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

Optimal. Leaf size=113 \[ -\frac {(b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{5/2}}+\frac {\sqrt {a+b x} (b c-3 a d)}{a c^2 \sqrt {c+d x}}-\frac {(a+b x)^{3/2}}{a c x \sqrt {c+d x}} \]

[Out]

-(-3*a*d+b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/c^(5/2)/a^(1/2)-(b*x+a)^(3/2)/a/c/x/(d*x+c)
^(1/2)+(-3*a*d+b*c)*(b*x+a)^(1/2)/a/c^2/(d*x+c)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {96, 94, 93, 208} \[ \frac {\sqrt {a+b x} (b c-3 a d)}{a c^2 \sqrt {c+d x}}-\frac {(b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{5/2}}-\frac {(a+b x)^{3/2}}{a c x \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - 3*a*d)*Sqrt[a + b*x])/(a*c^2*Sqrt[c + d*x]) - (a + b*x)^(3/2)/(a*c*x*Sqrt[c + d*x]) - ((b*c - 3*a*d)*A
rcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(Sqrt[a]*c^(5/2))

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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[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^2 (c+d x)^{3/2}} \, dx &=-\frac {(a+b x)^{3/2}}{a c x \sqrt {c+d x}}-\frac {\left (-\frac {b c}{2}+\frac {3 a d}{2}\right ) \int \frac {\sqrt {a+b x}}{x (c+d x)^{3/2}} \, dx}{a c}\\ &=\frac {(b c-3 a d) \sqrt {a+b x}}{a c^2 \sqrt {c+d x}}-\frac {(a+b x)^{3/2}}{a c x \sqrt {c+d x}}+\frac {(b c-3 a d) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 c^2}\\ &=\frac {(b c-3 a d) \sqrt {a+b x}}{a c^2 \sqrt {c+d x}}-\frac {(a+b x)^{3/2}}{a c x \sqrt {c+d x}}+\frac {(b c-3 a d) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{c^2}\\ &=\frac {(b c-3 a d) \sqrt {a+b x}}{a c^2 \sqrt {c+d x}}-\frac {(a+b x)^{3/2}}{a c x \sqrt {c+d x}}-\frac {(b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 83, normalized size = 0.73 \[ \frac {(3 a d-b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{5/2}}-\frac {\sqrt {a+b x} (c+3 d x)}{c^2 x \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.36, size = 330, normalized size = 2.92 \[ \left [-\frac {{\left ({\left (b c d - 3 \, a d^{2}\right )} x^{2} + {\left (b c^{2} - 3 \, a c d\right )} x\right )} \sqrt {a c} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (3 \, a c d x + a c^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, {\left (a c^{3} d x^{2} + a c^{4} x\right )}}, \frac {{\left ({\left (b c d - 3 \, a d^{2}\right )} x^{2} + {\left (b c^{2} - 3 \, a c d\right )} x\right )} \sqrt {-a c} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (3 \, a c d x + a c^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a c^{3} d x^{2} + a c^{4} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 6.21, size = 457, normalized size = 4.04 \[ -\frac {2 \, \sqrt {b x + a} b^{2} d}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} c^{2} {\left | b \right |}} - \frac {{\left (\sqrt {b d} b^{3} c - 3 \, \sqrt {b d} a b^{2} d\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b c^{2} {\left | b \right |}} - \frac {2 \, {\left (\sqrt {b d} b^{5} c^{2} - 2 \, \sqrt {b d} a b^{4} c d + \sqrt {b d} a^{2} b^{3} d^{2} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{3} c - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{2} d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}\right )} c^{2} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(b*x + a)*b^2*d/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*c^2*abs(b)) - (sqrt(b*d)*b^3*c - 3*sqrt(b*d)*a*b^2
*d)*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)*b*c^2*abs(b)) - 2*(sqrt(b*d)*b^5*c^2 - 2*sqrt(b*d)*a*b^4*c*d + sqrt(b*d)*a^2*b^3*d^2
- sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^3*c - sqrt(b*d)*(sqrt(b*d)*sq
rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^2*d)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(
b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)*c^2*abs(b
))

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maple [B]  time = 0.03, size = 267, normalized size = 2.36 \[ \frac {\sqrt {b x +a}\, \left (3 a \,d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-b c d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+3 a c d x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-b \,c^{2} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-6 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, d x -2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, c \right )}{2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {d x +c}\, c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x+a)^(1/2)/c^2*(3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a*d^2-ln((a*d*x+b
*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*b*c*d+3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*
(d*x+c))^(1/2))/x)*x*a*c*d-ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x*b*c^2-6*x*d*(a*c)
^(1/2)*((b*x+a)*(d*x+c))^(1/2)-2*c*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/(a*c)^(1/2)/x/((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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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