3.7.38 \(\int \frac {(a+b x)^{3/2}}{x^2 (c+d x)^{3/2}} \, dx\) [638]

Optimal. Leaf size=108 \[ \frac {3 (b c-a d) \sqrt {a+b x}}{c^2 \sqrt {c+d x}}-\frac {(a+b x)^{3/2}}{c x \sqrt {c+d x}}-\frac {3 \sqrt {a} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {96, 95, 214} \begin {gather*} -\frac {3 \sqrt {a} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {3 \sqrt {a+b x} (b c-a d)}{c^2 \sqrt {c+d x}}-\frac {(a+b x)^{3/2}}{c x \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 95

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 96

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[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 214

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

Rubi steps

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

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Mathematica [A]
time = 0.88, size = 135, normalized size = 1.25 \begin {gather*} \frac {\sqrt {a+b x} (2 b c x-a (c+3 d x))}{c^2 x \sqrt {c+d x}}-\frac {3 \sqrt {a} \sqrt {\frac {b}{d}} \sqrt {d} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \left (-b x+\sqrt {\frac {b}{d}} \sqrt {a+b x} \sqrt {c+d x}\right )}{\sqrt {a} \sqrt {b} \sqrt {c}}\right )}{\sqrt {b} c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(297\) vs. \(2(88)=176\).
time = 0.08, size = 298, normalized size = 2.76

method	result	size
default	\(\frac {\sqrt {b x +a}\, \left (3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{2} d^{2} x^{2}-3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a b c d \,x^{2}+3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{2} c d x -3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a b \,c^{2} x -6 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a d x +4 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b c x -2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a c \sqrt {a c}\right )}{2 c^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, x \sqrt {a c}\, \sqrt {d x +c}}\)	\(298\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/x^2/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2*(b*x+a)^(1/2)*(3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*a^2*d^2*x^2-3*ln((a*d*x+b
*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*a*b*c*d*x^2+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x
+a))^(1/2)+2*a*c)/x)*a^2*c*d*x-3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*a*b*c^2*x-6*(
a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*d*x+4*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*b*c*x-2*((d*x+c)*(b*x+a))^(1/2)
*a*c*(a*c)^(1/2))/c^2/((d*x+c)*(b*x+a))^(1/2)/x/(a*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)^(3/2)/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 detail

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Fricas [A]
time = 1.45, size = 343, normalized size = 3.18 \begin {gather*} \left [-\frac {3 \, {\left ({\left (b c d - a d^{2}\right )} x^{2} + {\left (b c^{2} - a c d\right )} x\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 \, {\left (a c - {\left (2 \, b c - 3 \, a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, {\left (c^{2} d x^{2} + c^{3} x\right )}}, \frac {3 \, {\left ({\left (b c d - a d^{2}\right )} x^{2} + {\left (b c^{2} - a c d\right )} x\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 \, {\left (a c - {\left (2 \, b c - 3 \, a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (c^{2} d x^{2} + c^{3} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
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)**(3/2)/x**2/(d*x+c)**(3/2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 475 vs. \(2 (88) = 176\).
time = 2.50, size = 475, normalized size = 4.40 \begin {gather*} \frac {2 \, {\left (b^{3} c - a b^{2} d\right )} \sqrt {b x + a}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} c^{2} {\left | b \right |}} - \frac {3 \, {\left (\sqrt {b d} a b^{3} c - \sqrt {b d} a^{2} 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} a b^{5} c^{2} - 2 \, \sqrt {b d} a^{2} b^{4} c d + \sqrt {b d} a^{3} 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} a 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^{2} 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 |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^3*c - a*b^2*d)*sqrt(b*x + a)/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*c^2*abs(b)) - 3*(sqrt(b*d)*a*b^3*c - sq
rt(b*d)*a^2*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)*a*b^5*c^2 - 2*sqrt(b*d)*a^2*b^4*c*d + sq
rt(b*d)*a^3*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*a*b^3*c - sq
rt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{3/2}}{x^2\,{\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)^(3/2)/(x^2*(c + d*x)^(3/2)),x)

[Out]

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

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