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

Optimal. Leaf size=119 \[ -\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{3/2}}+\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{3/2}}-\frac{2 \sqrt{a+b x} (b c-a d)}{c d \sqrt{c+d x}} \]

[Out]

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

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Rubi [A]  time = 0.0680015, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {98, 157, 63, 217, 206, 93, 208} \[ -\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{3/2}}+\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{3/2}}-\frac{2 \sqrt{a+b x} (b c-a d)}{c d \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 98

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

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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 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]

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

Mathematica [A]  time = 0.794246, size = 162, normalized size = 1.36 \[ \frac{2 \left (\frac{\sqrt{d} \left (\sqrt{c} \sqrt{a+b x} (a d-b c)-a^{3/2} d \sqrt{c+d x} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )}{c^{3/2}}+b \sqrt{b c-a d} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )\right )}{d^{3/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.022, size = 306, normalized size = 2.6 \begin{align*}{\frac{1}{dc}\sqrt{bx+a} \left ( \ln \left ({\frac{1}{2} \left ( 2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) x{b}^{2}cd\sqrt{ac}-\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac \right ) } \right ) x{a}^{2}{d}^{2}\sqrt{bd}+\ln \left ({\frac{1}{2} \left ( 2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) \sqrt{ac}{b}^{2}{c}^{2}-\sqrt{bd}\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac \right ) } \right ){a}^{2}cd+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}ad-2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}bc \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 6.69364, size = 2140, normalized size = 17.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.34973, size = 277, normalized size = 2.33 \begin{align*} -\frac{2 \, \sqrt{b d} a^{2} 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{\sqrt{b d} b^{2} \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 )}^{2}\right )}{d^{2}{\left | b \right |}} - \frac{2 \,{\left (b^{3} c{\left | b \right |} - a b^{2} d{\left | b \right |}\right )} \sqrt{b x + a}}{\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} b^{2} c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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