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

Optimal. Leaf size=126 \[ \frac {a^2 (6 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{7/2}}-\frac {a \sqrt {x} \sqrt {a+b x} (6 A b-5 a B)}{8 b^3}+\frac {x^{3/2} \sqrt {a+b x} (6 A b-5 a B)}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b} \]

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Rubi [A]  time = 0.05, antiderivative size = 126, 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 = {80, 50, 63, 217, 206} \begin {gather*} \frac {a^2 (6 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{7/2}}+\frac {x^{3/2} \sqrt {a+b x} (6 A b-5 a B)}{12 b^2}-\frac {a \sqrt {x} \sqrt {a+b x} (6 A b-5 a B)}{8 b^3}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(6*A*b - 5*a*B)*Sqrt[x]*Sqrt[a + b*x])/(8*b^3) + ((6*A*b - 5*a*B)*x^(3/2)*Sqrt[a + b*x])/(12*b^2) + (B*x^(
5/2)*Sqrt[a + b*x])/(3*b) + (a^2*(6*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(8*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 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 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)}{\sqrt {a+b x}} \, dx &=\frac {B x^{5/2} \sqrt {a+b x}}{3 b}+\frac {\left (3 A b-\frac {5 a B}{2}\right ) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{3 b}\\ &=\frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}-\frac {(a (6 A b-5 a B)) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{8 b^2}\\ &=-\frac {a (6 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{8 b^3}+\frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}+\frac {\left (a^2 (6 A b-5 a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{16 b^3}\\ &=-\frac {a (6 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{8 b^3}+\frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}+\frac {\left (a^2 (6 A b-5 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{8 b^3}\\ &=-\frac {a (6 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{8 b^3}+\frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}+\frac {\left (a^2 (6 A b-5 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^3}\\ &=-\frac {a (6 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{8 b^3}+\frac {(6 A b-5 a B) x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b}+\frac {a^2 (6 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 113, normalized size = 0.90 \begin {gather*} \frac {\sqrt {b} \sqrt {x} (a+b x) \left (15 a^2 B-2 a b (9 A+5 B x)+4 b^2 x (3 A+2 B x)\right )-3 a^{5/2} \sqrt {\frac {b x}{a}+1} (5 a B-6 A b) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{24 b^{7/2} \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*Sqrt[x]*(a + b*x)*(15*a^2*B + 4*b^2*x*(3*A + 2*B*x) - 2*a*b*(9*A + 5*B*x)) - 3*a^(5/2)*(-6*A*b + 5*a*
B)*Sqrt[1 + (b*x)/a]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(24*b^(7/2)*Sqrt[a + b*x])

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IntegrateAlgebraic [A]  time = 0.20, size = 117, normalized size = 0.93 \begin {gather*} \frac {\sqrt {a+b x} \left (15 a^2 B \sqrt {x}-18 a A b \sqrt {x}-10 a b B x^{3/2}+12 A b^2 x^{3/2}+8 b^2 B x^{5/2}\right )}{24 b^3}+\frac {\left (5 a^3 B-6 a^2 A b\right ) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{8 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(-18*a*A*b*Sqrt[x] + 15*a^2*B*Sqrt[x] + 12*A*b^2*x^(3/2) - 10*a*b*B*x^(3/2) + 8*b^2*B*x^(5/2)))
/(24*b^3) + ((-6*a^2*A*b + 5*a^3*B)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(8*b^(7/2))

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fricas [A]  time = 1.59, size = 200, normalized size = 1.59 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (8 \, B b^{3} x^{2} + 15 \, B a^{2} b - 18 \, A a b^{2} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b^{4}}, \frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (8 \, B b^{3} x^{2} + 15 \, B a^{2} b - 18 \, A a b^{2} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(5*B*a^3 - 6*A*a^2*b)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(8*B*b^3*x^2 + 15
*B*a^2*b - 18*A*a*b^2 - 2*(5*B*a*b^2 - 6*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4, 1/24*(3*(5*B*a^3 - 6*A*a^2*b)*s
qrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*B*b^3*x^2 + 15*B*a^2*b - 18*A*a*b^2 - 2*(5*B*a*b^2 - 6
*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4]

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giac [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^(3/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.02, size = 176, normalized size = 1.40 \begin {gather*} \frac {\sqrt {b x +a}\, \left (16 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {5}{2}} x^{2}+18 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 )+24 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {5}{2}} x -20 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {3}{2}} x -36 \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}}{48 \sqrt {\left (b x +a \right ) x}\, 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)^(1/2),x)

[Out]

1/48*x^(1/2)*(b*x+a)^(1/2)/b^(7/2)*(16*((b*x+a)*x)^(1/2)*B*b^(5/2)*x^2+24*((b*x+a)*x)^(1/2)*A*b^(5/2)*x-20*((b
*x+a)*x)^(1/2)*B*a*b^(3/2)*x+18*A*a^2*b*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))-36*((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*x+a)*x)^(1/2)

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maxima [A]  time = 0.91, size = 160, normalized size = 1.27 \begin {gather*} \frac {\sqrt {b x^{2} + a x} B x^{2}}{3 \, b} - \frac {5 \, \sqrt {b x^{2} + a x} B a x}{12 \, b^{2}} + \frac {\sqrt {b x^{2} + a x} A x}{2 \, b} - \frac {5 \, B a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {5}{2}}} + \frac {5 \, \sqrt {b x^{2} + a x} B a^{2}}{8 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a x} A a}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

1/3*sqrt(b*x^2 + a*x)*B*x^2/b - 5/12*sqrt(b*x^2 + a*x)*B*a*x/b^2 + 1/2*sqrt(b*x^2 + a*x)*A*x/b - 5/16*B*a^3*lo
g(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) + 3/8*A*a^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^
(5/2) + 5/8*sqrt(b*x^2 + a*x)*B*a^2/b^3 - 3/4*sqrt(b*x^2 + a*x)*A*a/b^2

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

[Out]

int((x^(3/2)*(A + B*x))/(a + b*x)^(1/2), x)

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sympy [B]  time = 36.74, size = 245, normalized size = 1.94 \begin {gather*} - \frac {3 A a^{\frac {3}{2}} \sqrt {x}}{4 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {A \sqrt {a} x^{\frac {3}{2}}}{4 b \sqrt {1 + \frac {b x}{a}}} + \frac {3 A a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {5}{2}}} + \frac {A x^{\frac {5}{2}}}{2 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} + \frac {5 B a^{\frac {5}{2}} \sqrt {x}}{8 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {5 B a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {B \sqrt {a} x^{\frac {5}{2}}}{12 b \sqrt {1 + \frac {b x}{a}}} - \frac {5 B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {7}{2}}} + \frac {B x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(B*x+A)/(b*x+a)**(1/2),x)

[Out]

-3*A*a**(3/2)*sqrt(x)/(4*b**2*sqrt(1 + b*x/a)) - A*sqrt(a)*x**(3/2)/(4*b*sqrt(1 + b*x/a)) + 3*A*a**2*asinh(sqr
t(b)*sqrt(x)/sqrt(a))/(4*b**(5/2)) + A*x**(5/2)/(2*sqrt(a)*sqrt(1 + b*x/a)) + 5*B*a**(5/2)*sqrt(x)/(8*b**3*sqr
t(1 + b*x/a)) + 5*B*a**(3/2)*x**(3/2)/(24*b**2*sqrt(1 + b*x/a)) - B*sqrt(a)*x**(5/2)/(12*b*sqrt(1 + b*x/a)) -
5*B*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(7/2)) + B*x**(7/2)/(3*sqrt(a)*sqrt(1 + b*x/a))

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