3.473 \(\int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^2} \, dx\)

Optimal. Leaf size=147 \[ -\frac {(b c-a d)^{3/2} (a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 b^{3/2}}+\frac {c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^3}-\frac {\sqrt {c+d x} (b c-a d)^2}{a^2 b (a+b x)}-\frac {c^2 \sqrt {c+d x}}{a^2 x} \]

[Out]

c^(3/2)*(-5*a*d+4*b*c)*arctanh((d*x+c)^(1/2)/c^(1/2))/a^3-(-a*d+b*c)^(3/2)*(a*d+4*b*c)*arctanh(b^(1/2)*(d*x+c)
^(1/2)/(-a*d+b*c)^(1/2))/a^3/b^(3/2)-c^2*(d*x+c)^(1/2)/a^2/x-(-a*d+b*c)^2*(d*x+c)^(1/2)/a^2/b/(b*x+a)

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Rubi [A]  time = 0.20, antiderivative size = 159, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {98, 149, 156, 63, 208} \[ -\frac {(b c-a d)^{3/2} (a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 b^{3/2}}+\frac {c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^3}-\frac {\sqrt {c+d x} (b c-a d) (2 b c-a d)}{a^2 b (a+b x)}-\frac {c (c+d x)^{3/2}}{a x (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(x^2*(a + b*x)^2),x]

[Out]

-(((b*c - a*d)*(2*b*c - a*d)*Sqrt[c + d*x])/(a^2*b*(a + b*x))) - (c*(c + d*x)^(3/2))/(a*x*(a + b*x)) + (c^(3/2
)*(4*b*c - 5*a*d)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/a^3 - ((b*c - a*d)^(3/2)*(4*b*c + a*d)*ArcTanh[(Sqrt[b]*Sqrt
[c + d*x])/Sqrt[b*c - a*d]])/(a^3*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 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(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 - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

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)^{5/2}}{x^2 (a+b x)^2} \, dx &=-\frac {c (c+d x)^{3/2}}{a x (a+b x)}-\frac {\int \frac {\sqrt {c+d x} \left (\frac {1}{2} c (4 b c-5 a d)+\frac {1}{2} d (b c-2 a d) x\right )}{x (a+b x)^2} \, dx}{a}\\ &=-\frac {(b c-a d) (2 b c-a d) \sqrt {c+d x}}{a^2 b (a+b x)}-\frac {c (c+d x)^{3/2}}{a x (a+b x)}+\frac {\int \frac {-\frac {1}{2} b c^2 (4 b c-5 a d)-\frac {1}{2} d \left (2 b^2 c^2-2 a b c d-a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}} \, dx}{a^2 b}\\ &=-\frac {(b c-a d) (2 b c-a d) \sqrt {c+d x}}{a^2 b (a+b x)}-\frac {c (c+d x)^{3/2}}{a x (a+b x)}-\frac {\left (c^2 (4 b c-5 a d)\right ) \int \frac {1}{x \sqrt {c+d x}} \, dx}{2 a^3}+\frac {\left ((b c-a d)^2 (4 b c+a d)\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{2 a^3 b}\\ &=-\frac {(b c-a d) (2 b c-a d) \sqrt {c+d x}}{a^2 b (a+b x)}-\frac {c (c+d x)^{3/2}}{a x (a+b x)}-\frac {\left (c^2 (4 b c-5 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{a^3 d}+\frac {\left ((b c-a d)^2 (4 b c+a d)\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{a^3 b d}\\ &=-\frac {(b c-a d) (2 b c-a d) \sqrt {c+d x}}{a^2 b (a+b x)}-\frac {c (c+d x)^{3/2}}{a x (a+b x)}+\frac {c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^3}-\frac {(b c-a d)^{3/2} (4 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 159, normalized size = 1.08 \[ \frac {-\frac {a \sqrt {c+d x} \left (a^2 d^2 x+a b c (c-2 d x)+2 b^2 c^2 x\right )}{b x (a+b x)}-\frac {\sqrt {b c-a d} \left (-a^2 d^2-3 a b c d+4 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2}}+c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/2)/(x^2*(a + b*x)^2),x]

[Out]

(-((a*Sqrt[c + d*x]*(2*b^2*c^2*x + a^2*d^2*x + a*b*c*(c - 2*d*x)))/(b*x*(a + b*x))) + c^(3/2)*(4*b*c - 5*a*d)*
ArcTanh[Sqrt[c + d*x]/Sqrt[c]] - (Sqrt[b*c - a*d]*(4*b^2*c^2 - 3*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c +
d*x])/Sqrt[b*c - a*d]])/b^(3/2))/a^3

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fricas [A]  time = 0.99, size = 1002, normalized size = 6.82 \[ \left [-\frac {{\left ({\left (4 \, b^{3} c^{2} - 3 \, a b^{2} c d - a^{2} b d^{2}\right )} x^{2} + {\left (4 \, a b^{2} c^{2} - 3 \, a^{2} b c d - a^{3} d^{2}\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 ) + {\left ({\left (4 \, b^{3} c^{2} - 5 \, a b^{2} c d\right )} x^{2} + {\left (4 \, a b^{2} c^{2} - 5 \, a^{2} b c d\right )} x\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (a^{2} b c^{2} + {\left (2 \, a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x\right )} \sqrt {d x + c}}{2 \, {\left (a^{3} b^{2} x^{2} + a^{4} b x\right )}}, -\frac {2 \, {\left ({\left (4 \, b^{3} c^{2} - 3 \, a b^{2} c d - a^{2} b d^{2}\right )} x^{2} + {\left (4 \, a b^{2} c^{2} - 3 \, a^{2} b c d - a^{3} d^{2}\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 ({\left (4 \, b^{3} c^{2} - 5 \, a b^{2} c d\right )} x^{2} + {\left (4 \, a b^{2} c^{2} - 5 \, a^{2} b c d\right )} x\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (a^{2} b c^{2} + {\left (2 \, a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x\right )} \sqrt {d x + c}}{2 \, {\left (a^{3} b^{2} x^{2} + a^{4} b x\right )}}, -\frac {2 \, {\left ({\left (4 \, b^{3} c^{2} - 5 \, a b^{2} c d\right )} x^{2} + {\left (4 \, a b^{2} c^{2} - 5 \, a^{2} b c d\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left ({\left (4 \, b^{3} c^{2} - 3 \, a b^{2} c d - a^{2} b d^{2}\right )} x^{2} + {\left (4 \, a b^{2} c^{2} - 3 \, a^{2} b c d - a^{3} d^{2}\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^{2} b c^{2} + {\left (2 \, a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x\right )} \sqrt {d x + c}}{2 \, {\left (a^{3} b^{2} x^{2} + a^{4} b x\right )}}, -\frac {{\left ({\left (4 \, b^{3} c^{2} - 3 \, a b^{2} c d - a^{2} b d^{2}\right )} x^{2} + {\left (4 \, a b^{2} c^{2} - 3 \, a^{2} b c d - a^{3} d^{2}\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 ({\left (4 \, b^{3} c^{2} - 5 \, a b^{2} c d\right )} x^{2} + {\left (4 \, a b^{2} c^{2} - 5 \, a^{2} b c d\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left (a^{2} b c^{2} + {\left (2 \, a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x\right )} \sqrt {d x + c}}{a^{3} b^{2} x^{2} + a^{4} b x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^2/(b*x+a)^2,x, algorithm="fricas")

[Out]

[-1/2*(((4*b^3*c^2 - 3*a*b^2*c*d - a^2*b*d^2)*x^2 + (4*a*b^2*c^2 - 3*a^2*b*c*d - a^3*d^2)*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)) + ((4*b^3*c^2 - 5*a*b^2*c*d)*x
^2 + (4*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(a^2*b*c^2 + (2*a
*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x)*sqrt(d*x + c))/(a^3*b^2*x^2 + a^4*b*x), -1/2*(2*((4*b^3*c^2 - 3*a*b^2*c*d
- a^2*b*d^2)*x^2 + (4*a*b^2*c^2 - 3*a^2*b*c*d - a^3*d^2)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt
(-(b*c - a*d)/b)/(b*c - a*d)) + ((4*b^3*c^2 - 5*a*b^2*c*d)*x^2 + (4*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(c)*log((d
*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(a^2*b*c^2 + (2*a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x)*sqrt(d*x + c)
)/(a^3*b^2*x^2 + a^4*b*x), -1/2*(2*((4*b^3*c^2 - 5*a*b^2*c*d)*x^2 + (4*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(-c)*ar
ctan(sqrt(d*x + c)*sqrt(-c)/c) + ((4*b^3*c^2 - 3*a*b^2*c*d - a^2*b*d^2)*x^2 + (4*a*b^2*c^2 - 3*a^2*b*c*d - a^3
*d^2)*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^2*b*c^2 + (2*a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x)*sqrt(d*x + c))/(a^3*b^2*x^2 + a^4*b*x), -(((4*b^3*c^2 -
3*a*b^2*c*d - a^2*b*d^2)*x^2 + (4*a*b^2*c^2 - 3*a^2*b*c*d - a^3*d^2)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x
+ c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + ((4*b^3*c^2 - 5*a*b^2*c*d)*x^2 + (4*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqr
t(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) + (a^2*b*c^2 + (2*a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x)*sqrt(d*x + c))/
(a^3*b^2*x^2 + a^4*b*x)]

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giac [B]  time = 1.39, size = 260, normalized size = 1.77 \[ -\frac {{\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a^{3} \sqrt {-c}} + \frac {{\left (4 \, b^{3} c^{3} - 7 \, a b^{2} c^{2} d + 2 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{3} b} - \frac {2 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c^{2} d - 2 \, \sqrt {d x + c} b^{2} c^{3} d - 2 \, {\left (d x + c\right )}^{\frac {3}{2}} a b c d^{2} + 3 \, \sqrt {d x + c} a b c^{2} d^{2} + {\left (d x + c\right )}^{\frac {3}{2}} a^{2} d^{3} - \sqrt {d x + c} a^{2} c d^{3}}{{\left ({\left (d x + c\right )}^{2} b - 2 \, {\left (d x + c\right )} b c + b c^{2} + {\left (d x + c\right )} a d - a c d\right )} a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^2/(b*x+a)^2,x, algorithm="giac")

[Out]

-(4*b*c^3 - 5*a*c^2*d)*arctan(sqrt(d*x + c)/sqrt(-c))/(a^3*sqrt(-c)) + (4*b^3*c^3 - 7*a*b^2*c^2*d + 2*a^2*b*c*
d^2 + a^3*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a^3*b) - (2*(d*x + c)^(3/2)*
b^2*c^2*d - 2*sqrt(d*x + c)*b^2*c^3*d - 2*(d*x + c)^(3/2)*a*b*c*d^2 + 3*sqrt(d*x + c)*a*b*c^2*d^2 + (d*x + c)^
(3/2)*a^2*d^3 - sqrt(d*x + c)*a^2*c*d^3)/(((d*x + c)^2*b - 2*(d*x + c)*b*c + b*c^2 + (d*x + c)*a*d - a*c*d)*a^
2*b)

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maple [B]  time = 0.02, size = 313, normalized size = 2.13 \[ \frac {2 c \,d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a}-\frac {7 b \,c^{2} d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a^{2}}+\frac {4 b^{2} c^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a^{3}}+\frac {d^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b}+\frac {2 \sqrt {d x +c}\, c \,d^{2}}{\left (b d x +a d \right ) a}-\frac {\sqrt {d x +c}\, b \,c^{2} d}{\left (b d x +a d \right ) a^{2}}-\frac {\sqrt {d x +c}\, d^{3}}{\left (b d x +a d \right ) b}-\frac {5 c^{\frac {3}{2}} d \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{2}}+\frac {4 b \,c^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{3}}-\frac {\sqrt {d x +c}\, c^{2}}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)/x^2/(b*x+a)^2,x)

[Out]

-d^3/b*(d*x+c)^(1/2)/(b*d*x+a*d)+2*d^2/a*(d*x+c)^(1/2)/(b*d*x+a*d)*c-d/a^2*b*(d*x+c)^(1/2)/(b*d*x+a*d)*c^2+d^3
/b/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)+2*d^2/a/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^
(1/2)/((a*d-b*c)*b)^(1/2)*b)*c-7*d/a^2*b/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*c^2+4
/a^3/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*c^3*b^2-c^2*(d*x+c)^(1/2)/a^2/x-5*d*c^(3/
2)/a^2*arctanh((d*x+c)^(1/2)/c^(1/2))+4*c^(5/2)/a^3*arctanh((d*x+c)^(1/2)/c^(1/2))*b

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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)^(5/2)/x^2/(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 = 0.40, size = 1127, normalized size = 7.67 \[ \frac {\frac {\sqrt {c+d\,x}\,\left (a^2\,c\,d^3-3\,a\,b\,c^2\,d^2+2\,b^2\,c^3\,d\right )}{a^2\,b}-\frac {d\,{\left (c+d\,x\right )}^{3/2}\,\left (a^2\,d^2-2\,a\,b\,c\,d+2\,b^2\,c^2\right )}{a^2\,b}}{\left (a\,d-2\,b\,c\right )\,\left (c+d\,x\right )+b\,{\left (c+d\,x\right )}^2+b\,c^2-a\,c\,d}-\frac {\mathrm {atanh}\left (\frac {10\,d^9\,\sqrt {c^3}\,\sqrt {c+d\,x}}{10\,c^2\,d^9+\frac {32\,b\,c^3\,d^8}{a}-\frac {132\,b^2\,c^4\,d^7}{a^2}+\frac {130\,b^3\,c^5\,d^6}{a^3}-\frac {40\,b^4\,c^6\,d^5}{a^4}}+\frac {32\,c\,d^8\,\sqrt {c^3}\,\sqrt {c+d\,x}}{32\,c^3\,d^8+\frac {10\,a\,c^2\,d^9}{b}-\frac {132\,b\,c^4\,d^7}{a}+\frac {130\,b^2\,c^5\,d^6}{a^2}-\frac {40\,b^3\,c^6\,d^5}{a^3}}-\frac {132\,b\,c^2\,d^7\,\sqrt {c^3}\,\sqrt {c+d\,x}}{32\,a\,c^3\,d^8-132\,b\,c^4\,d^7+\frac {130\,b^2\,c^5\,d^6}{a}+\frac {10\,a^2\,c^2\,d^9}{b}-\frac {40\,b^3\,c^6\,d^5}{a^2}}+\frac {130\,b^2\,c^3\,d^6\,\sqrt {c^3}\,\sqrt {c+d\,x}}{32\,a^2\,c^3\,d^8+130\,b^2\,c^5\,d^6-\frac {40\,b^3\,c^6\,d^5}{a}+\frac {10\,a^3\,c^2\,d^9}{b}-132\,a\,b\,c^4\,d^7}-\frac {40\,b^3\,c^4\,d^5\,\sqrt {c^3}\,\sqrt {c+d\,x}}{32\,a^3\,c^3\,d^8-40\,b^3\,c^6\,d^5+130\,a\,b^2\,c^5\,d^6-132\,a^2\,b\,c^4\,d^7+\frac {10\,a^4\,c^2\,d^9}{b}}\right )\,\left (5\,a\,d-4\,b\,c\right )\,\sqrt {c^3}}{a^3}-\frac {\mathrm {atanh}\left (\frac {30\,c^3\,d^6\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{14\,a^3\,c^2\,d^9+110\,b^3\,c^5\,d^6-82\,a\,b^2\,c^4\,d^7-4\,a^2\,b\,c^3\,d^8+\frac {2\,a^4\,c\,d^{10}}{b}-\frac {40\,b^4\,c^6\,d^5}{a}}-\frac {2\,c\,d^8\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{4\,b^3\,c^3\,d^8-14\,a\,b^2\,c^2\,d^9+\frac {82\,b^4\,c^4\,d^7}{a}-\frac {110\,b^5\,c^5\,d^6}{a^2}+\frac {40\,b^6\,c^6\,d^5}{a^3}-2\,a^2\,b\,c\,d^{10}}+\frac {18\,c^2\,d^7\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{2\,a^3\,c\,d^{10}-82\,b^3\,c^4\,d^7-4\,a\,b^2\,c^3\,d^8+14\,a^2\,b\,c^2\,d^9+\frac {110\,b^4\,c^5\,d^6}{a}-\frac {40\,b^5\,c^6\,d^5}{a^2}}+\frac {40\,c^4\,d^5\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{4\,a^3\,c^3\,d^8+40\,b^3\,c^6\,d^5-110\,a\,b^2\,c^5\,d^6+82\,a^2\,b\,c^4\,d^7-\frac {2\,a^5\,c\,d^{10}}{b^2}-\frac {14\,a^4\,c^2\,d^9}{b}}\right )\,\sqrt {-b^3\,{\left (a\,d-b\,c\right )}^3}\,\left (a\,d+4\,b\,c\right )}{a^3\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(x^2*(a + b*x)^2),x)

[Out]

(((c + d*x)^(1/2)*(a^2*c*d^3 + 2*b^2*c^3*d - 3*a*b*c^2*d^2))/(a^2*b) - (d*(c + d*x)^(3/2)*(a^2*d^2 + 2*b^2*c^2
- 2*a*b*c*d))/(a^2*b))/((a*d - 2*b*c)*(c + d*x) + b*(c + d*x)^2 + b*c^2 - a*c*d) - (atanh((10*d^9*(c^3)^(1/2)
*(c + d*x)^(1/2))/(10*c^2*d^9 + (32*b*c^3*d^8)/a - (132*b^2*c^4*d^7)/a^2 + (130*b^3*c^5*d^6)/a^3 - (40*b^4*c^6
*d^5)/a^4) + (32*c*d^8*(c^3)^(1/2)*(c + d*x)^(1/2))/(32*c^3*d^8 + (10*a*c^2*d^9)/b - (132*b*c^4*d^7)/a + (130*
b^2*c^5*d^6)/a^2 - (40*b^3*c^6*d^5)/a^3) - (132*b*c^2*d^7*(c^3)^(1/2)*(c + d*x)^(1/2))/(32*a*c^3*d^8 - 132*b*c
^4*d^7 + (130*b^2*c^5*d^6)/a + (10*a^2*c^2*d^9)/b - (40*b^3*c^6*d^5)/a^2) + (130*b^2*c^3*d^6*(c^3)^(1/2)*(c +
d*x)^(1/2))/(32*a^2*c^3*d^8 + 130*b^2*c^5*d^6 - (40*b^3*c^6*d^5)/a + (10*a^3*c^2*d^9)/b - 132*a*b*c^4*d^7) - (
40*b^3*c^4*d^5*(c^3)^(1/2)*(c + d*x)^(1/2))/(32*a^3*c^3*d^8 - 40*b^3*c^6*d^5 + 130*a*b^2*c^5*d^6 - 132*a^2*b*c
^4*d^7 + (10*a^4*c^2*d^9)/b))*(5*a*d - 4*b*c)*(c^3)^(1/2))/a^3 - (atanh((30*c^3*d^6*(c + d*x)^(1/2)*(b^6*c^3 -
a^3*b^3*d^3 + 3*a^2*b^4*c*d^2 - 3*a*b^5*c^2*d)^(1/2))/(14*a^3*c^2*d^9 + 110*b^3*c^5*d^6 - 82*a*b^2*c^4*d^7 -
4*a^2*b*c^3*d^8 + (2*a^4*c*d^10)/b - (40*b^4*c^6*d^5)/a) - (2*c*d^8*(c + d*x)^(1/2)*(b^6*c^3 - a^3*b^3*d^3 + 3
*a^2*b^4*c*d^2 - 3*a*b^5*c^2*d)^(1/2))/(4*b^3*c^3*d^8 - 14*a*b^2*c^2*d^9 + (82*b^4*c^4*d^7)/a - (110*b^5*c^5*d
^6)/a^2 + (40*b^6*c^6*d^5)/a^3 - 2*a^2*b*c*d^10) + (18*c^2*d^7*(c + d*x)^(1/2)*(b^6*c^3 - a^3*b^3*d^3 + 3*a^2*
b^4*c*d^2 - 3*a*b^5*c^2*d)^(1/2))/(2*a^3*c*d^10 - 82*b^3*c^4*d^7 - 4*a*b^2*c^3*d^8 + 14*a^2*b*c^2*d^9 + (110*b
^4*c^5*d^6)/a - (40*b^5*c^6*d^5)/a^2) + (40*c^4*d^5*(c + d*x)^(1/2)*(b^6*c^3 - a^3*b^3*d^3 + 3*a^2*b^4*c*d^2 -
3*a*b^5*c^2*d)^(1/2))/(4*a^3*c^3*d^8 + 40*b^3*c^6*d^5 - 110*a*b^2*c^5*d^6 + 82*a^2*b*c^4*d^7 - (2*a^5*c*d^10)
/b^2 - (14*a^4*c^2*d^9)/b))*(-b^3*(a*d - b*c)^3)^(1/2)*(a*d + 4*b*c))/(a^3*b^3)

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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)**(5/2)/x**2/(b*x+a)**2,x)

[Out]

Timed out

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