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]

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

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Rubi [A]  time = 0.09, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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 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*}

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Mathematica [A]  time = 0.20, 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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fricas [A]  time = 1.10, size = 624, normalized size = 5.43 \[ \left [\frac {{\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (b^{2} c x + a b c\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (a b c - a^{2} d\right )} \sqrt {d x + c}}{2 \, {\left (a^{2} b^{2} x + a^{3} b\right )}}, \frac {{\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (b^{2} c x + a b c\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + {\left (a b c - a^{2} d\right )} \sqrt {d x + c}}{a^{2} b^{2} x + a^{3} b}, \frac {4 \, {\left (b^{2} c x + a b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (a b c - a^{2} d\right )} \sqrt {d x + c}}{2 \, {\left (a^{2} b^{2} x + a^{3} b\right )}}, \frac {{\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + 2 \, {\left (b^{2} c x + a b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left (a b c - a^{2} d\right )} \sqrt {d x + c}}{a^{2} b^{2} x + a^{3} b}\right ] \]

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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giac [A]  time = 1.29, size = 144, normalized size = 1.25 \[ \frac {2 \, c^{2} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c}} - \frac {{\left (2 \, b^{2} c^{2} - a b c d - a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{2} b} + \frac {\sqrt {d x + c} b c d - \sqrt {d x + c} a d^{2}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} a b} \]

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)

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maple [A]  time = 0.02, size = 194, normalized size = 1.69 \[ \frac {c d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a}-\frac {2 b \,c^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b}+\frac {\sqrt {d x +c}\, c d}{\left (b d x +a d \right ) a}-\frac {\sqrt {d x +c}\, d^{2}}{\left (b d x +a d \right ) b}-\frac {2 c^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.60, size = 482, normalized size = 4.19 \[ -\frac {2\,\mathrm {atanh}\left (\frac {4\,d^6\,\sqrt {c^3}\,\sqrt {c+d\,x}}{4\,c^2\,d^6+\frac {8\,b\,c^3\,d^5}{a}-\frac {12\,b^2\,c^4\,d^4}{a^2}}+\frac {8\,c\,d^5\,\sqrt {c^3}\,\sqrt {c+d\,x}}{8\,c^3\,d^5+\frac {4\,a\,c^2\,d^6}{b}-\frac {12\,b\,c^4\,d^4}{a}}-\frac {12\,b\,c^2\,d^4\,\sqrt {c^3}\,\sqrt {c+d\,x}}{8\,a\,c^3\,d^5-12\,b\,c^4\,d^4+\frac {4\,a^2\,c^2\,d^6}{b}}\right )\,\sqrt {c^3}}{a^2}-\frac {\mathrm {atanh}\left (\frac {10\,c^2\,d^5\,\sqrt {b^4\,c-a\,b^3\,d}\,\sqrt {c+d\,x}}{2\,a^2\,c\,d^7+2\,b^2\,c^3\,d^5-\frac {12\,b^3\,c^4\,d^4}{a}+8\,a\,b\,c^2\,d^6}+\frac {12\,c^3\,d^4\,\sqrt {b^4\,c-a\,b^3\,d}\,\sqrt {c+d\,x}}{8\,a^2\,c^2\,d^6-12\,b^2\,c^4\,d^4+\frac {2\,a^3\,c\,d^7}{b}+2\,a\,b\,c^3\,d^5}+\frac {2\,c\,d^6\,\sqrt {b^4\,c-a\,b^3\,d}\,\sqrt {c+d\,x}}{8\,b^2\,c^2\,d^6+2\,a\,b\,c\,d^7+\frac {2\,b^3\,c^3\,d^5}{a}-\frac {12\,b^4\,c^4\,d^4}{a^2}}\right )\,\left (a\,d+2\,b\,c\right )\,\sqrt {-b^3\,\left (a\,d-b\,c\right )}}{a^2\,b^3}-\frac {d\,\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}{a\,b\,\left (a\,d-b\,c+b\,\left (c+d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*atanh((4*d^6*(c^3)^(1/2)*(c + d*x)^(1/2))/(4*c^2*d^6 + (8*b*c^3*d^5)/a - (12*b^2*c^4*d^4)/a^2) + (8*c*d^5
*(c^3)^(1/2)*(c + d*x)^(1/2))/(8*c^3*d^5 + (4*a*c^2*d^6)/b - (12*b*c^4*d^4)/a) - (12*b*c^2*d^4*(c^3)^(1/2)*(c
+ d*x)^(1/2))/(8*a*c^3*d^5 - 12*b*c^4*d^4 + (4*a^2*c^2*d^6)/b))*(c^3)^(1/2))/a^2 - (atanh((10*c^2*d^5*(b^4*c -
a*b^3*d)^(1/2)*(c + d*x)^(1/2))/(2*a^2*c*d^7 + 2*b^2*c^3*d^5 - (12*b^3*c^4*d^4)/a + 8*a*b*c^2*d^6) + (12*c^3*
d^4*(b^4*c - a*b^3*d)^(1/2)*(c + d*x)^(1/2))/(8*a^2*c^2*d^6 - 12*b^2*c^4*d^4 + (2*a^3*c*d^7)/b + 2*a*b*c^3*d^5
) + (2*c*d^6*(b^4*c - a*b^3*d)^(1/2)*(c + d*x)^(1/2))/(8*b^2*c^2*d^6 + 2*a*b*c*d^7 + (2*b^3*c^3*d^5)/a - (12*b
^4*c^4*d^4)/a^2))*(a*d + 2*b*c)*(-b^3*(a*d - b*c))^(1/2))/(a^2*b^3) - (d*(a*d - b*c)*(c + d*x)^(1/2))/(a*b*(a*
d - b*c + b*(c + d*x)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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