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

Optimal. Leaf size=286 \[ \frac {2 x^{7/2} (a+b x) (A b-a B)}{7 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^2 x^{3/2} (a+b x) (A b-a B)}{3 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a x^{5/2} (a+b x) (A b-a B)}{5 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{9/2} (a+b x)}{9 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^{7/2} (a+b x) (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a^3 \sqrt {x} (a+b x) (A b-a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.13, antiderivative size = 286, 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, 80, 50, 63, 205} \begin {gather*} \frac {2 x^{7/2} (a+b x) (A b-a B)}{7 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a x^{5/2} (a+b x) (A b-a B)}{5 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^2 x^{3/2} (a+b x) (A b-a B)}{3 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a^3 \sqrt {x} (a+b x) (A b-a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^{7/2} (a+b x) (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{9/2} (a+b x)}{9 b \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))/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(-2*a^3*(A*b - a*B)*Sqrt[x]*(a + b*x))/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (2*a^2*(A*b - a*B)*x^(3/2)*(a + b
*x))/(3*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (2*a*(A*b - a*B)*x^(5/2)*(a + b*x))/(5*b^3*Sqrt[a^2 + 2*a*b*x + b
^2*x^2]) + (2*(A*b - a*B)*x^(7/2)*(a + b*x))/(7*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (2*B*x^(9/2)*(a + b*x))/(
9*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (2*a^(7/2)*(A*b - a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(b^(1
1/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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 80

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

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)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {x^{7/2} (A+B x)}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 B x^{9/2} (a+b x)}{9 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (\frac {9 A b^2}{2}-\frac {9 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac {x^{7/2}}{a b+b^2 x} \, dx}{9 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) x^{7/2} (a+b x)}{7 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{9/2} (a+b x)}{9 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (2 a \left (\frac {9 A b^2}{2}-\frac {9 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac {x^{5/2}}{a b+b^2 x} \, dx}{9 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 a (A b-a B) x^{5/2} (a+b x)}{5 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{7/2} (a+b x)}{7 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{9/2} (a+b x)}{9 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 a^2 \left (\frac {9 A b^2}{2}-\frac {9 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac {x^{3/2}}{a b+b^2 x} \, dx}{9 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 a^2 (A b-a B) x^{3/2} (a+b x)}{3 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a (A b-a B) x^{5/2} (a+b x)}{5 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{7/2} (a+b x)}{7 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{9/2} (a+b x)}{9 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (2 a^3 \left (\frac {9 A b^2}{2}-\frac {9 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {x}}{a b+b^2 x} \, dx}{9 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 a^3 (A b-a B) \sqrt {x} (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^2 (A b-a B) x^{3/2} (a+b x)}{3 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a (A b-a B) x^{5/2} (a+b x)}{5 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{7/2} (a+b x)}{7 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{9/2} (a+b x)}{9 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 a^4 \left (\frac {9 A b^2}{2}-\frac {9 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac {1}{\sqrt {x} \left (a b+b^2 x\right )} \, dx}{9 b^6 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 a^3 (A b-a B) \sqrt {x} (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^2 (A b-a B) x^{3/2} (a+b x)}{3 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a (A b-a B) x^{5/2} (a+b x)}{5 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{7/2} (a+b x)}{7 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{9/2} (a+b x)}{9 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (4 a^4 \left (\frac {9 A b^2}{2}-\frac {9 a b B}{2}\right ) \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 )}{9 b^6 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 a^3 (A b-a B) \sqrt {x} (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^2 (A b-a B) x^{3/2} (a+b x)}{3 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a (A b-a B) x^{5/2} (a+b x)}{5 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{7/2} (a+b x)}{7 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{9/2} (a+b x)}{9 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^{7/2} (A b-a B) (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 139, normalized size = 0.49 \begin {gather*} \frac {2 (a+b x) \left (\sqrt {b} \sqrt {x} \left (315 a^4 B-105 a^3 b (3 A+B x)+21 a^2 b^2 x (5 A+3 B x)-9 a b^3 x^2 (7 A+5 B x)+5 b^4 x^3 (9 A+7 B x)\right )-315 a^{7/2} (a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{315 b^{11/2} \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*(Sqrt[b]*Sqrt[x]*(315*a^4*B - 105*a^3*b*(3*A + B*x) + 21*a^2*b^2*x*(5*A + 3*B*x) - 9*a*b^3*x^2*(7
*A + 5*B*x) + 5*b^4*x^3*(9*A + 7*B*x)) - 315*a^(7/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(315*b
^(11/2)*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [A]  time = 17.42, size = 153, normalized size = 0.53 \begin {gather*} \frac {(a+b x) \left (\frac {2 \sqrt {x} \left (315 a^4 B-315 a^3 A b-105 a^3 b B x+105 a^2 A b^2 x+63 a^2 b^2 B x^2-63 a A b^3 x^2-45 a b^3 B x^3+45 A b^4 x^3+35 b^4 B x^4\right )}{315 b^5}-\frac {2 \left (a^{9/2} B-a^{7/2} A b\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{11/2}}\right )}{\sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*((2*Sqrt[x]*(-315*a^3*A*b + 315*a^4*B + 105*a^2*A*b^2*x - 105*a^3*b*B*x - 63*a*A*b^3*x^2 + 63*a^2*b
^2*B*x^2 + 45*A*b^4*x^3 - 45*a*b^3*B*x^3 + 35*b^4*B*x^4))/(315*b^5) - (2*(-(a^(7/2)*A*b) + a^(9/2)*B)*ArcTan[(
Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(11/2)))/Sqrt[(a + b*x)^2]

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fricas [A]  time = 0.47, size = 276, normalized size = 0.97 \begin {gather*} \left [-\frac {315 \, {\left (B a^{4} - A a^{3} b\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 (35 \, B b^{4} x^{4} + 315 \, B a^{4} - 315 \, A a^{3} b - 45 \, {\left (B a b^{3} - A b^{4}\right )} x^{3} + 63 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 105 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{315 \, b^{5}}, -\frac {2 \, {\left (315 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (35 \, B b^{4} x^{4} + 315 \, B a^{4} - 315 \, A a^{3} b - 45 \, {\left (B a b^{3} - A b^{4}\right )} x^{3} + 63 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 105 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}\right )}}{315 \, b^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/315*(315*(B*a^4 - A*a^3*b)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(35*B*b^4*x^4
+ 315*B*a^4 - 315*A*a^3*b - 45*(B*a*b^3 - A*b^4)*x^3 + 63*(B*a^2*b^2 - A*a*b^3)*x^2 - 105*(B*a^3*b - A*a^2*b^2
)*x)*sqrt(x))/b^5, -2/315*(315*(B*a^4 - A*a^3*b)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) - (35*B*b^4*x^4 + 315
*B*a^4 - 315*A*a^3*b - 45*(B*a*b^3 - A*b^4)*x^3 + 63*(B*a^2*b^2 - A*a*b^3)*x^2 - 105*(B*a^3*b - A*a^2*b^2)*x)*
sqrt(x))/b^5]

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giac [A]  time = 0.37, size = 205, normalized size = 0.72 \begin {gather*} -\frac {2 \, {\left (B a^{5} \mathrm {sgn}\left (b x + a\right ) - A a^{4} b \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{5}} + \frac {2 \, {\left (35 \, B b^{8} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) - 45 \, B a b^{7} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + 45 \, A b^{8} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + 63 \, B a^{2} b^{6} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) - 63 \, A a b^{7} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) - 105 \, B a^{3} b^{5} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 105 \, A a^{2} b^{6} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 315 \, B a^{4} b^{4} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - 315 \, A a^{3} b^{5} \sqrt {x} \mathrm {sgn}\left (b x + a\right )\right )}}{315 \, b^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

-2*(B*a^5*sgn(b*x + a) - A*a^4*b*sgn(b*x + a))*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) + 2/315*(35*B*b^8*x
^(9/2)*sgn(b*x + a) - 45*B*a*b^7*x^(7/2)*sgn(b*x + a) + 45*A*b^8*x^(7/2)*sgn(b*x + a) + 63*B*a^2*b^6*x^(5/2)*s
gn(b*x + a) - 63*A*a*b^7*x^(5/2)*sgn(b*x + a) - 105*B*a^3*b^5*x^(3/2)*sgn(b*x + a) + 105*A*a^2*b^6*x^(3/2)*sgn
(b*x + a) + 315*B*a^4*b^4*sqrt(x)*sgn(b*x + a) - 315*A*a^3*b^5*sqrt(x)*sgn(b*x + a))/b^9

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maple [A]  time = 0.06, size = 197, normalized size = 0.69 \begin {gather*} \frac {2 \left (b x +a \right ) \left (35 \sqrt {a b}\, B \,b^{4} x^{\frac {9}{2}}+45 \sqrt {a b}\, A \,b^{4} x^{\frac {7}{2}}-45 \sqrt {a b}\, B a \,b^{3} x^{\frac {7}{2}}-63 \sqrt {a b}\, A a \,b^{3} x^{\frac {5}{2}}+63 \sqrt {a b}\, B \,a^{2} b^{2} x^{\frac {5}{2}}+315 A \,a^{4} b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-315 B \,a^{5} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+105 \sqrt {a b}\, A \,a^{2} b^{2} x^{\frac {3}{2}}-105 \sqrt {a b}\, B \,a^{3} b \,x^{\frac {3}{2}}-315 \sqrt {a b}\, A \,a^{3} b \sqrt {x}+315 \sqrt {a b}\, B \,a^{4} \sqrt {x}\right )}{315 \sqrt {\left (b x +a \right )^{2}}\, \sqrt {a b}\, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x+A)/((b*x+a)^2)^(1/2),x)

[Out]

2/315*(b*x+a)*(35*B*x^(9/2)*(a*b)^(1/2)*b^4+45*A*x^(7/2)*(a*b)^(1/2)*b^4-45*B*x^(7/2)*(a*b)^(1/2)*a*b^3-63*A*x
^(5/2)*(a*b)^(1/2)*a*b^3+63*B*x^(5/2)*(a*b)^(1/2)*a^2*b^2+105*A*x^(3/2)*(a*b)^(1/2)*a^2*b^2-105*B*x^(3/2)*(a*b
)^(1/2)*a^3*b-315*A*x^(1/2)*(a*b)^(1/2)*a^3*b+315*A*arctan(1/(a*b)^(1/2)*b*x^(1/2))*a^4*b+315*B*x^(1/2)*(a*b)^
(1/2)*a^4-315*B*arctan(1/(a*b)^(1/2)*b*x^(1/2))*a^5)/((b*x+a)^2)^(1/2)/b^5/(a*b)^(1/2)

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maxima [A]  time = 1.52, size = 257, normalized size = 0.90 \begin {gather*} \frac {10 \, {\left (7 \, B b^{4} x^{2} + 9 \, B a b^{3} x\right )} x^{\frac {7}{2}} - 2 \, {\left (5 \, {\left (11 \, B a b^{3} - 9 \, A b^{4}\right )} x^{2} + 9 \, {\left (9 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x\right )} x^{\frac {5}{2}} + 6 \, {\left (3 \, {\left (11 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{2} + 7 \, {\left (9 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x\right )} x^{\frac {3}{2}} + 21 \, {\left (3 \, {\left (11 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 5 \, {\left (9 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {x}}{315 \, {\left (b^{5} x + a b^{4}\right )}} - \frac {2 \, {\left (B a^{5} - A a^{4} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{5}} - \frac {{\left (11 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {x}}{3 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/315*(10*(7*B*b^4*x^2 + 9*B*a*b^3*x)*x^(7/2) - 2*(5*(11*B*a*b^3 - 9*A*b^4)*x^2 + 9*(9*B*a^2*b^2 - 7*A*a*b^3)*
x)*x^(5/2) + 6*(3*(11*B*a^2*b^2 - 9*A*a*b^3)*x^2 + 7*(9*B*a^3*b - 7*A*a^2*b^2)*x)*x^(3/2) + 21*(3*(11*B*a^3*b
- 9*A*a^2*b^2)*x^2 + 5*(9*B*a^4 - 7*A*a^3*b)*x)*sqrt(x))/(b^5*x + a*b^4) - 2*(B*a^5 - A*a^4*b)*arctan(b*sqrt(x
)/sqrt(a*b))/(sqrt(a*b)*b^5) - 1/3*((11*B*a^3*b - 9*A*a^2*b^2)*x^(3/2) - 6*(B*a^4 - A*a^3*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 )}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(7/2)*(A + B*x))/((a + b*x)^2)^(1/2),x)

[Out]

int((x^(7/2)*(A + B*x))/((a + b*x)^2)^(1/2), 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*x+a)**2)**(1/2),x)

[Out]

Timed out

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