3.6.63 \(\int \frac {\sqrt {a+b x}}{x (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=66 \[ \frac {2 \sqrt {a+b x}}{c \sqrt {c+d x}}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {94, 93, 208} \begin {gather*} \frac {2 \sqrt {a+b x}}{c \sqrt {c+d x}}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x])/(c*Sqrt[c + d*x]) - (2*Sqrt[a]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/c^(
3/2)

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 66, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {a+b x}}{c \sqrt {c+d x}}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x])/(c*Sqrt[c + d*x]) - (2*Sqrt[a]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/c^(
3/2)

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IntegrateAlgebraic [A]  time = 0.10, size = 66, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {a+b x}}{c \sqrt {c+d x}}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x])/(c*Sqrt[c + d*x]) - (2*Sqrt[a]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/c^(
3/2)

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fricas [B]  time = 1.59, size = 248, normalized size = 3.76 \begin {gather*} \left [\frac {{\left (d x + c\right )} \sqrt {\frac {a}{c}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (c d x + c^{2}\right )}}, \frac {{\left (d x + c\right )} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) + 2 \, \sqrt {b x + a} \sqrt {d x + c}}{c d x + c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((d*x + c)*sqrt(a/c)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a*c^2 + (b*c^2 + a*c*d)*
x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*sqrt(b*x + a)*sqrt(d*x + c))/(c*d
*x + c^2), ((d*x + c)*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-a/c)/(a*
b*d*x^2 + a^2*c + (a*b*c + a^2*d)*x)) + 2*sqrt(b*x + a)*sqrt(d*x + c))/(c*d*x + c^2)]

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giac [B]  time = 1.27, size = 130, normalized size = 1.97 \begin {gather*} -\frac {2 \, \sqrt {b d} a b \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} c {\left | b \right |}} + \frac {2 \, \sqrt {b x + a} b^{2}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} c {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(b*d)*a*b*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^
2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*c*abs(b)) + 2*sqrt(b*x + a)*b^2/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*c*
abs(b))

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maple [B]  time = 0.02, size = 143, normalized size = 2.17 \begin {gather*} \frac {\left (-a d x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-a c \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right ) \sqrt {b x +a}}{\sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {d x +c}\, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)/x/(d*x+c)^(3/2),x)

[Out]

(-a*d*x*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)-ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((
b*x+a)*(d*x+c))^(1/2))/x)*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))*(b*x+a)^(1/2)/c/(a*c)^(1/2)/((b*x+a)*(d*x
+c))^(1/2)/(d*x+c)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {a+b\,x}}{x\,{\left (c+d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(1/2)/(x*(c + d*x)^(3/2)),x)

[Out]

int((a + b*x)^(1/2)/(x*(c + d*x)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x}}{x \left (c + d x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)/x/(d*x+c)**(3/2),x)

[Out]

Integral(sqrt(a + b*x)/(x*(c + d*x)**(3/2)), x)

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