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

Optimal. Leaf size=134 \[ -\frac {3 a (4 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}}+\frac {3 \sqrt {x} \sqrt {a+b x} (4 A b-5 a B)}{4 b^3}-\frac {x^{3/2} \sqrt {a+b x} (4 A b-5 a B)}{2 a b^2}+\frac {2 x^{5/2} (A b-a B)}{a b \sqrt {a+b x}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {78, 50, 63, 217, 206} \begin {gather*} -\frac {x^{3/2} \sqrt {a+b x} (4 A b-5 a B)}{2 a b^2}+\frac {3 \sqrt {x} \sqrt {a+b x} (4 A b-5 a B)}{4 b^3}-\frac {3 a (4 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}}+\frac {2 x^{5/2} (A b-a B)}{a b \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 106, normalized size = 0.79 \begin {gather*} \frac {3 a^{3/2} \sqrt {\frac {b x}{a}+1} (5 a B-4 A b) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+\sqrt {b} \sqrt {x} \left (-15 a^2 B+a b (12 A-5 B x)+2 b^2 x (2 A+B x)\right )}{4 b^{7/2} \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*Sqrt[x]*(-15*a^2*B + a*b*(12*A - 5*B*x) + 2*b^2*x*(2*A + B*x)) + 3*a^(3/2)*(-4*A*b + 5*a*B)*Sqrt[1 +
(b*x)/a]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(7/2)*Sqrt[a + b*x])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.23, size = 115, normalized size = 0.86 \begin {gather*} \frac {-15 a^2 B \sqrt {x}+12 a A b \sqrt {x}-5 a b B x^{3/2}+4 A b^2 x^{3/2}+2 b^2 B x^{5/2}}{4 b^3 \sqrt {a+b x}}-\frac {3 \left (5 a^2 B-4 a A b\right ) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{4 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a*A*b*Sqrt[x] - 15*a^2*B*Sqrt[x] + 4*A*b^2*x^(3/2) - 5*a*b*B*x^(3/2) + 2*b^2*B*x^(5/2))/(4*b^3*Sqrt[a + b*
x]) - (3*(-4*a*A*b + 5*a^2*B)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(4*b^(7/2))

________________________________________________________________________________________

fricas [A]  time = 1.44, size = 255, normalized size = 1.90 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B a^{3} - 4 \, A a^{2} b + {\left (5 \, B a^{2} b - 4 \, A a 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 (2 \, B b^{3} x^{2} - 15 \, B a^{2} b + 12 \, A a b^{2} - {\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{8 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {3 \, {\left (5 \, B a^{3} - 4 \, A a^{2} b + {\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, B b^{3} x^{2} - 15 \, B a^{2} b + 12 \, A a b^{2} - {\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{4 \, {\left (b^{5} x + a 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*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(3*(5*B*a^3 - 4*A*a^2*b + (5*B*a^2*b - 4*A*a*b^2)*x)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x)
+ a) - 2*(2*B*b^3*x^2 - 15*B*a^2*b + 12*A*a*b^2 - (5*B*a*b^2 - 4*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^5*x + a*
b^4), -1/4*(3*(5*B*a^3 - 4*A*a^2*b + (5*B*a^2*b - 4*A*a*b^2)*x)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt
(x))) - (2*B*b^3*x^2 - 15*B*a^2*b + 12*A*a*b^2 - (5*B*a*b^2 - 4*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^5*x + a*b^
4)]

________________________________________________________________________________________

giac [A]  time = 107.23, size = 181, normalized size = 1.35 \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^{5}} - \frac {9 \, B a b^{9} {\left | b \right |} - 4 \, A b^{10} {\left | b \right |}}{b^{14}}\right )} - \frac {3 \, {\left (5 \, 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^{5}} - \frac {4 \, {\left (B a^{3} \sqrt {b} {\left | b \right |} - A a^{2} b^{\frac {3}{2}} {\left | b \right |}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)/(b*x+a)^(3/2),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^5 - (9*B*a*b^9*abs(b) - 4*A*b^10*abs(b))/b^1
4) - 3/8*(5*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^5 - 4*(B*a^3*sqrt(b)*abs(b) - A*a^2*b^(3/2)*abs(b))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^
2 + a*b)*b^4)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 244, normalized size = 1.82 \begin {gather*} -\frac {\left (12 A a \,b^{2} x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-15 B \,a^{2} b x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-4 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {5}{2}} x^{2}+12 A \,a^{2} b \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-15 B \,a^{3} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-8 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {5}{2}} x +10 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {3}{2}} x -24 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {3}{2}}+30 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} \sqrt {b}\right ) \sqrt {x}}{8 \sqrt {\left (b x +a \right ) x}\, \sqrt {b x +a}\, b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(-4*((b*x+a)*x)^(1/2)*B*b^(5/2)*x^2+12*A*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x*a*b^2-8*
((b*x+a)*x)^(1/2)*A*b^(5/2)*x-15*B*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x*a^2*b+10*((b*x+a)*x
)^(1/2)*B*a*b^(3/2)*x+12*A*a^2*b*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))-24*((b*x+a)*x)^(1/2)*A*
a*b^(3/2)-15*B*a^3*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))+30*((b*x+a)*x)^(1/2)*B*a^2*b^(1/2))/b
^(7/2)*x^(1/2)/((b*x+a)*x)^(1/2)/(b*x+a)^(1/2)

________________________________________________________________________________________

maxima [B]  time = 1.00, size = 246, normalized size = 1.84 \begin {gather*} -\frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}} - \frac {3 \, \sqrt {b x^{2} + a x} B a^{2}}{b^{4} x + a b^{3}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{2 \, {\left (b^{3} x + a b^{2}\right )}} + \frac {3 \, \sqrt {b x^{2} + a x} A a}{b^{3} x + a b^{2}} + \frac {15 \, B a^{2} \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{8 \, b^{\frac {7}{2}}} - \frac {3 \, A a \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{2 \, b^{\frac {5}{2}}} - \frac {3 \, \sqrt {b x^{2} + a x} B 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*x+a)^(3/2),x, algorithm="maxima")

[Out]

-(b*x^2 + a*x)^(3/2)*B*a/(b^4*x^2 + 2*a*b^3*x + a^2*b^2) - 3*sqrt(b*x^2 + a*x)*B*a^2/(b^4*x + a*b^3) + (b*x^2
+ a*x)^(3/2)*A/(b^3*x^2 + 2*a*b^2*x + a^2*b) + 1/2*(b*x^2 + a*x)^(3/2)*B/(b^3*x + a*b^2) + 3*sqrt(b*x^2 + a*x)
*A*a/(b^3*x + a*b^2) + 15/8*B*a^2*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(7/2) - 3/2*A*a*log(2*x + a/b
+ 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(5/2) - 3/4*sqrt(b*x^2 + a*x)*B*a/b^3

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{3/2}\,\left (A+B\,x\right )}{{\left (a+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 59.06, size = 182, normalized size = 1.36 \begin {gather*} A \left (\frac {3 \sqrt {a} \sqrt {x}}{b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {5}{2}}} + \frac {x^{\frac {3}{2}}}{\sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) + B \left (- \frac {15 a^{\frac {3}{2}} \sqrt {x}}{4 b^{3} \sqrt {1 + \frac {b x}{a}}} - \frac {5 \sqrt {a} x^{\frac {3}{2}}}{4 b^{2} \sqrt {1 + \frac {b x}{a}}} + \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {7}{2}}} + \frac {x^{\frac {5}{2}}}{2 \sqrt {a} b \sqrt {1 + \frac {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*x+a)**(3/2),x)

[Out]

A*(3*sqrt(a)*sqrt(x)/(b**2*sqrt(1 + b*x/a)) - 3*a*asinh(sqrt(b)*sqrt(x)/sqrt(a))/b**(5/2) + x**(3/2)/(sqrt(a)*
b*sqrt(1 + b*x/a))) + B*(-15*a**(3/2)*sqrt(x)/(4*b**3*sqrt(1 + b*x/a)) - 5*sqrt(a)*x**(3/2)/(4*b**2*sqrt(1 + b
*x/a)) + 15*a**2*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(4*b**(7/2)) + x**(5/2)/(2*sqrt(a)*b*sqrt(1 + b*x/a)))

________________________________________________________________________________________