3.8.57 \(\int \frac {x^{3/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=262 \[ \frac {x^{5/2} (A b-a B)}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {x^{3/2} (5 a B+3 A b)}{24 a b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\sqrt {x} (5 a B+3 A b)}{64 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {x} (5 a B+3 A b)}{32 a b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (5 a B+3 A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{5/2} b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {770, 78, 47, 51, 63, 205} \begin {gather*} \frac {x^{5/2} (A b-a B)}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {x^{3/2} (5 a B+3 A b)}{24 a b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\sqrt {x} (5 a B+3 A b)}{64 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {x} (5 a B+3 A b)}{32 a b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (5 a B+3 A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{5/2} b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {x^{3/2} (A+B x)}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(A b-a B) x^{5/2}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^2 (3 A b+5 a B) \left (a b+b^2 x\right )\right ) \int \frac {x^{3/2}}{\left (a b+b^2 x\right )^4} \, dx}{8 a \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(A b-a B) x^{5/2}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b+5 a B) x^{3/2}}{24 a b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((3 A b+5 a B) \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {x}}{\left (a b+b^2 x\right )^3} \, dx}{16 a \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(A b-a B) x^{5/2}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b+5 a B) x^{3/2}}{24 a b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b+5 a B) \sqrt {x}}{32 a b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((3 A b+5 a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\sqrt {x} \left (a b+b^2 x\right )^2} \, dx}{64 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(3 A b+5 a B) \sqrt {x}}{64 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{5/2}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b+5 a B) x^{3/2}}{24 a b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b+5 a B) \sqrt {x}}{32 a b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((3 A b+5 a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\sqrt {x} \left (a b+b^2 x\right )} \, dx}{128 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(3 A b+5 a B) \sqrt {x}}{64 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{5/2}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b+5 a B) x^{3/2}}{24 a b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b+5 a B) \sqrt {x}}{32 a b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((3 A b+5 a B) \left (a b+b^2 x\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a b+b^2 x^2} \, dx,x,\sqrt {x}\right )}{64 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(3 A b+5 a B) \sqrt {x}}{64 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{5/2}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b+5 a B) x^{3/2}}{24 a b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b+5 a B) \sqrt {x}}{32 a b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(3 A b+5 a B) (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{5/2} b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 79, normalized size = 0.30 \begin {gather*} \frac {x^{5/2} \left (5 a^4 (A b-a B)+(a+b x)^4 (5 a B+3 A b) \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};-\frac {b x}{a}\right )\right )}{20 a^5 b (a+b x)^3 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(5/2)*(5*a^4*(A*b - a*B) + (3*A*b + 5*a*B)*(a + b*x)^4*Hypergeometric2F1[5/2, 4, 7/2, -((b*x)/a)]))/(20*a^5
*b*(a + b*x)^3*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [A]  time = 20.95, size = 153, normalized size = 0.58 \begin {gather*} \frac {(a+b x) \left (\frac {(5 a B+3 A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{5/2} b^{7/2}}-\frac {\sqrt {x} \left (15 a^4 B+9 a^3 A b+55 a^3 b B x+33 a^2 A b^2 x+73 a^2 b^2 B x^2-33 a A b^3 x^2-15 a b^3 B x^3-9 A b^4 x^3\right )}{192 a^2 b^3 (a+b x)^4}\right )}{\sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(-1/192*(Sqrt[x]*(9*a^3*A*b + 15*a^4*B + 33*a^2*A*b^2*x + 55*a^3*b*B*x - 33*a*A*b^3*x^2 + 73*a^2*b^
2*B*x^2 - 9*A*b^4*x^3 - 15*a*b^3*B*x^3))/(a^2*b^3*(a + b*x)^4) + ((3*A*b + 5*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqr
t[a]])/(64*a^(5/2)*b^(7/2))))/Sqrt[(a + b*x)^2]

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fricas [A]  time = 0.45, size = 537, normalized size = 2.05 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B a^{5} + 3 \, A a^{4} b + {\left (5 \, B a b^{4} + 3 \, A b^{5}\right )} x^{4} + 4 \, {\left (5 \, B a^{2} b^{3} + 3 \, A a b^{4}\right )} x^{3} + 6 \, {\left (5 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (5 \, B a^{4} b + 3 \, A a^{3} b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (15 \, B a^{5} b + 9 \, A a^{4} b^{2} - 3 \, {\left (5 \, B a^{2} b^{4} + 3 \, A a b^{5}\right )} x^{3} + {\left (73 \, B a^{3} b^{3} - 33 \, A a^{2} b^{4}\right )} x^{2} + 11 \, {\left (5 \, B a^{4} b^{2} + 3 \, A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{384 \, {\left (a^{3} b^{8} x^{4} + 4 \, a^{4} b^{7} x^{3} + 6 \, a^{5} b^{6} x^{2} + 4 \, a^{6} b^{5} x + a^{7} b^{4}\right )}}, -\frac {3 \, {\left (5 \, B a^{5} + 3 \, A a^{4} b + {\left (5 \, B a b^{4} + 3 \, A b^{5}\right )} x^{4} + 4 \, {\left (5 \, B a^{2} b^{3} + 3 \, A a b^{4}\right )} x^{3} + 6 \, {\left (5 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (5 \, B a^{4} b + 3 \, A a^{3} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (15 \, B a^{5} b + 9 \, A a^{4} b^{2} - 3 \, {\left (5 \, B a^{2} b^{4} + 3 \, A a b^{5}\right )} x^{3} + {\left (73 \, B a^{3} b^{3} - 33 \, A a^{2} b^{4}\right )} x^{2} + 11 \, {\left (5 \, B a^{4} b^{2} + 3 \, A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{192 \, {\left (a^{3} b^{8} x^{4} + 4 \, a^{4} b^{7} x^{3} + 6 \, a^{5} b^{6} x^{2} + 4 \, a^{6} b^{5} x + a^{7} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/384*(3*(5*B*a^5 + 3*A*a^4*b + (5*B*a*b^4 + 3*A*b^5)*x^4 + 4*(5*B*a^2*b^3 + 3*A*a*b^4)*x^3 + 6*(5*B*a^3*b^2
+ 3*A*a^2*b^3)*x^2 + 4*(5*B*a^4*b + 3*A*a^3*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)
) + 2*(15*B*a^5*b + 9*A*a^4*b^2 - 3*(5*B*a^2*b^4 + 3*A*a*b^5)*x^3 + (73*B*a^3*b^3 - 33*A*a^2*b^4)*x^2 + 11*(5*
B*a^4*b^2 + 3*A*a^3*b^3)*x)*sqrt(x))/(a^3*b^8*x^4 + 4*a^4*b^7*x^3 + 6*a^5*b^6*x^2 + 4*a^6*b^5*x + a^7*b^4), -1
/192*(3*(5*B*a^5 + 3*A*a^4*b + (5*B*a*b^4 + 3*A*b^5)*x^4 + 4*(5*B*a^2*b^3 + 3*A*a*b^4)*x^3 + 6*(5*B*a^3*b^2 +
3*A*a^2*b^3)*x^2 + 4*(5*B*a^4*b + 3*A*a^3*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (15*B*a^5*b + 9*A*
a^4*b^2 - 3*(5*B*a^2*b^4 + 3*A*a*b^5)*x^3 + (73*B*a^3*b^3 - 33*A*a^2*b^4)*x^2 + 11*(5*B*a^4*b^2 + 3*A*a^3*b^3)
*x)*sqrt(x))/(a^3*b^8*x^4 + 4*a^4*b^7*x^3 + 6*a^5*b^6*x^2 + 4*a^6*b^5*x + a^7*b^4)]

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giac [A]  time = 0.27, size = 148, normalized size = 0.56 \begin {gather*} \frac {{\left (5 \, B a + 3 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{2} b^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {15 \, B a b^{3} x^{\frac {7}{2}} + 9 \, A b^{4} x^{\frac {7}{2}} - 73 \, B a^{2} b^{2} x^{\frac {5}{2}} + 33 \, A a b^{3} x^{\frac {5}{2}} - 55 \, B a^{3} b x^{\frac {3}{2}} - 33 \, A a^{2} b^{2} x^{\frac {3}{2}} - 15 \, B a^{4} \sqrt {x} - 9 \, A a^{3} b \sqrt {x}}{192 \, {\left (b x + a\right )}^{4} a^{2} b^{3} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

1/64*(5*B*a + 3*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^2*b^3*sgn(b*x + a)) + 1/192*(15*B*a*b^3*x^(7/2)
+ 9*A*b^4*x^(7/2) - 73*B*a^2*b^2*x^(5/2) + 33*A*a*b^3*x^(5/2) - 55*B*a^3*b*x^(3/2) - 33*A*a^2*b^2*x^(3/2) - 15
*B*a^4*sqrt(x) - 9*A*a^3*b*sqrt(x))/((b*x + a)^4*a^2*b^3*sgn(b*x + a))

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maple [A]  time = 0.07, size = 357, normalized size = 1.36 \begin {gather*} \frac {\left (9 A \,b^{5} x^{4} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+15 B a \,b^{4} x^{4} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+36 A a \,b^{4} x^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+60 B \,a^{2} b^{3} x^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+54 A \,a^{2} b^{3} x^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+90 B \,a^{3} b^{2} x^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+9 \sqrt {a b}\, A \,b^{4} x^{\frac {7}{2}}+15 \sqrt {a b}\, B a \,b^{3} x^{\frac {7}{2}}+36 A \,a^{3} b^{2} x \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+60 B \,a^{4} b x \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+33 \sqrt {a b}\, A a \,b^{3} x^{\frac {5}{2}}-73 \sqrt {a b}\, B \,a^{2} b^{2} x^{\frac {5}{2}}+9 A \,a^{4} b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+15 B \,a^{5} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-33 \sqrt {a b}\, A \,a^{2} b^{2} x^{\frac {3}{2}}-55 \sqrt {a b}\, B \,a^{3} b \,x^{\frac {3}{2}}-9 \sqrt {a b}\, A \,a^{3} b \sqrt {x}-15 \sqrt {a b}\, B \,a^{4} \sqrt {x}\right ) \left (b x +a \right )}{192 \sqrt {a b}\, \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} a^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/192*(9*(a*b)^(1/2)*A*b^4*x^(7/2)+15*(a*b)^(1/2)*B*a*b^3*x^(7/2)+33*(a*b)^(1/2)*A*a*b^3*x^(5/2)+9*A*b^5*x^4*a
rctan(1/(a*b)^(1/2)*b*x^(1/2))-73*(a*b)^(1/2)*B*a^2*b^2*x^(5/2)+15*B*a*b^4*x^4*arctan(1/(a*b)^(1/2)*b*x^(1/2))
+36*A*a*b^4*x^3*arctan(1/(a*b)^(1/2)*b*x^(1/2))+60*B*a^2*b^3*x^3*arctan(1/(a*b)^(1/2)*b*x^(1/2))-33*(a*b)^(1/2
)*A*a^2*b^2*x^(3/2)+54*A*a^2*b^3*x^2*arctan(1/(a*b)^(1/2)*b*x^(1/2))-55*(a*b)^(1/2)*B*a^3*b*x^(3/2)+90*B*a^3*b
^2*x^2*arctan(1/(a*b)^(1/2)*b*x^(1/2))+36*A*a^3*b^2*x*arctan(1/(a*b)^(1/2)*b*x^(1/2))+60*B*a^4*b*x*arctan(1/(a
*b)^(1/2)*b*x^(1/2))-9*(a*b)^(1/2)*A*a^3*b*x^(1/2)+9*A*a^4*b*arctan(1/(a*b)^(1/2)*b*x^(1/2))-15*(a*b)^(1/2)*B*
a^4*x^(1/2)+15*B*a^5*arctan(1/(a*b)^(1/2)*b*x^(1/2)))*(b*x+a)/(a*b)^(1/2)/b^3/a^2/((b*x+a)^2)^(5/2)

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maxima [B]  time = 1.80, size = 372, normalized size = 1.42 \begin {gather*} -\frac {5 \, {\left ({\left (7 \, B a b^{5} + 3 \, A b^{6}\right )} x^{2} - 9 \, {\left (B a^{2} b^{4} + A a b^{5}\right )} x\right )} x^{\frac {9}{2}} + 10 \, {\left ({\left (7 \, B a^{2} b^{4} + 3 \, A a b^{5}\right )} x^{2} - 27 \, {\left (B a^{3} b^{3} + A a^{2} b^{4}\right )} x\right )} x^{\frac {7}{2}} - 20 \, {\left (2 \, {\left (7 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{2} + 33 \, {\left (B a^{4} b^{2} + A a^{3} b^{3}\right )} x\right )} x^{\frac {5}{2}} + 6 \, {\left (5 \, {\left (3 \, B a^{4} b^{2} - 17 \, A a^{3} b^{3}\right )} x^{2} - {\left (11 \, B a^{5} b + 139 \, A a^{4} b^{2}\right )} x\right )} x^{\frac {3}{2}} + 3 \, {\left ({\left (7 \, B a^{5} b + 3 \, A a^{4} b^{2}\right )} x^{2} - 5 \, {\left (B a^{6} + A a^{5} b\right )} x\right )} \sqrt {x}}{1920 \, {\left (a^{4} b^{7} x^{5} + 5 \, a^{5} b^{6} x^{4} + 10 \, a^{6} b^{5} x^{3} + 10 \, a^{7} b^{4} x^{2} + 5 \, a^{8} b^{3} x + a^{9} b^{2}\right )}} + \frac {{\left (5 \, B a + 3 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{2} b^{3}} + \frac {{\left (7 \, B a b + 3 \, A b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (5 \, B a^{2} + 3 \, A a b\right )} \sqrt {x}}{384 \, a^{4} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

-1/1920*(5*((7*B*a*b^5 + 3*A*b^6)*x^2 - 9*(B*a^2*b^4 + A*a*b^5)*x)*x^(9/2) + 10*((7*B*a^2*b^4 + 3*A*a*b^5)*x^2
- 27*(B*a^3*b^3 + A*a^2*b^4)*x)*x^(7/2) - 20*(2*(7*B*a^3*b^3 + 3*A*a^2*b^4)*x^2 + 33*(B*a^4*b^2 + A*a^3*b^3)*
x)*x^(5/2) + 6*(5*(3*B*a^4*b^2 - 17*A*a^3*b^3)*x^2 - (11*B*a^5*b + 139*A*a^4*b^2)*x)*x^(3/2) + 3*((7*B*a^5*b +
3*A*a^4*b^2)*x^2 - 5*(B*a^6 + A*a^5*b)*x)*sqrt(x))/(a^4*b^7*x^5 + 5*a^5*b^6*x^4 + 10*a^6*b^5*x^3 + 10*a^7*b^4
*x^2 + 5*a^8*b^3*x + a^9*b^2) + 1/64*(5*B*a + 3*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^2*b^3) + 1/384*(
(7*B*a*b + 3*A*b^2)*x^(3/2) - 6*(5*B*a^2 + 3*A*a*b)*sqrt(x))/(a^4*b^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^{3/2}\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(3/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

int((x^(3/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)

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sympy [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(x**(3/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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