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

Optimal. Leaf size=98 \[ \frac {(2 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{5/2}}-\frac {\sqrt {x} \sqrt {a+b x} (2 A b-3 a B)}{a b^2}+\frac {2 x^{3/2} (A b-a B)}{a b \sqrt {a+b x}} \]

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Rubi [A]  time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {\sqrt {x} \sqrt {a+b x} (2 A b-3 a B)}{a b^2}+\frac {(2 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{5/2}}+\frac {2 x^{3/2} (A b-a B)}{a b \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.09, size = 85, normalized size = 0.87 \begin {gather*} \frac {\sqrt {b} \sqrt {x} (3 a B-2 A b+b B x)-\sqrt {a} \sqrt {\frac {b x}{a}+1} (3 a B-2 A b) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2} \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.17, size = 79, normalized size = 0.81 \begin {gather*} \frac {(3 a B-2 A b) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{b^{5/2}}+\frac {3 a B \sqrt {x}-2 A b \sqrt {x}+b B x^{3/2}}{b^2 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.44, size = 195, normalized size = 1.99 \begin {gather*} \left [-\frac {{\left (3 \, B a^{2} - 2 \, A a b + {\left (3 \, B a b - 2 \, 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 (B b^{2} x + 3 \, B a b - 2 \, A b^{2}\right )} \sqrt {b x + a} \sqrt {x}}{2 \, {\left (b^{4} x + a b^{3}\right )}}, \frac {{\left (3 \, B a^{2} - 2 \, A a b + {\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (B b^{2} x + 3 \, B a b - 2 \, A b^{2}\right )} \sqrt {b x + a} \sqrt {x}}{b^{4} x + a b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 86.24, size = 144, normalized size = 1.47 \begin {gather*} \frac {\sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} B {\left | b \right |}}{b^{4}} + \frac {{\left (3 \, B a \sqrt {b} {\left | b \right |} - 2 \, 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 )}{2 \, b^{4}} + \frac {4 \, {\left (B a^{2} \sqrt {b} {\left | b \right |} - A a 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^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*B*abs(b)/b^4 + 1/2*(3*B*a*sqrt(b)*abs(b) - 2*A*b^(3/2)*abs(b))*log((sqrt
(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^4 + 4*(B*a^2*sqrt(b)*abs(b) - A*a*b^(3/2)*abs(b))/(((sqrt(b*
x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)*b^3)

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maple [B]  time = 0.02, size = 201, normalized size = 2.05 \begin {gather*} \frac {\left (2 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 )-3 B a b x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+2 A a b \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-3 B \,a^{2} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+2 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {3}{2}} x -4 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {3}{2}}+6 \sqrt {\left (b x +a \right ) x}\, B a \sqrt {b}\right ) \sqrt {x}}{2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b x +a}\, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*A*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x*b^2-3*B*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b
^(1/2))/b^(1/2))*x*a*b+2*((b*x+a)*x)^(1/2)*B*b^(3/2)*x+2*A*a*b*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^
(1/2))-4*((b*x+a)*x)^(1/2)*A*b^(3/2)-3*B*a^2*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))+6*((b*x+a)*
x)^(1/2)*B*a*b^(1/2))*x^(1/2)/((b*x+a)*x)^(1/2)/b^(5/2)/(b*x+a)^(1/2)

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maxima [A]  time = 1.01, size = 134, normalized size = 1.37 \begin {gather*} \frac {2 \, \sqrt {b x^{2} + a x} B a}{b^{3} x + a b^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} A}{b^{2} x + a b} - \frac {3 \, B 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 {A \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}} + \frac {\sqrt {b x^{2} + a x} B}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b*x^2 + a*x)*B*a/(b^3*x + a*b^2) - 2*sqrt(b*x^2 + a*x)*A/(b^2*x + a*b) - 3/2*B*a*log(2*x + a/b + 2*sqrt
(b*x^2 + a*x)/sqrt(b))/b^(5/2) + A*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(3/2) + sqrt(b*x^2 + a*x)*B/
b^2

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

[Out]

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

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sympy [A]  time = 9.10, size = 122, normalized size = 1.24 \begin {gather*} A \left (\frac {2 \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {2 \sqrt {x}}{\sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) + B \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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