3.8.45 \(\int \frac {x^{7/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=302 \[ \frac {x^{9/2} (A b-a B)}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^{7/2} (5 A b-9 a B)}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 a \sqrt {x} (a+b x) (5 A b-9 a B)}{4 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 x^{3/2} (a+b x) (5 A b-9 a B)}{12 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 x^{5/2} (a+b x) (5 A b-9 a B)}{20 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 a^{3/2} (a+b x) (5 A b-9 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 302, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {770, 78, 47, 50, 63, 205} \begin {gather*} \frac {x^{9/2} (A b-a B)}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^{7/2} (5 A b-9 a B)}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 x^{5/2} (a+b x) (5 A b-9 a B)}{20 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 x^{3/2} (a+b x) (5 A b-9 a B)}{12 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 a \sqrt {x} (a+b x) (5 A b-9 a B)}{4 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 a^{3/2} (a+b x) (5 A b-9 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

((5*A*b - 9*a*B)*x^(7/2))/(4*a*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*x^(9/2))/(2*a*b*(a + b*x)*Sqr
t[a^2 + 2*a*b*x + b^2*x^2]) - (7*a*(5*A*b - 9*a*B)*Sqrt[x]*(a + b*x))/(4*b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
(7*(5*A*b - 9*a*B)*x^(3/2)*(a + b*x))/(12*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (7*(5*A*b - 9*a*B)*x^(5/2)*(a +
b*x))/(20*a*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (7*a^(3/2)*(5*A*b - 9*a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x]
)/Sqrt[a]])/(4*b^(11/2)*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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {x^{7/2} (A+B x)}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(A b-a B) x^{9/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left ((5 A b-9 a B) \left (a b+b^2 x\right )\right ) \int \frac {x^{7/2}}{\left (a b+b^2 x\right )^2} \, dx}{4 a \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 (5 A b-9 a B) \left (a b+b^2 x\right )\right ) \int \frac {x^{5/2}}{a b+b^2 x} \, dx}{8 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (5 A b-9 a B) x^{5/2} (a+b x)}{20 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (7 (5 A b-9 a B) \left (a b+b^2 x\right )\right ) \int \frac {x^{3/2}}{a b+b^2 x} \, dx}{8 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 (5 A b-9 a B) x^{3/2} (a+b x)}{12 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (5 A b-9 a B) x^{5/2} (a+b x)}{20 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 a (5 A b-9 a B) \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {x}}{a b+b^2 x} \, dx}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 a (5 A b-9 a B) \sqrt {x} (a+b x)}{4 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 (5 A b-9 a B) x^{3/2} (a+b x)}{12 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (5 A b-9 a B) x^{5/2} (a+b x)}{20 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (7 a^2 (5 A b-9 a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\sqrt {x} \left (a b+b^2 x\right )} \, dx}{8 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 a (5 A b-9 a B) \sqrt {x} (a+b x)}{4 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 (5 A b-9 a B) x^{3/2} (a+b x)}{12 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (5 A b-9 a B) x^{5/2} (a+b x)}{20 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (7 a^2 (5 A b-9 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 )}{4 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 a (5 A b-9 a B) \sqrt {x} (a+b x)}{4 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 (5 A b-9 a B) x^{3/2} (a+b x)}{12 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (5 A b-9 a B) x^{5/2} (a+b x)}{20 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 a^{3/2} (5 A b-9 a B) (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{11/2} \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 = 0.26 \begin {gather*} \frac {x^{9/2} \left (9 a^2 (A b-a B)+(a+b x)^2 (9 a B-5 A b) \, _2F_1\left (2,\frac {9}{2};\frac {11}{2};-\frac {b x}{a}\right )\right )}{18 a^3 b (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(7/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(x^(9/2)*(9*a^2*(A*b - a*B) + (-5*A*b + 9*a*B)*(a + b*x)^2*Hypergeometric2F1[2, 9/2, 11/2, -((b*x)/a)]))/(18*a
^3*b*(a + b*x)*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [A]  time = 28.48, size = 163, normalized size = 0.54 \begin {gather*} \frac {(a+b x) \left (\frac {\sqrt {x} \left (945 a^4 B-525 a^3 A b+1575 a^3 b B x-875 a^2 A b^2 x+504 a^2 b^2 B x^2-280 a A b^3 x^2-72 a b^3 B x^3+40 A b^4 x^3+24 b^4 B x^4\right )}{60 b^5 (a+b x)^2}-\frac {7 \left (9 a^{5/2} B-5 a^{3/2} A b\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{11/2}}\right )}{\sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(7/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

((a + b*x)*((Sqrt[x]*(-525*a^3*A*b + 945*a^4*B - 875*a^2*A*b^2*x + 1575*a^3*b*B*x - 280*a*A*b^3*x^2 + 504*a^2*
b^2*B*x^2 + 40*A*b^4*x^3 - 72*a*b^3*B*x^3 + 24*b^4*B*x^4))/(60*b^5*(a + b*x)^2) - (7*(-5*a^(3/2)*A*b + 9*a^(5/
2)*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(11/2))))/Sqrt[(a + b*x)^2]

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fricas [A]  time = 0.47, size = 408, normalized size = 1.35 \begin {gather*} \left [-\frac {105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (24 \, B b^{4} x^{4} + 945 \, B a^{4} - 525 \, A a^{3} b - 8 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{3} + 56 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 175 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{120 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac {105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (24 \, B b^{4} x^{4} + 945 \, B a^{4} - 525 \, A a^{3} b - 8 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{3} + 56 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 175 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{60 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/120*(105*(9*B*a^4 - 5*A*a^3*b + (9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 2*(9*B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(-a/b)*
log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(24*B*b^4*x^4 + 945*B*a^4 - 525*A*a^3*b - 8*(9*B*a*b^3 -
5*A*b^4)*x^3 + 56*(9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 175*(9*B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(x))/(b^7*x^2 + 2*a*b^
6*x + a^2*b^5), -1/60*(105*(9*B*a^4 - 5*A*a^3*b + (9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 2*(9*B*a^3*b - 5*A*a^2*b^2)*
x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) - (24*B*b^4*x^4 + 945*B*a^4 - 525*A*a^3*b - 8*(9*B*a*b^3 - 5*A*b^4)
*x^3 + 56*(9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 175*(9*B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(x))/(b^7*x^2 + 2*a*b^6*x + a^2
*b^5)]

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giac [A]  time = 0.22, size = 170, normalized size = 0.56 \begin {gather*} -\frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {17 \, B a^{3} b x^{\frac {3}{2}} - 13 \, A a^{2} b^{2} x^{\frac {3}{2}} + 15 \, B a^{4} \sqrt {x} - 11 \, A a^{3} b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, {\left (3 \, B b^{12} x^{\frac {5}{2}} - 15 \, B a b^{11} x^{\frac {3}{2}} + 5 \, A b^{12} x^{\frac {3}{2}} + 90 \, B a^{2} b^{10} \sqrt {x} - 45 \, A a b^{11} \sqrt {x}\right )}}{15 \, b^{15} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

-7/4*(9*B*a^3 - 5*A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5*sgn(b*x + a)) + 1/4*(17*B*a^3*b*x^(3/2)
- 13*A*a^2*b^2*x^(3/2) + 15*B*a^4*sqrt(x) - 11*A*a^3*b*sqrt(x))/((b*x + a)^2*b^5*sgn(b*x + a)) + 2/15*(3*B*b^1
2*x^(5/2) - 15*B*a*b^11*x^(3/2) + 5*A*b^12*x^(3/2) + 90*B*a^2*b^10*sqrt(x) - 45*A*a*b^11*sqrt(x))/(b^15*sgn(b*
x + a))

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maple [A]  time = 0.07, size = 283, normalized size = 0.94 \begin {gather*} \frac {\left (24 \sqrt {a b}\, B \,b^{4} x^{\frac {9}{2}}+525 A \,a^{2} b^{3} x^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-945 B \,a^{3} b^{2} x^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+40 \sqrt {a b}\, A \,b^{4} x^{\frac {7}{2}}-72 \sqrt {a b}\, B a \,b^{3} x^{\frac {7}{2}}+1050 A \,a^{3} b^{2} x \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-1890 B \,a^{4} b x \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-280 \sqrt {a b}\, A a \,b^{3} x^{\frac {5}{2}}+504 \sqrt {a b}\, B \,a^{2} b^{2} x^{\frac {5}{2}}+525 A \,a^{4} b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-945 B \,a^{5} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-875 \sqrt {a b}\, A \,a^{2} b^{2} x^{\frac {3}{2}}+1575 \sqrt {a b}\, B \,a^{3} b \,x^{\frac {3}{2}}-525 \sqrt {a b}\, A \,a^{3} b \sqrt {x}+945 \sqrt {a b}\, B \,a^{4} \sqrt {x}\right ) \left (b x +a \right )}{60 \sqrt {a b}\, \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/60*(24*(a*b)^(1/2)*B*b^4*x^(9/2)-72*(a*b)^(1/2)*B*a*b^3*x^(7/2)+40*(a*b)^(1/2)*A*b^4*x^(7/2)+504*(a*b)^(1/2)
*B*a^2*b^2*x^(5/2)-280*(a*b)^(1/2)*A*a*b^3*x^(5/2)-875*(a*b)^(1/2)*A*a^2*b^2*x^(3/2)+525*A*arctan(1/(a*b)^(1/2
)*b*x^(1/2))*x^2*a^2*b^3+1575*(a*b)^(1/2)*B*a^3*b*x^(3/2)-945*B*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x^2*a^3*b^2+10
50*A*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x*a^3*b^2-1890*B*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x*a^4*b-525*(a*b)^(1/2)*
A*a^3*b*x^(1/2)+525*A*a^4*b*arctan(1/(a*b)^(1/2)*b*x^(1/2))+945*(a*b)^(1/2)*B*a^4*x^(1/2)-945*B*a^5*arctan(1/(
a*b)^(1/2)*b*x^(1/2)))*(b*x+a)/(a*b)^(1/2)/b^5/((b*x+a)^2)^(3/2)

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maxima [A]  time = 1.86, size = 273, normalized size = 0.90 \begin {gather*} \frac {16 \, {\left (3 \, B b^{4} x^{2} + 5 \, B a b^{3} x\right )} x^{\frac {7}{2}} + {\left (89 \, {\left (11 \, B a b^{3} - 5 \, A b^{4}\right )} x^{2} + 285 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} x^{\frac {5}{2}} + 12 \, {\left (12 \, {\left (11 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 35 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} x^{\frac {3}{2}} + 7 \, {\left (9 \, {\left (11 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + 25 \, {\left (3 \, B a^{4} - A a^{3} b\right )} x\right )} \sqrt {x}}{120 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} - \frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{5}} - \frac {7 \, {\left ({\left (11 \, B a b - 5 \, A b^{2}\right )} x^{\frac {3}{2}} - 2 \, {\left (9 \, B a^{2} - 5 \, A a b\right )} \sqrt {x}\right )}}{8 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/120*(16*(3*B*b^4*x^2 + 5*B*a*b^3*x)*x^(7/2) + (89*(11*B*a*b^3 - 5*A*b^4)*x^2 + 285*(3*B*a^2*b^2 - A*a*b^3)*x
)*x^(5/2) + 12*(12*(11*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 35*(3*B*a^3*b - A*a^2*b^2)*x)*x^(3/2) + 7*(9*(11*B*a^3*b -
5*A*a^2*b^2)*x^2 + 25*(3*B*a^4 - A*a^3*b)*x)*sqrt(x))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4) - 7/4*(
9*B*a^3 - 5*A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) - 7/8*((11*B*a*b - 5*A*b^2)*x^(3/2) - 2*(9*B*
a^2 - 5*A*a*b)*sqrt(x))/b^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^{7/2}\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(7/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int((x^(7/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), 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**(7/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Timed out

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