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

Optimal. Leaf size=142 \[ \frac {(b c-a d)^{3/2} (3 a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^2 b^{5/2}}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}-\frac {d \sqrt {c+d x} (b c-3 a d)}{a b^2}+\frac {(c+d x)^{3/2} (b c-a d)}{a b (a+b x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 142, normalized size of antiderivative = 1.00, 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, 154, 156, 63, 208} \begin {gather*} \frac {(b c-a d)^{3/2} (3 a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^2 b^{5/2}}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}-\frac {d \sqrt {c+d x} (b c-3 a d)}{a b^2}+\frac {(c+d x)^{3/2} (b c-a d)}{a b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.31, size = 145, normalized size = 1.02 \begin {gather*} \frac {\frac {a \sqrt {c+d x} \left (3 a^2 d^2+2 a b d (d x-c)+b^2 c^2\right )}{b^2 (a+b x)}+\frac {\sqrt {b c-a d} \left (-3 a^2 d^2+a b c d+2 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2}}-2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.56, size = 175, normalized size = 1.23 \begin {gather*} \frac {d \sqrt {c+d x} \left (3 a^2 d^2+2 a b d (c+d x)-4 a b c d+b^2 c^2\right )}{a b^2 (a d+b (c+d x)-b c)}-\frac {\sqrt {a d-b c} \left (-3 a^2 d^2+a b c d+2 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{a^2 b^{5/2}}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.92, size = 865, normalized size = 6.09 \begin {gather*} \left [-\frac {{\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b 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 (b^{3} c^{2} x + a b^{2} c^{2}\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, a^{2} b d^{2} x + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x + c}}{2 \, {\left (a^{2} b^{3} x + a^{3} b^{2}\right )}}, \frac {{\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b 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 (b^{3} c^{2} x + a b^{2} c^{2}\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + {\left (2 \, a^{2} b d^{2} x + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x + c}}{a^{2} b^{3} x + a^{3} b^{2}}, \frac {4 \, {\left (b^{3} c^{2} x + a b^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - {\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b 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 (2 \, a^{2} b d^{2} x + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x + c}}{2 \, {\left (a^{2} b^{3} x + a^{3} b^{2}\right )}}, \frac {{\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b 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 ) + 2 \, {\left (b^{3} c^{2} x + a b^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left (2 \, a^{2} b d^{2} x + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x + c}}{a^{2} b^{3} x + a^{3} b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.30, size = 193, normalized size = 1.36 \begin {gather*} \frac {2 \, c^{3} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c}} + \frac {2 \, \sqrt {d x + c} d^{2}}{b^{2}} - \frac {{\left (2 \, b^{3} c^{3} - a b^{2} c^{2} d - 4 \, a^{2} b c d^{2} + 3 \, 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^{2} b^{2}} + \frac {\sqrt {d x + c} b^{2} c^{2} d - 2 \, \sqrt {d x + c} a b c d^{2} + \sqrt {d x + c} a^{2} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} a b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.02, size = 284, normalized size = 2.00 \begin {gather*} -\frac {3 a \,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^{2}}+\frac {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}-\frac {2 b \,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^{2}}+\frac {4 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}\, b}+\frac {\sqrt {d x +c}\, a \,d^{3}}{\left (b d x +a d \right ) b^{2}}+\frac {\sqrt {d x +c}\, c^{2} d}{\left (b d x +a d \right ) a}-\frac {2 \sqrt {d x +c}\, c \,d^{2}}{\left (b d x +a d \right ) b}-\frac {2 c^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{2}}+\frac {2 \sqrt {d x +c}\, d^{2}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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((d*x+c)^(5/2)/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?

________________________________________________________________________________________

mupad [B]  time = 0.72, size = 1295, normalized size = 9.12 \begin {gather*} \frac {2\,d^2\,\sqrt {c+d\,x}}{b^2}+\frac {\mathrm {atan}\left (\frac {a^2\,d^8\,\sqrt {c^5}\,\sqrt {c+d\,x}\,36{}\mathrm {i}}{36\,a^2\,c^3\,d^8+40\,b^2\,c^5\,d^6+\frac {80\,b^3\,c^6\,d^5}{a}-\frac {60\,b^4\,c^7\,d^4}{a^2}-96\,a\,b\,c^4\,d^7}+\frac {c^2\,d^6\,\sqrt {c^5}\,\sqrt {c+d\,x}\,40{}\mathrm {i}}{40\,c^5\,d^6-\frac {96\,a\,c^4\,d^7}{b}+\frac {80\,b\,c^6\,d^5}{a}-\frac {60\,b^2\,c^7\,d^4}{a^2}+\frac {36\,a^2\,c^3\,d^8}{b^2}}+\frac {c^3\,d^5\,\sqrt {c^5}\,\sqrt {c+d\,x}\,80{}\mathrm {i}}{80\,c^6\,d^5+\frac {40\,a\,c^5\,d^6}{b}-\frac {60\,b\,c^7\,d^4}{a}-\frac {96\,a^2\,c^4\,d^7}{b^2}+\frac {36\,a^3\,c^3\,d^8}{b^3}}-\frac {b\,c^4\,d^4\,\sqrt {c^5}\,\sqrt {c+d\,x}\,60{}\mathrm {i}}{80\,a\,c^6\,d^5-60\,b\,c^7\,d^4+\frac {40\,a^2\,c^5\,d^6}{b}-\frac {96\,a^3\,c^4\,d^7}{b^2}+\frac {36\,a^4\,c^3\,d^8}{b^3}}-\frac {a\,c\,d^7\,\sqrt {c^5}\,\sqrt {c+d\,x}\,96{}\mathrm {i}}{40\,b\,c^5\,d^6-96\,a\,c^4\,d^7+\frac {80\,b^2\,c^6\,d^5}{a}+\frac {36\,a^2\,c^3\,d^8}{b}-\frac {60\,b^3\,c^7\,d^4}{a^2}}\right )\,\sqrt {c^5}\,2{}\mathrm {i}}{a^2}+\frac {\sqrt {c+d\,x}\,\left (a^2\,d^3-2\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{a\,\left (b^3\,\left (c+d\,x\right )-b^3\,c+a\,b^2\,d\right )}-\frac {\mathrm {atan}\left (\frac {c^4\,d^5\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^5\,d^3+3\,a^2\,b^6\,c\,d^2-3\,a\,b^7\,c^2\,d+b^8\,c^3}\,70{}\mathrm {i}}{72\,a^3\,b\,c^3\,d^8-50\,b^4\,c^6\,d^5-170\,a\,b^3\,c^5\,d^6-162\,a^4\,c^2\,d^9+\frac {54\,a^5\,c\,d^{10}}{b}+196\,a^2\,b^2\,c^4\,d^7+\frac {60\,b^5\,c^7\,d^4}{a}}-\frac {c^3\,d^6\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^5\,d^3+3\,a^2\,b^6\,c\,d^2-3\,a\,b^7\,c^2\,d+b^8\,c^3}\,90{}\mathrm {i}}{54\,a^4\,c\,d^{10}-170\,b^4\,c^5\,d^6+196\,a\,b^3\,c^4\,d^7-162\,a^3\,b\,c^2\,d^9+72\,a^2\,b^2\,c^3\,d^8-\frac {50\,b^5\,c^6\,d^5}{a}+\frac {60\,b^6\,c^7\,d^4}{a^2}}+\frac {c^5\,d^4\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^5\,d^3+3\,a^2\,b^6\,c\,d^2-3\,a\,b^7\,c^2\,d+b^8\,c^3}\,60{}\mathrm {i}}{72\,a^4\,c^3\,d^8+60\,b^4\,c^7\,d^4-50\,a\,b^3\,c^6\,d^5+196\,a^3\,b\,c^4\,d^7+\frac {54\,a^6\,c\,d^{10}}{b^2}-170\,a^2\,b^2\,c^5\,d^6-\frac {162\,a^5\,c^2\,d^9}{b}}-\frac {a\,c^2\,d^7\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^5\,d^3+3\,a^2\,b^6\,c\,d^2-3\,a\,b^7\,c^2\,d+b^8\,c^3}\,54{}\mathrm {i}}{196\,a\,b^4\,c^4\,d^7-170\,b^5\,c^5\,d^6+72\,a^2\,b^3\,c^3\,d^8-162\,a^3\,b^2\,c^2\,d^9-\frac {50\,b^6\,c^6\,d^5}{a}+\frac {60\,b^7\,c^7\,d^4}{a^2}+54\,a^4\,b\,c\,d^{10}}+\frac {a^2\,c\,d^8\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b^5\,d^3+3\,a^2\,b^6\,c\,d^2-3\,a\,b^7\,c^2\,d+b^8\,c^3}\,54{}\mathrm {i}}{196\,a\,b^5\,c^4\,d^7-170\,b^6\,c^5\,d^6+54\,a^4\,b^2\,c\,d^{10}+72\,a^2\,b^4\,c^3\,d^8-162\,a^3\,b^3\,c^2\,d^9-\frac {50\,b^7\,c^6\,d^5}{a}+\frac {60\,b^8\,c^7\,d^4}{a^2}}\right )\,\sqrt {-b^5\,{\left (a\,d-b\,c\right )}^3}\,\left (3\,a\,d+2\,b\,c\right )\,1{}\mathrm {i}}{a^2\,b^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d^2*(c + d*x)^(1/2))/b^2 + (atan((a^2*d^8*(c^5)^(1/2)*(c + d*x)^(1/2)*36i)/(36*a^2*c^3*d^8 + 40*b^2*c^5*d^6
+ (80*b^3*c^6*d^5)/a - (60*b^4*c^7*d^4)/a^2 - 96*a*b*c^4*d^7) + (c^2*d^6*(c^5)^(1/2)*(c + d*x)^(1/2)*40i)/(40
*c^5*d^6 - (96*a*c^4*d^7)/b + (80*b*c^6*d^5)/a - (60*b^2*c^7*d^4)/a^2 + (36*a^2*c^3*d^8)/b^2) + (c^3*d^5*(c^5)
^(1/2)*(c + d*x)^(1/2)*80i)/(80*c^6*d^5 + (40*a*c^5*d^6)/b - (60*b*c^7*d^4)/a - (96*a^2*c^4*d^7)/b^2 + (36*a^3
*c^3*d^8)/b^3) - (b*c^4*d^4*(c^5)^(1/2)*(c + d*x)^(1/2)*60i)/(80*a*c^6*d^5 - 60*b*c^7*d^4 + (40*a^2*c^5*d^6)/b
- (96*a^3*c^4*d^7)/b^2 + (36*a^4*c^3*d^8)/b^3) - (a*c*d^7*(c^5)^(1/2)*(c + d*x)^(1/2)*96i)/(40*b*c^5*d^6 - 96
*a*c^4*d^7 + (80*b^2*c^6*d^5)/a + (36*a^2*c^3*d^8)/b - (60*b^3*c^7*d^4)/a^2))*(c^5)^(1/2)*2i)/a^2 + ((c + d*x)
^(1/2)*(a^2*d^3 + b^2*c^2*d - 2*a*b*c*d^2))/(a*(b^3*(c + d*x) - b^3*c + a*b^2*d)) - (atan((c^4*d^5*(c + d*x)^(
1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*c^2*d)^(1/2)*70i)/(72*a^3*b*c^3*d^8 - 50*b^4*c^6*d^5 -
170*a*b^3*c^5*d^6 - 162*a^4*c^2*d^9 + (54*a^5*c*d^10)/b + 196*a^2*b^2*c^4*d^7 + (60*b^5*c^7*d^4)/a) - (c^3*d^
6*(c + d*x)^(1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*c^2*d)^(1/2)*90i)/(54*a^4*c*d^10 - 170*b^
4*c^5*d^6 + 196*a*b^3*c^4*d^7 - 162*a^3*b*c^2*d^9 + 72*a^2*b^2*c^3*d^8 - (50*b^5*c^6*d^5)/a + (60*b^6*c^7*d^4)
/a^2) + (c^5*d^4*(c + d*x)^(1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*c^2*d)^(1/2)*60i)/(72*a^4*
c^3*d^8 + 60*b^4*c^7*d^4 - 50*a*b^3*c^6*d^5 + 196*a^3*b*c^4*d^7 + (54*a^6*c*d^10)/b^2 - 170*a^2*b^2*c^5*d^6 -
(162*a^5*c^2*d^9)/b) - (a*c^2*d^7*(c + d*x)^(1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*c^2*d)^(1
/2)*54i)/(196*a*b^4*c^4*d^7 - 170*b^5*c^5*d^6 + 72*a^2*b^3*c^3*d^8 - 162*a^3*b^2*c^2*d^9 - (50*b^6*c^6*d^5)/a
+ (60*b^7*c^7*d^4)/a^2 + 54*a^4*b*c*d^10) + (a^2*c*d^8*(c + d*x)^(1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^
2 - 3*a*b^7*c^2*d)^(1/2)*54i)/(196*a*b^5*c^4*d^7 - 170*b^6*c^5*d^6 + 54*a^4*b^2*c*d^10 + 72*a^2*b^4*c^3*d^8 -
162*a^3*b^3*c^2*d^9 - (50*b^7*c^6*d^5)/a + (60*b^8*c^7*d^4)/a^2))*(-b^5*(a*d - b*c)^3)^(1/2)*(3*a*d + 2*b*c)*1
i)/(a^2*b^5)

________________________________________________________________________________________

sympy [F(-1)]  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((d*x+c)**(5/2)/x/(b*x+a)**2,x)

[Out]

Timed out

________________________________________________________________________________________