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

Optimal. Leaf size=82 \[ \frac {2 x^{3/2} (A b-a B)}{3 a b (a+b x)^{3/2}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{5/2}}-\frac {2 B \sqrt {x}}{b^2 \sqrt {a+b x}} \]

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Rubi [A]  time = 0.03, antiderivative size = 82, 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 x^{3/2} (A b-a B)}{3 a b (a+b x)^{3/2}}-\frac {2 B \sqrt {x}}{b^2 \sqrt {a+b x}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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)^{5/2}} \, dx &=\frac {2 (A b-a B) x^{3/2}}{3 a b (a+b x)^{3/2}}+\frac {B \int \frac {\sqrt {x}}{(a+b x)^{3/2}} \, dx}{b}\\ &=\frac {2 (A b-a B) x^{3/2}}{3 a b (a+b x)^{3/2}}-\frac {2 B \sqrt {x}}{b^2 \sqrt {a+b x}}+\frac {B \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{b^2}\\ &=\frac {2 (A b-a B) x^{3/2}}{3 a b (a+b x)^{3/2}}-\frac {2 B \sqrt {x}}{b^2 \sqrt {a+b x}}+\frac {(2 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}}{3 a b (a+b x)^{3/2}}-\frac {2 B \sqrt {x}}{b^2 \sqrt {a+b x}}+\frac {(2 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}}{3 a b (a+b x)^{3/2}}-\frac {2 B \sqrt {x}}{b^2 \sqrt {a+b x}}+\frac {2 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 = 0.12, size = 95, normalized size = 1.16 \begin {gather*} \frac {6 a^{3/2} B (a+b x) \sqrt {\frac {b x}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+2 \sqrt {b} \sqrt {x} \left (-3 a^2 B-4 a b B x+A b^2 x\right )}{3 a b^{5/2} (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.18, size = 84, normalized size = 1.02 \begin {gather*} -\frac {2 \left (3 a^2 B \sqrt {x}+4 a b B x^{3/2}-A b^2 x^{3/2}\right )}{3 a b^2 (a+b x)^{3/2}}-\frac {2 B \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.87, size = 232, normalized size = 2.83 \begin {gather*} \left [\frac {3 \, {\left (B a b^{2} x^{2} + 2 \, B a^{2} b x + B a^{3}\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (3 \, B a^{2} b + {\left (4 \, B a b^{2} - A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (a b^{5} x^{2} + 2 \, a^{2} b^{4} x + a^{3} b^{3}\right )}}, -\frac {2 \, {\left (3 \, {\left (B a b^{2} x^{2} + 2 \, B a^{2} b x + B a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (3 \, B a^{2} b + {\left (4 \, B a b^{2} - A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}\right )}}{3 \, {\left (a b^{5} x^{2} + 2 \, a^{2} b^{4} x + a^{3} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 108.64, size = 222, normalized size = 2.71 \begin {gather*} -\frac {B {\left | b \right |} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{b^{\frac {7}{2}}} - \frac {4 \, {\left (6 \, B a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} \sqrt {b} {\left | b \right |} + 6 \, B a^{2} {\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 |} - 3 \, A {\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 |} + 4 \, B a^{3} b^{\frac {5}{2}} {\left | b \right |} - A a^{2} 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^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*abs(b)*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^(7/2) - 4/3*(6*B*a*(sqrt(b*x + a)*sqrt(b)
- sqrt((b*x + a)*b - a*b))^4*sqrt(b)*abs(b) + 6*B*a^2*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(
3/2)*abs(b) - 3*A*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b^(3/2)*abs(b) + 4*B*a^3*b^(5/2)*abs(b)
- A*a^2*b^(7/2)*abs(b))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)^3*b^3)

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maple [B]  time = 0.02, size = 182, normalized size = 2.22 \begin {gather*} \frac {\left (3 B a \,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 )+6 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 )+3 B \,a^{3} \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 {5}{2}} x -8 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {3}{2}} x -6 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} \sqrt {b}\right ) \sqrt {x}}{3 \sqrt {\left (b x +a \right ) x}\, \left (b x +a \right )^{\frac {3}{2}} 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)^(5/2),x)

[Out]

1/3*(3*B*a*b^2*x^2*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))+2*((b*x+a)*x)^(1/2)*A*b^(5/2)*x+6*B*a
^2*b*x*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))-8*((b*x+a)*x)^(1/2)*B*a*b^(3/2)*x+3*B*a^3*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^2*b^(1/2))*x^(1/2)/a/((b*x+a)*x)^(1/2)/
b^(5/2)/(b*x+a)^(3/2)

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maxima [A]  time = 2.00, size = 89, normalized size = 1.09 \begin {gather*} -\frac {1}{3} \, B {\left (\frac {2 \, {\left (b + \frac {3 \, {\left (b x + a\right )}}{x}\right )} x^{\frac {3}{2}}}{{\left (b x + a\right )}^{\frac {3}{2}} b^{2}} + \frac {3 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{b^{\frac {5}{2}}}\right )} + \frac {2 \, A x^{\frac {3}{2}}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{5/2}} \,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)^(5/2),x)

[Out]

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

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sympy [B]  time = 17.43, size = 376, normalized size = 4.59 \begin {gather*} \frac {2 A x^{\frac {3}{2}}}{3 a^{\frac {5}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {3}{2}} b x \sqrt {1 + \frac {b x}{a}}} + B \left (\frac {6 a^{\frac {39}{2}} b^{11} x^{\frac {27}{2}} \sqrt {1 + \frac {b x}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{\frac {29}{2}} \sqrt {1 + \frac {b x}{a}}} + \frac {6 a^{\frac {37}{2}} b^{12} x^{\frac {29}{2}} \sqrt {1 + \frac {b x}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{\frac {29}{2}} \sqrt {1 + \frac {b x}{a}}} - \frac {6 a^{19} b^{\frac {23}{2}} x^{14}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{\frac {29}{2}} \sqrt {1 + \frac {b x}{a}}} - \frac {8 a^{18} b^{\frac {25}{2}} x^{15}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{\frac {29}{2}} \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)**(5/2),x)

[Out]

2*A*x**(3/2)/(3*a**(5/2)*sqrt(1 + b*x/a) + 3*a**(3/2)*b*x*sqrt(1 + b*x/a)) + B*(6*a**(39/2)*b**11*x**(27/2)*sq
rt(1 + b*x/a)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/a) + 3*a**(37/2)*b*
*(29/2)*x**(29/2)*sqrt(1 + b*x/a)) + 6*a**(37/2)*b**12*x**(29/2)*sqrt(1 + b*x/a)*asinh(sqrt(b)*sqrt(x)/sqrt(a)
)/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/a) + 3*a**(37/2)*b**(29/2)*x**(29/2)*sqrt(1 + b*x/a)) - 6*a**1
9*b**(23/2)*x**14/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/a) + 3*a**(37/2)*b**(29/2)*x**(29/2)*sqrt(1 +
b*x/a)) - 8*a**18*b**(25/2)*x**15/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/a) + 3*a**(37/2)*b**(29/2)*x**
(29/2)*sqrt(1 + b*x/a)))

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