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

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

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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 = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {770, 78, 51, 63, 205} \begin {gather*} -\frac {7 b (a+b x) (9 A b-5 a B)}{4 a^5 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 (a+b x) (9 A b-5 a B)}{12 a^4 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (a+b x) (9 A b-5 a B)}{20 a^3 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b^{3/2} (a+b x) (9 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 {A+B x}{x^{7/2} \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 {A+B x}{x^{7/2} \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}{2 a b x^{5/2} (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((9 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{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 {9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (7 (9 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{x^{7/2} \left (a b+b^2 x\right )} \, dx}{8 a^2 b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (9 A b-5 a B) (a+b x)}{20 a^3 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 (9 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{x^{5/2} \left (a b+b^2 x\right )} \, dx}{8 a^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (9 A b-5 a B) (a+b x)}{20 a^3 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 (9 A b-5 a B) (a+b x)}{12 a^4 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (7 b (9 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{x^{3/2} \left (a b+b^2 x\right )} \, dx}{8 a^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (9 A b-5 a B) (a+b x)}{20 a^3 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 (9 A b-5 a B) (a+b x)}{12 a^4 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b (9 A b-5 a B) (a+b x)}{4 a^5 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 b^2 (9 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}{8 a^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (9 A b-5 a B) (a+b x)}{20 a^3 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 (9 A b-5 a B) (a+b x)}{12 a^4 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b (9 A b-5 a B) (a+b x)}{4 a^5 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 b^2 (9 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 )}{4 a^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A b-a B}{2 a b x^{5/2} (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (9 A b-5 a B) (a+b x)}{20 a^3 b x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 (9 A b-5 a B) (a+b x)}{12 a^4 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b (9 A b-5 a B) (a+b x)}{4 a^5 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b^{3/2} (9 A b-5 a B) (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{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 {5 a^2 (A b-a B)+(a+b x)^2 (5 a B-9 A b) \, _2F_1\left (-\frac {5}{2},2;-\frac {3}{2};-\frac {b x}{a}\right )}{10 a^3 b x^{5/2} (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^2*(A*b - a*B) + (-9*A*b + 5*a*B)*(a + b*x)^2*Hypergeometric2F1[-5/2, 2, -3/2, -((b*x)/a)])/(10*a^3*b*x^(5
/2)*(a + b*x)*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [A]  time = 33.99, size = 166, normalized size = 0.55 \begin {gather*} \frac {(a+b x) \left (\frac {7 \left (5 a b^{3/2} B-9 A b^{5/2}\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{11/2}}+\frac {-24 a^4 A-40 a^4 B x+72 a^3 A b x+280 a^3 b B x^2-504 a^2 A b^2 x^2+875 a^2 b^2 B x^3-1575 a A b^3 x^3+525 a b^3 B x^4-945 A b^4 x^4}{60 a^5 x^{5/2} (a+b x)^2}\right )}{\sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 437, normalized size = 1.45 \begin {gather*} \left [-\frac {105 \, {\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + 2 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4} + {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (24 \, A a^{4} - 105 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} - 175 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 56 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt {x}}{120 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}, -\frac {105 \, {\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + 2 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4} + {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (24 \, A a^{4} - 105 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} - 175 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 56 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt {x}}{60 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 159, normalized size = 0.53 \begin {gather*} \frac {7 \, {\left (5 \, B a b^{2} - 9 \, A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {11 \, B a b^{3} x^{\frac {3}{2}} - 15 \, A b^{4} x^{\frac {3}{2}} + 13 \, B a^{2} b^{2} \sqrt {x} - 17 \, A a b^{3} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, {\left (45 \, B a b x^{2} - 90 \, A b^{2} x^{2} - 5 \, B a^{2} x + 15 \, A a b x - 3 \, A a^{2}\right )}}{15 \, a^{5} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 289, normalized size = 0.96 \begin {gather*} -\frac {\left (945 A \,b^{5} x^{\frac {9}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-525 B a \,b^{4} x^{\frac {9}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+1890 A a \,b^{4} x^{\frac {7}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-1050 B \,a^{2} b^{3} x^{\frac {7}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+945 A \,a^{2} b^{3} x^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-525 B \,a^{3} b^{2} x^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+945 \sqrt {a b}\, A \,b^{4} x^{4}-525 \sqrt {a b}\, B a \,b^{3} x^{4}+1575 \sqrt {a b}\, A a \,b^{3} x^{3}-875 \sqrt {a b}\, B \,a^{2} b^{2} x^{3}+504 \sqrt {a b}\, A \,a^{2} b^{2} x^{2}-280 \sqrt {a b}\, B \,a^{3} b \,x^{2}-72 \sqrt {a b}\, A \,a^{3} b x +40 \sqrt {a b}\, B \,a^{4} x +24 \sqrt {a b}\, A \,a^{4}\right ) \left (b x +a \right )}{60 \sqrt {a b}\, \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} a^{5} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(-1050*B*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x^(7/2)*a^2*b^3+1890*A*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x^(7/2)*
a*b^4-525*B*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x^(9/2)*a*b^4-280*B*(a*b)^(1/2)*x^2*a^3*b-72*A*(a*b)^(1/2)*x*a^3*b
+945*A*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x^(5/2)*a^2*b^3-875*B*(a*b)^(1/2)*x^3*a^2*b^2-525*B*arctan(1/(a*b)^(1/2
)*b*x^(1/2))*x^(5/2)*a^3*b^2+504*A*(a*b)^(1/2)*x^2*a^2*b^2+24*A*(a*b)^(1/2)*a^4+945*A*(a*b)^(1/2)*x^4*b^4+40*B
*(a*b)^(1/2)*x*a^4+945*A*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x^(9/2)*b^5-525*B*(a*b)^(1/2)*x^4*a*b^3+1575*A*(a*b)^
(1/2)*x^3*a*b^3)*(b*x+a)/x^(5/2)/(a*b)^(1/2)/a^5/((b*x+a)^2)^(3/2)

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maxima [A]  time = 1.68, size = 404, normalized size = 1.34 \begin {gather*} \frac {1260 \, {\left (5 \, B a^{3} b^{4} - 11 \, A a^{2} b^{5}\right )} x^{\frac {5}{2}} + 35 \, {\left (5 \, {\left (B a b^{6} - 3 \, A b^{7}\right )} x^{2} + 9 \, {\left (5 \, B a^{2} b^{5} - 11 \, A a b^{6}\right )} x\right )} x^{\frac {5}{2}} - 105 \, {\left (15 \, {\left (B a^{3} b^{4} - 3 \, A a^{2} b^{5}\right )} x^{2} - 17 \, {\left (5 \, B a^{4} b^{3} - 11 \, A a^{3} b^{4}\right )} x\right )} \sqrt {x} - \frac {112 \, {\left (25 \, {\left (B a^{4} b^{3} - 3 \, A a^{3} b^{4}\right )} x^{2} - 9 \, {\left (5 \, B a^{5} b^{2} - 11 \, A a^{4} b^{3}\right )} x\right )}}{\sqrt {x}} - \frac {48 \, {\left (35 \, {\left (B a^{5} b^{2} - 3 \, A a^{4} b^{3}\right )} x^{2} - 3 \, {\left (5 \, B a^{6} b - 11 \, A a^{5} b^{2}\right )} x\right )}}{x^{\frac {3}{2}}} - \frac {16 \, {\left (15 \, {\left (B a^{6} b - 3 \, A a^{5} b^{2}\right )} x^{2} + {\left (5 \, B a^{7} - 11 \, A a^{6} b\right )} x\right )}}{x^{\frac {5}{2}}} - \frac {16 \, {\left (5 \, A a^{6} b x^{2} + 3 \, A a^{7} x\right )}}{x^{\frac {7}{2}}}}{120 \, {\left (a^{7} b^{3} x^{3} + 3 \, a^{8} b^{2} x^{2} + 3 \, a^{9} b x + a^{10}\right )}} + \frac {7 \, {\left (5 \, B a b^{2} - 9 \, A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{5}} - \frac {7 \, {\left (5 \, {\left (B a b^{3} - 3 \, A b^{4}\right )} x^{\frac {3}{2}} + 6 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} \sqrt {x}\right )}}{24 \, a^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(1260*(5*B*a^3*b^4 - 11*A*a^2*b^5)*x^(5/2) + 35*(5*(B*a*b^6 - 3*A*b^7)*x^2 + 9*(5*B*a^2*b^5 - 11*A*a*b^6
)*x)*x^(5/2) - 105*(15*(B*a^3*b^4 - 3*A*a^2*b^5)*x^2 - 17*(5*B*a^4*b^3 - 11*A*a^3*b^4)*x)*sqrt(x) - 112*(25*(B
*a^4*b^3 - 3*A*a^3*b^4)*x^2 - 9*(5*B*a^5*b^2 - 11*A*a^4*b^3)*x)/sqrt(x) - 48*(35*(B*a^5*b^2 - 3*A*a^4*b^3)*x^2
- 3*(5*B*a^6*b - 11*A*a^5*b^2)*x)/x^(3/2) - 16*(15*(B*a^6*b - 3*A*a^5*b^2)*x^2 + (5*B*a^7 - 11*A*a^6*b)*x)/x^
(5/2) - 16*(5*A*a^6*b*x^2 + 3*A*a^7*x)/x^(7/2))/(a^7*b^3*x^3 + 3*a^8*b^2*x^2 + 3*a^9*b*x + a^10) + 7/4*(5*B*a*
b^2 - 9*A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) - 7/24*(5*(B*a*b^3 - 3*A*b^4)*x^(3/2) + 6*(5*B*a^2*
b^2 - 9*A*a*b^3)*sqrt(x))/a^7

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

[Out]

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

[Out]

Timed out

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