2/c^2*(c*x^2+b*x+a)^(1/2)*a^2+5/128*A*b^4/c^3*(c*x^
2+b*x+a)^(1/2)*a-5/32*A*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^3-1/24*A*b^2/c^2*(c*x^2+b*x+a)
^(5/2)+5/384*A*b^4/c^3*(c*x^2+b*x+a)^(3/2)-5/1024*A*b^6/c^4*(c*x^2+b*x+a)^(1/2)+5/2048*A*b^7/c^(9/2)*ln((c*x+1
/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-15/2048*B*b^5/c^4*(c*x^2+b*x+a)^(3/2)

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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