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

Optimal. Leaf size=115 \[ \frac{\sqrt{b c-a d} (a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^2 b^{3/2}}-\frac{2 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^2}+\frac{\sqrt{c+d x} (b c-a d)}{a b (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0915105, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {98, 156, 63, 208} \[ \frac{\sqrt{b c-a d} (a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^2 b^{3/2}}-\frac{2 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^2}+\frac{\sqrt{c+d x} (b c-a d)}{a b (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*Sqrt[c + d*x])/(a*b*(a + b*x)) - (2*c^(3/2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/a^2 + (Sqrt[b*c - a*d
]*(2*b*c + a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(a^2*b^(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 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 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{(c+d x)^{3/2}}{x (a+b x)^2} \, dx &=\frac{(b c-a d) \sqrt{c+d x}}{a b (a+b x)}+\frac{\int \frac{b c^2+\frac{1}{2} d (b c+a d) x}{x (a+b x) \sqrt{c+d x}} \, dx}{a b}\\ &=\frac{(b c-a d) \sqrt{c+d x}}{a b (a+b x)}+\frac{c^2 \int \frac{1}{x \sqrt{c+d x}} \, dx}{a^2}-\frac{((b c-a d) (2 b c+a d)) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{2 a^2 b}\\ &=\frac{(b c-a d) \sqrt{c+d x}}{a b (a+b x)}+\frac{\left (2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{a^2 d}-\frac{((b c-a d) (2 b c+a d)) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{a^2 b d}\\ &=\frac{(b c-a d) \sqrt{c+d x}}{a b (a+b x)}-\frac{2 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^2}+\frac{\sqrt{b c-a d} (2 b c+a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^2 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.154862, size = 111, normalized size = 0.97 \[ \frac{\frac{\sqrt{b c-a d} (a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{3/2}}+\frac{a \sqrt{c+d x} (b c-a d)}{b (a+b x)}-2 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 194, normalized size = 1.7 \begin{align*} -2\,{\frac{{c}^{3/2}}{{a}^{2}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }-{\frac{{d}^{2}}{b \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{dc}{a \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{{d}^{2}}{b}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}}+{\frac{dc}{a}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}}-2\,{\frac{{c}^{2}b}{{a}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.29862, size = 1388, normalized size = 12.07 \begin{align*} \left [\frac{{\left (2 \, a b c + a^{2} d +{\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x + 2 \, b c - a d + 2 \, \sqrt{d x + c} b \sqrt{\frac{b c - a d}{b}}}{b x + a}\right ) + 2 \,{\left (b^{2} c x + a b c\right )} \sqrt{c} \log \left (\frac{d x - 2 \, \sqrt{d x + c} \sqrt{c} + 2 \, c}{x}\right ) + 2 \,{\left (a b c - a^{2} d\right )} \sqrt{d x + c}}{2 \,{\left (a^{2} b^{2} x + a^{3} b\right )}}, \frac{{\left (2 \, a b c + a^{2} d +{\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) +{\left (b^{2} c x + a b c\right )} \sqrt{c} \log \left (\frac{d x - 2 \, \sqrt{d x + c} \sqrt{c} + 2 \, c}{x}\right ) +{\left (a b c - a^{2} d\right )} \sqrt{d x + c}}{a^{2} b^{2} x + a^{3} b}, \frac{4 \,{\left (b^{2} c x + a b c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x + c} \sqrt{-c}}{c}\right ) +{\left (2 \, a b c + a^{2} d +{\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x + 2 \, b c - a d + 2 \, \sqrt{d x + c} b \sqrt{\frac{b c - a d}{b}}}{b x + a}\right ) + 2 \,{\left (a b c - a^{2} d\right )} \sqrt{d x + c}}{2 \,{\left (a^{2} b^{2} x + a^{3} b\right )}}, \frac{{\left (2 \, a b c + a^{2} d +{\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) + 2 \,{\left (b^{2} c x + a b c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x + c} \sqrt{-c}}{c}\right ) +{\left (a b c - a^{2} d\right )} \sqrt{d x + c}}{a^{2} b^{2} x + a^{3} b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 121.318, size = 928, normalized size = 8.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*d**3*sqrt(c + d*x)/(2*a**2*b*d**2 - 2*a*b**2*c*d + 2*a*b**2*d**2*x - 2*b**3*c*d*x) + a*d**3*sqrt(-1/(b*(a
*d - b*c)**3))*log(-a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) + 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) - b**2*c**2*
sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/(2*b) - a*d**3*sqrt(-1/(b*(a*d - b*c)**3))*log(a**2*d**2*sqrt(-1/
(b*(a*d - b*c)**3)) - 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) + b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c +
d*x))/(2*b) - 2*b*c**2*d*sqrt(c + d*x)/(2*a**3*d**2 - 2*a**2*b*c*d + 2*a**2*b*d**2*x - 2*a*b**2*c*d*x) - c*d*
*2*sqrt(-1/(b*(a*d - b*c)**3))*log(-a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) + 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**
3)) - b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x)) + c*d**2*sqrt(-1/(b*(a*d - b*c)**3))*log(a**2*d**
2*sqrt(-1/(b*(a*d - b*c)**3)) - 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) + b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3))
+ sqrt(c + d*x)) + 4*c*d**2*sqrt(c + d*x)/(2*a**2*d**2 - 2*a*b*c*d + 2*a*b*d**2*x - 2*b**2*c*d*x) + 2*d**2*ata
n(sqrt(c + d*x)/sqrt(a*d/b - c))/(b**2*sqrt(a*d/b - c)) + b*c**2*d*sqrt(-1/(b*(a*d - b*c)**3))*log(-a**2*d**2*
sqrt(-1/(b*(a*d - b*c)**3)) + 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) - b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) +
sqrt(c + d*x))/(2*a) - b*c**2*d*sqrt(-1/(b*(a*d - b*c)**3))*log(a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) - 2*a*b*
c*d*sqrt(-1/(b*(a*d - b*c)**3)) + b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/(2*a) - 2*c**2*atan(s
qrt(c + d*x)/sqrt(a*d/b - c))/(a**2*sqrt(a*d/b - c)) + 2*c**2*atan(sqrt(c + d*x)/sqrt(-c))/(a**2*sqrt(-c))

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Giac [A]  time = 1.23026, size = 194, normalized size = 1.69 \begin{align*} \frac{2 \, c^{2} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c}} - \frac{{\left (2 \, b^{2} c^{2} - a b c d - a^{2} d^{2}\right )} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} a^{2} b} + \frac{\sqrt{d x + c} b c d - \sqrt{d x + c} a d^{2}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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