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

Optimal. Leaf size=169 \[ -\frac {5 a (4 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{9/2}}+\frac {5 \sqrt {x} \sqrt {a+b x} (4 A b-7 a B)}{4 b^4}-\frac {5 x^{3/2} \sqrt {a+b x} (4 A b-7 a B)}{6 a b^3}+\frac {2 x^{5/2} (4 A b-7 a B)}{3 a b^2 \sqrt {a+b x}}+\frac {2 x^{7/2} (A b-a B)}{3 a b (a+b x)^{3/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {78, 47, 50, 63, 217, 206} \begin {gather*} \frac {2 x^{5/2} (4 A b-7 a B)}{3 a b^2 \sqrt {a+b x}}-\frac {5 x^{3/2} \sqrt {a+b x} (4 A b-7 a B)}{6 a b^3}+\frac {5 \sqrt {x} \sqrt {a+b x} (4 A b-7 a B)}{4 b^4}-\frac {5 a (4 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{9/2}}+\frac {2 x^{7/2} (A b-a B)}{3 a b (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - a*B)*x^(7/2))/(3*a*b*(a + b*x)^(3/2)) + (2*(4*A*b - 7*a*B)*x^(5/2))/(3*a*b^2*Sqrt[a + b*x]) + (5*(4*
A*b - 7*a*B)*Sqrt[x]*Sqrt[a + b*x])/(4*b^4) - (5*(4*A*b - 7*a*B)*x^(3/2)*Sqrt[a + b*x])/(6*a*b^3) - (5*a*(4*A*
b - 7*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(4*b^(9/2))

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 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {x^{5/2} (A+B x)}{(a+b x)^{5/2}} \, dx &=\frac {2 (A b-a B) x^{7/2}}{3 a b (a+b x)^{3/2}}-\frac {\left (2 \left (2 A b-\frac {7 a B}{2}\right )\right ) \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx}{3 a b}\\ &=\frac {2 (A b-a B) x^{7/2}}{3 a b (a+b x)^{3/2}}+\frac {2 (4 A b-7 a B) x^{5/2}}{3 a b^2 \sqrt {a+b x}}-\frac {(5 (4 A b-7 a B)) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{3 a b^2}\\ &=\frac {2 (A b-a B) x^{7/2}}{3 a b (a+b x)^{3/2}}+\frac {2 (4 A b-7 a B) x^{5/2}}{3 a b^2 \sqrt {a+b x}}-\frac {5 (4 A b-7 a B) x^{3/2} \sqrt {a+b x}}{6 a b^3}+\frac {(5 (4 A b-7 a B)) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{4 b^3}\\ &=\frac {2 (A b-a B) x^{7/2}}{3 a b (a+b x)^{3/2}}+\frac {2 (4 A b-7 a B) x^{5/2}}{3 a b^2 \sqrt {a+b x}}+\frac {5 (4 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{4 b^4}-\frac {5 (4 A b-7 a B) x^{3/2} \sqrt {a+b x}}{6 a b^3}-\frac {(5 a (4 A b-7 a B)) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{8 b^4}\\ &=\frac {2 (A b-a B) x^{7/2}}{3 a b (a+b x)^{3/2}}+\frac {2 (4 A b-7 a B) x^{5/2}}{3 a b^2 \sqrt {a+b x}}+\frac {5 (4 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{4 b^4}-\frac {5 (4 A b-7 a B) x^{3/2} \sqrt {a+b x}}{6 a b^3}-\frac {(5 a (4 A b-7 a B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^4}\\ &=\frac {2 (A b-a B) x^{7/2}}{3 a b (a+b x)^{3/2}}+\frac {2 (4 A b-7 a B) x^{5/2}}{3 a b^2 \sqrt {a+b x}}+\frac {5 (4 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{4 b^4}-\frac {5 (4 A b-7 a B) x^{3/2} \sqrt {a+b x}}{6 a b^3}-\frac {(5 a (4 A b-7 a B)) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^4}\\ &=\frac {2 (A b-a B) x^{7/2}}{3 a b (a+b x)^{3/2}}+\frac {2 (4 A b-7 a B) x^{5/2}}{3 a b^2 \sqrt {a+b x}}+\frac {5 (4 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{4 b^4}-\frac {5 (4 A b-7 a B) x^{3/2} \sqrt {a+b x}}{6 a b^3}-\frac {5 a (4 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 80, normalized size = 0.47 \begin {gather*} \frac {2 x^{7/2} \left ((a+b x) \sqrt {\frac {b x}{a}+1} (7 a B-4 A b) \, _2F_1\left (\frac {3}{2},\frac {7}{2};\frac {9}{2};-\frac {b x}{a}\right )+7 a (A b-a B)\right )}{21 a^2 b (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(7/2)*(7*a*(A*b - a*B) + (-4*A*b + 7*a*B)*(a + b*x)*Sqrt[1 + (b*x)/a]*Hypergeometric2F1[3/2, 7/2, 9/2, -(
(b*x)/a)]))/(21*a^2*b*(a + b*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.32, size = 143, normalized size = 0.85 \begin {gather*} \frac {-105 a^3 B \sqrt {x}+60 a^2 A b \sqrt {x}-140 a^2 b B x^{3/2}+80 a A b^2 x^{3/2}-21 a b^2 B x^{5/2}+12 A b^3 x^{5/2}+6 b^3 B x^{7/2}}{12 b^4 (a+b x)^{3/2}}-\frac {5 \left (7 a^2 B-4 a A b\right ) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{4 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(60*a^2*A*b*Sqrt[x] - 105*a^3*B*Sqrt[x] + 80*a*A*b^2*x^(3/2) - 140*a^2*b*B*x^(3/2) + 12*A*b^3*x^(5/2) - 21*a*b
^2*B*x^(5/2) + 6*b^3*B*x^(7/2))/(12*b^4*(a + b*x)^(3/2)) - (5*(-4*a*A*b + 7*a^2*B)*Log[-(Sqrt[b]*Sqrt[x]) + Sq
rt[a + b*x]])/(4*b^(9/2))

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fricas [A]  time = 1.70, size = 373, normalized size = 2.21 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B a^{4} - 4 \, A a^{3} b + {\left (7 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{3} b - 4 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (6 \, B b^{4} x^{3} - 105 \, B a^{3} b + 60 \, A a^{2} b^{2} - 3 \, {\left (7 \, B a b^{3} - 4 \, A b^{4}\right )} x^{2} - 20 \, {\left (7 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac {15 \, {\left (7 \, B a^{4} - 4 \, A a^{3} b + {\left (7 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{3} b - 4 \, A a^{2} b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (6 \, B b^{4} x^{3} - 105 \, B a^{3} b + 60 \, A a^{2} b^{2} - 3 \, {\left (7 \, B a b^{3} - 4 \, A b^{4}\right )} x^{2} - 20 \, {\left (7 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{12 \, {\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^(5/2)*(B*x+A)/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/24*(15*(7*B*a^4 - 4*A*a^3*b + (7*B*a^2*b^2 - 4*A*a*b^3)*x^2 + 2*(7*B*a^3*b - 4*A*a^2*b^2)*x)*sqrt(b)*log(2
*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(6*B*b^4*x^3 - 105*B*a^3*b + 60*A*a^2*b^2 - 3*(7*B*a*b^3 - 4*A
*b^4)*x^2 - 20*(7*B*a^2*b^2 - 4*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5), -1/12*(15*
(7*B*a^4 - 4*A*a^3*b + (7*B*a^2*b^2 - 4*A*a*b^3)*x^2 + 2*(7*B*a^3*b - 4*A*a^2*b^2)*x)*sqrt(-b)*arctan(sqrt(b*x
+ a)*sqrt(-b)/(b*sqrt(x))) - (6*B*b^4*x^3 - 105*B*a^3*b + 60*A*a^2*b^2 - 3*(7*B*a*b^3 - 4*A*b^4)*x^2 - 20*(7*
B*a^2*b^2 - 4*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)]

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giac [B]  time = 107.31, size = 346, normalized size = 2.05 \begin {gather*} \frac {1}{4} \, \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} B {\left | b \right |}}{b^{6}} - \frac {13 \, B a b^{11} {\left | b \right |} - 4 \, A b^{12} {\left | b \right |}}{b^{17}}\right )} - \frac {5 \, {\left (7 \, B a^{2} \sqrt {b} {\left | b \right |} - 4 \, A a b^{\frac {3}{2}} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{8 \, b^{6}} - \frac {4 \, {\left (12 \, B a^{3} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} \sqrt {b} {\left | b \right |} + 18 \, B a^{4} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {3}{2}} {\left | b \right |} - 9 \, A a^{2} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac {3}{2}} {\left | b \right |} + 10 \, B a^{5} b^{\frac {5}{2}} {\left | b \right |} - 12 \, A a^{3} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {5}{2}} {\left | b \right |} - 7 \, A a^{4} b^{\frac {7}{2}} {\left | b \right |}\right )}}{3 \, {\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)*B*abs(b)/b^6 - (13*B*a*b^11*abs(b) - 4*A*b^12*abs(b))/b
^17) - 5/8*(7*B*a^2*sqrt(b)*abs(b) - 4*A*a*b^(3/2)*abs(b))*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b
))^2)/b^6 - 4/3*(12*B*a^3*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*sqrt(b)*abs(b) + 18*B*a^4*(sqrt(
b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(3/2)*abs(b) - 9*A*a^2*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)
*b - a*b))^4*b^(3/2)*abs(b) + 10*B*a^5*b^(5/2)*abs(b) - 12*A*a^3*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a
*b))^2*b^(5/2)*abs(b) - 7*A*a^4*b^(7/2)*abs(b))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)^3
*b^5)

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maple [B]  time = 0.02, size = 362, normalized size = 2.14 \begin {gather*} -\frac {\left (60 A a \,b^{3} x^{2} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-105 B \,a^{2} b^{2} x^{2} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-12 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {7}{2}} x^{3}+120 A \,a^{2} b^{2} x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-210 B \,a^{3} b x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-24 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {7}{2}} x^{2}+42 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {5}{2}} x^{2}+60 A \,a^{3} b \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-105 B \,a^{4} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-160 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {5}{2}} x +280 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} b^{\frac {3}{2}} x -120 \sqrt {\left (b x +a \right ) x}\, A \,a^{2} b^{\frac {3}{2}}+210 \sqrt {\left (b x +a \right ) x}\, B \,a^{3} \sqrt {b}\right ) \sqrt {x}}{24 \sqrt {\left (b x +a \right ) x}\, \left (b x +a \right )^{\frac {3}{2}} b^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(-12*((b*x+a)*x)^(1/2)*B*b^(7/2)*x^3+60*A*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x^2*a*b^
3-24*((b*x+a)*x)^(1/2)*A*b^(7/2)*x^2-105*B*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x^2*a^2*b^2+4
2*((b*x+a)*x)^(1/2)*B*a*b^(5/2)*x^2+120*A*a^2*b^2*x*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))-160*
((b*x+a)*x)^(1/2)*A*a*b^(5/2)*x-210*B*a^3*b*x*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))+280*((b*x+
a)*x)^(1/2)*B*a^2*b^(3/2)*x+60*A*a^3*b*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))-120*((b*x+a)*x)^(
1/2)*A*a^2*b^(3/2)-105*B*a^4*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))+210*((b*x+a)*x)^(1/2)*B*a^3
*b^(1/2))/b^(9/2)*x^(1/2)/((b*x+a)*x)^(1/2)/(b*x+a)^(3/2)

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maxima [B]  time = 1.09, size = 517, normalized size = 3.06 \begin {gather*} -\frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B a}{b^{6} x^{4} + 4 \, a b^{5} x^{3} + 6 \, a^{2} b^{4} x^{2} + 4 \, a^{3} b^{3} x + a^{4} b^{2}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a^{2}}{6 \, {\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}} + \frac {5 \, \sqrt {b x^{2} + a x} B a^{3}}{6 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} A}{b^{5} x^{4} + 4 \, a b^{4} x^{3} + 6 \, a^{2} b^{3} x^{2} + 4 \, a^{3} b^{2} x + a^{4} b} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{2 \, {\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{6 \, {\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} - \frac {5 \, \sqrt {b x^{2} + a x} A a^{2}}{6 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} - \frac {115 \, \sqrt {b x^{2} + a x} B a^{2}}{12 \, {\left (b^{5} x + a b^{4}\right )}} + \frac {35 \, \sqrt {b x^{2} + a x} A a}{6 \, {\left (b^{4} x + a b^{3}\right )}} + \frac {35 \, B a^{2} \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{8 \, b^{\frac {9}{2}}} - \frac {5 \, A a \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{2 \, b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x^2 + a*x)^(5/2)*B*a/(b^6*x^4 + 4*a*b^5*x^3 + 6*a^2*b^4*x^2 + 4*a^3*b^3*x + a^4*b^2) - 5/6*(b*x^2 + a*x)^(
3/2)*B*a^2/(b^6*x^3 + 3*a*b^5*x^2 + 3*a^2*b^4*x + a^3*b^3) + 5/6*sqrt(b*x^2 + a*x)*B*a^3/(b^6*x^2 + 2*a*b^5*x
+ a^2*b^4) + (b*x^2 + a*x)^(5/2)*A/(b^5*x^4 + 4*a*b^4*x^3 + 6*a^2*b^3*x^2 + 4*a^3*b^2*x + a^4*b) + 1/2*(b*x^2
+ a*x)^(5/2)*B/(b^5*x^3 + 3*a*b^4*x^2 + 3*a^2*b^3*x + a^3*b^2) + 5/6*(b*x^2 + a*x)^(3/2)*A*a/(b^5*x^3 + 3*a*b^
4*x^2 + 3*a^2*b^3*x + a^3*b^2) - 5/4*(b*x^2 + a*x)^(3/2)*B*a/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) - 5/6*sqrt(b*x^2
+ a*x)*A*a^2/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) - 115/12*sqrt(b*x^2 + a*x)*B*a^2/(b^5*x + a*b^4) + 35/6*sqrt(b*x^
2 + a*x)*A*a/(b^4*x + a*b^3) + 35/8*B*a^2*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(9/2) - 5/2*A*a*log(2
*x + a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(7/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{5/2}\,\left (A+B\,x\right )}{{\left (a+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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