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

Optimal. Leaf size=69 \[ -\frac {2 A (a+b x)^{3/2}}{3 a x^{3/2}}-\frac {2 B \sqrt {a+b x}}{\sqrt {x}}+2 \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 69, 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, 47, 63, 217, 206} \begin {gather*} -\frac {2 A (a+b x)^{3/2}}{3 a x^{3/2}}-\frac {2 B \sqrt {a+b x}}{\sqrt {x}}+2 \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.10, size = 85, normalized size = 1.23 \begin {gather*} -\frac {2 (a+b x)^{3/2} (A b-a B)}{3 a b x^{3/2}}-\frac {2 a B \sqrt {a+b x} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};-\frac {b x}{a}\right )}{3 b x^{3/2} \sqrt {\frac {b x}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.15, size = 66, normalized size = 0.96 \begin {gather*} -\frac {2 \sqrt {a+b x} (a A+3 a B x+A b x)}{3 a x^{3/2}}-2 \sqrt {b} B \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.44, size = 134, normalized size = 1.94 \begin {gather*} \left [\frac {3 \, B a \sqrt {b} x^{2} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (A a + {\left (3 \, B a + A b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3 \, a x^{2}}, -\frac {2 \, {\left (3 \, B a \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (A a + {\left (3 \, B a + A b\right )} x\right )} \sqrt {b x + a} \sqrt {x}\right )}}{3 \, a x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(5/2),x, algorithm="fricas")

[Out]

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

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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((B*x+A)*(b*x+a)^(1/2)/x^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.02, size = 112, normalized size = 1.62 \begin {gather*} -\frac {\sqrt {b x +a}\, \left (-3 B a b \,x^{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}\, A \,b^{\frac {3}{2}} x +6 \sqrt {\left (b x +a \right ) x}\, B a \sqrt {b}\, x +2 \sqrt {\left (b x +a \right ) x}\, A a \sqrt {b}\right )}{3 \sqrt {\left (b x +a \right ) x}\, a \sqrt {b}\, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x+a)^(1/2)/x^(5/2),x)

[Out]

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

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maxima [A]  time = 2.01, size = 73, normalized size = 1.06 \begin {gather*} -{\left (\sqrt {b} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right ) + \frac {2 \, \sqrt {b x + a}}{\sqrt {x}}\right )} B - \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} A}{3 \, a x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(1/2))/x^(5/2),x)

[Out]

int(((A + B*x)*(a + b*x)^(1/2))/x^(5/2), x)

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sympy [A]  time = 24.27, size = 114, normalized size = 1.65 \begin {gather*} A \left (- \frac {2 \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {2 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3 a}\right ) + B \left (- \frac {2 \sqrt {a}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} + 2 \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} - \frac {2 b \sqrt {x}}{\sqrt {a} \sqrt {1 + \frac {b x}{a}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)**(1/2)/x**(5/2),x)

[Out]

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

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