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

Optimal. Leaf size=77 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}-\frac {2 c \sqrt {a+b x}}{d \sqrt {c+d x} (b c-a d)} \]

[Out]

2*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(3/2)/b^(1/2)-2*c*(b*x+a)^(1/2)/d/(-a*d+b*c)/(d*x+c)^
(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {78, 63, 217, 206} \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}-\frac {2 c \sqrt {a+b x}}{d \sqrt {c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.18, size = 110, normalized size = 1.43 \[ \frac {2 \left (b c \sqrt {d} \sqrt {a+b x}-(b c-a d)^{3/2} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )\right )}{b d^{3/2} \sqrt {c+d x} (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.05, size = 335, normalized size = 4.35 \[ \left [-\frac {4 \, \sqrt {b x + a} \sqrt {d x + c} b c d - {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right )}{2 \, {\left (b^{2} c^{2} d^{2} - a b c d^{3} + {\left (b^{2} c d^{3} - a b d^{4}\right )} x\right )}}, -\frac {2 \, \sqrt {b x + a} \sqrt {d x + c} b c d + {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right )}{b^{2} c^{2} d^{2} - a b c d^{3} + {\left (b^{2} c d^{3} - a b d^{4}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(4*sqrt(b*x + a)*sqrt(d*x + c)*b*c*d - (b*c^2 - a*c*d + (b*c*d - a*d^2)*x)*sqrt(b*d)*log(8*b^2*d^2*x^2 +
b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d +
a*b*d^2)*x))/(b^2*c^2*d^2 - a*b*c*d^3 + (b^2*c*d^3 - a*b*d^4)*x), -(2*sqrt(b*x + a)*sqrt(d*x + c)*b*c*d + (b*c
^2 - a*c*d + (b*c*d - a*d^2)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x
+ c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)))/(b^2*c^2*d^2 - a*b*c*d^3 + (b^2*c*d^3 - a*b*d^4)*x)]

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giac [A]  time = 1.26, size = 108, normalized size = 1.40 \[ -\frac {2 \, {\left (\frac {\sqrt {b x + a} b^{3} c {\left | b \right |}}{{\left (b^{3} c d - a b^{2} d^{2}\right )} \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {{\left | b \right |} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d}\right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(sqrt(b*x + a)*b^3*c*abs(b)/((b^3*c*d - a*b^2*d^2)*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) + abs(b)*log(abs(-s
qrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d))/b

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maple [B]  time = 0.02, size = 251, normalized size = 3.26 \[ \frac {\sqrt {b x +a}\, \left (a \,d^{2} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-b c d x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+a c d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-b \,c^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, c \right )}{\sqrt {b d}\, \left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {d x +c}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(d*x+c)^(3/2)/(b*x+a)^(1/2),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 = -0.01 \[ \int \frac {x}{\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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