\(\int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx\) [462]

Optimal result

Integrand size = 20, antiderivative size = 207 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\frac {c (2 b c-3 a d) \sqrt {c+d x}}{4 a^2 x^2}-\frac {\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt {c+d x}}{8 a^3 x}-\frac {c (c+d x)^{3/2}}{3 a x^3}+\frac {\left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 a^4 \sqrt {c}}-\frac {2 \sqrt {b} (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^4} \]

[Out]

-1/3*c*(d*x+c)^(3/2)/a/x^3-2*(-a*d+b*c)^(5/2)*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))*b^(1/2)/a^4+1/8*
(-5*a^3*d^3+30*a^2*b*c*d^2-40*a*b^2*c^2*d+16*b^3*c^3)*arctanh((d*x+c)^(1/2)/c^(1/2))/a^4/c^(1/2)+1/4*c*(-3*a*d
+2*b*c)*(d*x+c)^(1/2)/a^2/x^2-1/8*(11*a^2*d^2-18*a*b*c*d+8*b^2*c^2)*(d*x+c)^(1/2)/a^3/x

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {100, 154, 156, 162, 65, 214} \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=-\frac {2 \sqrt {b} (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^4}+\frac {c \sqrt {c+d x} (2 b c-3 a d)}{4 a^2 x^2}-\frac {\sqrt {c+d x} \left (11 a^2 d^2-18 a b c d+8 b^2 c^2\right )}{8 a^3 x}+\frac {\left (-5 a^3 d^3+30 a^2 b c d^2-40 a b^2 c^2 d+16 b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 a^4 \sqrt {c}}-\frac {c (c+d x)^{3/2}}{3 a x^3} \]

[In]

Int[(c + d*x)^(5/2)/(x^4*(a + b*x)),x]

[Out]

(c*(2*b*c - 3*a*d)*Sqrt[c + d*x])/(4*a^2*x^2) - ((8*b^2*c^2 - 18*a*b*c*d + 11*a^2*d^2)*Sqrt[c + d*x])/(8*a^3*x
) - (c*(c + d*x)^(3/2))/(3*a*x^3) + ((16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[Sqrt[c
+ d*x]/Sqrt[c]])/(8*a^4*Sqrt[c]) - (2*Sqrt[b]*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*
d]])/a^4

Rule 65

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 100

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[(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] && ILtQ[m, -1] && GtQ[n, 0]

Rule 156

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 + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

Rule 162

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps \begin{align*} \text {integral}& = -\frac {c (c+d x)^{3/2}}{3 a x^3}-\frac {\int \frac {\sqrt {c+d x} \left (\frac {3}{2} c (2 b c-3 a d)+\frac {3}{2} d (b c-2 a d) x\right )}{x^3 (a+b x)} \, dx}{3 a} \\ & = \frac {c (2 b c-3 a d) \sqrt {c+d x}}{4 a^2 x^2}-\frac {c (c+d x)^{3/2}}{3 a x^3}-\frac {\int \frac {-\frac {3}{4} c \left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right )-\frac {3}{4} d \left (6 b^2 c^2-13 a b c d+8 a^2 d^2\right ) x}{x^2 (a+b x) \sqrt {c+d x}} \, dx}{6 a^2} \\ & = \frac {c (2 b c-3 a d) \sqrt {c+d x}}{4 a^2 x^2}-\frac {\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt {c+d x}}{8 a^3 x}-\frac {c (c+d x)^{3/2}}{3 a x^3}+\frac {\int \frac {-\frac {3}{8} c \left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right )-\frac {3}{8} b c d \left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}} \, dx}{6 a^3 c} \\ & = \frac {c (2 b c-3 a d) \sqrt {c+d x}}{4 a^2 x^2}-\frac {\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt {c+d x}}{8 a^3 x}-\frac {c (c+d x)^{3/2}}{3 a x^3}+\frac {\left (b (b c-a d)^3\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{a^4}-\frac {\left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \int \frac {1}{x \sqrt {c+d x}} \, dx}{16 a^4} \\ & = \frac {c (2 b c-3 a d) \sqrt {c+d x}}{4 a^2 x^2}-\frac {\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt {c+d x}}{8 a^3 x}-\frac {c (c+d x)^{3/2}}{3 a x^3}+\frac {\left (2 b (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{a^4 d}-\frac {\left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{8 a^4 d} \\ & = \frac {c (2 b c-3 a d) \sqrt {c+d x}}{4 a^2 x^2}-\frac {\left (8 b^2 c^2-18 a b c d+11 a^2 d^2\right ) \sqrt {c+d x}}{8 a^3 x}-\frac {c (c+d x)^{3/2}}{3 a x^3}+\frac {\left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 a^4 \sqrt {c}}-\frac {2 \sqrt {b} (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=-\frac {\frac {a \sqrt {c+d x} \left (24 b^2 c^2 x^2-6 a b c x (2 c+9 d x)+a^2 \left (8 c^2+26 c d x+33 d^2 x^2\right )\right )}{x^3}+48 \sqrt {b} (-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )-\frac {3 \left (16 b^3 c^3-40 a b^2 c^2 d+30 a^2 b c d^2-5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c}}}{24 a^4} \]

[In]

Integrate[(c + d*x)^(5/2)/(x^4*(a + b*x)),x]

[Out]

-1/24*((a*Sqrt[c + d*x]*(24*b^2*c^2*x^2 - 6*a*b*c*x*(2*c + 9*d*x) + a^2*(8*c^2 + 26*c*d*x + 33*d^2*x^2)))/x^3
+ 48*Sqrt[b]*(-(b*c) + a*d)^(5/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]] - (3*(16*b^3*c^3 - 40*a*b
^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/Sqrt[c])/a^4

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.85

method	result	size
pseudoelliptic	\(\frac {-\frac {2 \left (a d -b c \right )^{3} b \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\frac {\sqrt {d x +c}\, a \left (33 a^{2} d^{2} x^{2}-54 a b c d \,x^{2}+24 b^{2} c^{2} x^{2}+26 a^{2} c d x -12 a b \,c^{2} x +8 a^{2} c^{2}\right )}{24 x^{3}}-\frac {\left (5 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 \sqrt {c}}}{a^{4}}\)	\(176\)
risch	\(-\frac {\sqrt {d x +c}\, \left (33 a^{2} d^{2} x^{2}-54 a b c d \,x^{2}+24 b^{2} c^{2} x^{2}+26 a^{2} c d x -12 a b \,c^{2} x +8 a^{2} c^{2}\right )}{24 a^{3} x^{3}}-\frac {d \left (\frac {16 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) b \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a d \sqrt {\left (a d -b c \right ) b}}-\frac {\left (-5 a^{3} d^{3}+30 a^{2} b c \,d^{2}-40 a \,b^{2} c^{2} d +16 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a d \sqrt {c}}\right )}{8 a^{3}}\)	\(219\)
derivativedivides	\(2 d^{4} \left (\frac {-\frac {\left (\frac {11}{16} a^{3} d^{3}-\frac {9}{8} a^{2} b c \,d^{2}+\frac {1}{2} a \,b^{2} c^{2} d \right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {5}{6} a^{3} c \,d^{3}+2 a^{2} d^{2} b \,c^{2}-a \,b^{2} c^{3} d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} b \,c^{3} d^{2}+\frac {1}{2} a \,b^{2} c^{4} d +\frac {5}{16} a^{3} c^{2} d^{3}\right ) \sqrt {d x +c}}{d^{3} x^{3}}-\frac {\left (5 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}}{a^{4} d^{4}}-\frac {\left (a d -b c \right )^{3} b \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a^{4} d^{4} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(250\)
default	\(2 d^{4} \left (\frac {-\frac {\left (\frac {11}{16} a^{3} d^{3}-\frac {9}{8} a^{2} b c \,d^{2}+\frac {1}{2} a \,b^{2} c^{2} d \right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {5}{6} a^{3} c \,d^{3}+2 a^{2} d^{2} b \,c^{2}-a \,b^{2} c^{3} d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} b \,c^{3} d^{2}+\frac {1}{2} a \,b^{2} c^{4} d +\frac {5}{16} a^{3} c^{2} d^{3}\right ) \sqrt {d x +c}}{d^{3} x^{3}}-\frac {\left (5 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}}{a^{4} d^{4}}-\frac {\left (a d -b c \right )^{3} b \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a^{4} d^{4} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(250\)

[In]

int((d*x+c)^(5/2)/x^4/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/a^4*(-2*(a*d-b*c)^3*b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))-1/24*(d*x+c)^(1/2)*a*(
33*a^2*d^2*x^2-54*a*b*c*d*x^2+24*b^2*c^2*x^2+26*a^2*c*d*x-12*a*b*c^2*x+8*a^2*c^2)/x^3-1/8*(5*a^3*d^3-30*a^2*b*
c*d^2+40*a*b^2*c^2*d-16*b^3*c^3)/c^(1/2)*arctanh((d*x+c)^(1/2)/c^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 930, normalized size of antiderivative = 4.49 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\left [\frac {48 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {b^{2} c - a b d} x^{3} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) - 3 \, {\left (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {c} x^{3} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (8 \, a^{3} c^{3} + 3 \, {\left (8 \, a b^{2} c^{3} - 18 \, a^{2} b c^{2} d + 11 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (6 \, a^{2} b c^{3} - 13 \, a^{3} c^{2} d\right )} x\right )} \sqrt {d x + c}}{48 \, a^{4} c x^{3}}, \frac {96 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {-b^{2} c + a b d} x^{3} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) - 3 \, {\left (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {c} x^{3} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (8 \, a^{3} c^{3} + 3 \, {\left (8 \, a b^{2} c^{3} - 18 \, a^{2} b c^{2} d + 11 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (6 \, a^{2} b c^{3} - 13 \, a^{3} c^{2} d\right )} x\right )} \sqrt {d x + c}}{48 \, a^{4} c x^{3}}, -\frac {3 \, {\left (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - 24 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {b^{2} c - a b d} x^{3} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) + {\left (8 \, a^{3} c^{3} + 3 \, {\left (8 \, a b^{2} c^{3} - 18 \, a^{2} b c^{2} d + 11 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (6 \, a^{2} b c^{3} - 13 \, a^{3} c^{2} d\right )} x\right )} \sqrt {d x + c}}{24 \, a^{4} c x^{3}}, \frac {48 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {-b^{2} c + a b d} x^{3} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) - 3 \, {\left (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - {\left (8 \, a^{3} c^{3} + 3 \, {\left (8 \, a b^{2} c^{3} - 18 \, a^{2} b c^{2} d + 11 \, a^{3} c d^{2}\right )} x^{2} - 2 \, {\left (6 \, a^{2} b c^{3} - 13 \, a^{3} c^{2} d\right )} x\right )} \sqrt {d x + c}}{24 \, a^{4} c x^{3}}\right ] \]

[In]

integrate((d*x+c)^(5/2)/x^4/(b*x+a),x, algorithm="fricas")

[Out]

[1/48*(48*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(b^2*c - a*b*d)*x^3*log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c
- a*b*d)*sqrt(d*x + c))/(b*x + a)) - 3*(16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(c)*x^3*
log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(8*a^3*c^3 + 3*(8*a*b^2*c^3 - 18*a^2*b*c^2*d + 11*a^3*c*d^2)*
x^2 - 2*(6*a^2*b*c^3 - 13*a^3*c^2*d)*x)*sqrt(d*x + c))/(a^4*c*x^3), 1/48*(96*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^
2)*sqrt(-b^2*c + a*b*d)*x^3*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - 3*(16*b^3*c^3 - 40*a*b^
2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(c)*x^3*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(8*a^3*c^3
+ 3*(8*a*b^2*c^3 - 18*a^2*b*c^2*d + 11*a^3*c*d^2)*x^2 - 2*(6*a^2*b*c^3 - 13*a^3*c^2*d)*x)*sqrt(d*x + c))/(a^4*
c*x^3), -1/24*(3*(16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(-c)*x^3*arctan(sqrt(d*x + c)*
sqrt(-c)/c) - 24*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(b^2*c - a*b*d)*x^3*log((b*d*x + 2*b*c - a*d - 2*sqrt
(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) + (8*a^3*c^3 + 3*(8*a*b^2*c^3 - 18*a^2*b*c^2*d + 11*a^3*c*d^2)*x^2 -
2*(6*a^2*b*c^3 - 13*a^3*c^2*d)*x)*sqrt(d*x + c))/(a^4*c*x^3), 1/24*(48*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sq
rt(-b^2*c + a*b*d)*x^3*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - 3*(16*b^3*c^3 - 40*a*b^2*c^2
*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(-c)*x^3*arctan(sqrt(d*x + c)*sqrt(-c)/c) - (8*a^3*c^3 + 3*(8*a*b^2*c^3 -
18*a^2*b*c^2*d + 11*a^3*c*d^2)*x^2 - 2*(6*a^2*b*c^3 - 13*a^3*c^2*d)*x)*sqrt(d*x + c))/(a^4*c*x^3)]

Sympy [F]

\[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x^{4} \left (a + b x\right )}\, dx \]

[In]

integrate((d*x+c)**(5/2)/x**4/(b*x+a),x)

[Out]

Integral((c + d*x)**(5/2)/(x**4*(a + b*x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*x+c)^(5/2)/x^4/(b*x+a),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 detail

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.45 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\frac {2 \, {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{4}} - \frac {{\left (16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{8 \, a^{4} \sqrt {-c}} - \frac {24 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} c^{2} d - 48 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c^{3} d + 24 \, \sqrt {d x + c} b^{2} c^{4} d - 54 \, {\left (d x + c\right )}^{\frac {5}{2}} a b c d^{2} + 96 \, {\left (d x + c\right )}^{\frac {3}{2}} a b c^{2} d^{2} - 42 \, \sqrt {d x + c} a b c^{3} d^{2} + 33 \, {\left (d x + c\right )}^{\frac {5}{2}} a^{2} d^{3} - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} c d^{3} + 15 \, \sqrt {d x + c} a^{2} c^{2} d^{3}}{24 \, a^{3} d^{3} x^{3}} \]

[In]

integrate((d*x+c)^(5/2)/x^4/(b*x+a),x, algorithm="giac")

[Out]

2*(b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-
b^2*c + a*b*d)*a^4) - 1/8*(16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^3)*arctan(sqrt(d*x + c)/sqrt
(-c))/(a^4*sqrt(-c)) - 1/24*(24*(d*x + c)^(5/2)*b^2*c^2*d - 48*(d*x + c)^(3/2)*b^2*c^3*d + 24*sqrt(d*x + c)*b^
2*c^4*d - 54*(d*x + c)^(5/2)*a*b*c*d^2 + 96*(d*x + c)^(3/2)*a*b*c^2*d^2 - 42*sqrt(d*x + c)*a*b*c^3*d^2 + 33*(d
*x + c)^(5/2)*a^2*d^3 - 40*(d*x + c)^(3/2)*a^2*c*d^3 + 15*sqrt(d*x + c)*a^2*c^2*d^3)/(a^3*d^3*x^3)

Mupad [B] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 2147, normalized size of antiderivative = 10.37 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)} \, dx=\text {Too large to display} \]

[In]

int((c + d*x)^(5/2)/(x^4*(a + b*x)),x)

[Out]

(((c + d*x)^(1/2)*(8*b^2*c^4*d + 5*a^2*c^2*d^3 - 14*a*b*c^3*d^2))/(8*a^3) - ((c + d*x)^(3/2)*(5*a^2*c*d^3 + 6*
b^2*c^3*d - 12*a*b*c^2*d^2))/(3*a^3) + (d*(c + d*x)^(5/2)*(11*a^2*d^2 + 8*b^2*c^2 - 18*a*b*c*d))/(8*a^3))/(3*c
*(c + d*x)^2 - 3*c^2*(c + d*x) - (c + d*x)^3 + c^3) + (2*atanh((25*b^3*d^8*(c + d*x)^(1/2)*(b^6*c^5 - a^5*b*d^
5 + 5*a^4*b^2*c*d^4 + 10*a^2*b^4*c^3*d^2 - 10*a^3*b^3*c^2*d^3 - 5*a*b^5*c^4*d)^(1/2))/(16*((25*a^3*b^3*d^11)/1
6 - (217*b^6*c^3*d^8)/16 + (227*a*b^5*c^2*d^9)/16 - (119*a^2*b^4*c*d^10)/16 + (13*b^7*c^4*d^7)/(2*a) - (5*b^8*
c^5*d^6)/(4*a^2))) + (5*b^5*c^2*d^6*(c + d*x)^(1/2)*(b^6*c^5 - a^5*b*d^5 + 5*a^4*b^2*c*d^4 + 10*a^2*b^4*c^3*d^
2 - 10*a^3*b^3*c^2*d^3 - 5*a*b^5*c^4*d)^(1/2))/(4*((25*a^5*b^3*d^11)/16 - (5*b^8*c^5*d^6)/4 + (13*a*b^7*c^4*d^
7)/2 - (119*a^4*b^4*c*d^10)/16 - (217*a^2*b^6*c^3*d^8)/16 + (227*a^3*b^5*c^2*d^9)/16)) - (11*b^4*c*d^7*(c + d*
x)^(1/2)*(b^6*c^5 - a^5*b*d^5 + 5*a^4*b^2*c*d^4 + 10*a^2*b^4*c^3*d^2 - 10*a^3*b^3*c^2*d^3 - 5*a*b^5*c^4*d)^(1/
2))/(4*((25*a^4*b^3*d^11)/16 + (13*b^7*c^4*d^7)/2 - (217*a*b^6*c^3*d^8)/16 - (119*a^3*b^4*c*d^10)/16 + (227*a^
2*b^5*c^2*d^9)/16 - (5*b^8*c^5*d^6)/(4*a))))*(-b*(a*d - b*c)^5)^(1/2))/a^4 + (atan((((((c + d*x)^(1/2)*(281*a^
6*b^3*d^8 + 512*b^9*c^6*d^2 - 2816*a*b^8*c^5*d^3 - 1836*a^5*b^4*c*d^7 + 6400*a^2*b^7*c^4*d^4 - 7680*a^3*b^6*c^
3*d^5 + 5140*a^4*b^5*c^2*d^6))/(32*a^6) - (((80*a^11*b^2*d^6 - 304*a^10*b^3*c*d^5 - 128*a^8*b^5*c^3*d^3 + 352*
a^9*b^4*c^2*d^4)/(32*a^9) - ((256*a^9*b^2*d^3 - 512*a^8*b^3*c*d^2)*(c + d*x)^(1/2)*(5*a^3*d^3 - 16*b^3*c^3 + 4
0*a*b^2*c^2*d - 30*a^2*b*c*d^2))/(512*a^10*c^(1/2)))*(5*a^3*d^3 - 16*b^3*c^3 + 40*a*b^2*c^2*d - 30*a^2*b*c*d^2
))/(16*a^4*c^(1/2)))*(5*a^3*d^3 - 16*b^3*c^3 + 40*a*b^2*c^2*d - 30*a^2*b*c*d^2)*1i)/(16*a^4*c^(1/2)) + ((((c +
d*x)^(1/2)*(281*a^6*b^3*d^8 + 512*b^9*c^6*d^2 - 2816*a*b^8*c^5*d^3 - 1836*a^5*b^4*c*d^7 + 6400*a^2*b^7*c^4*d^
4 - 7680*a^3*b^6*c^3*d^5 + 5140*a^4*b^5*c^2*d^6))/(32*a^6) + (((80*a^11*b^2*d^6 - 304*a^10*b^3*c*d^5 - 128*a^8
*b^5*c^3*d^3 + 352*a^9*b^4*c^2*d^4)/(32*a^9) + ((256*a^9*b^2*d^3 - 512*a^8*b^3*c*d^2)*(c + d*x)^(1/2)*(5*a^3*d
^3 - 16*b^3*c^3 + 40*a*b^2*c^2*d - 30*a^2*b*c*d^2))/(512*a^10*c^(1/2)))*(5*a^3*d^3 - 16*b^3*c^3 + 40*a*b^2*c^2
*d - 30*a^2*b*c*d^2))/(16*a^4*c^(1/2)))*(5*a^3*d^3 - 16*b^3*c^3 + 40*a*b^2*c^2*d - 30*a^2*b*c*d^2)*1i)/(16*a^4
*c^(1/2)))/((55*a^8*b^3*d^11 + 128*b^11*c^8*d^3 - 992*a*b^10*c^7*d^4 - 585*a^7*b^4*c*d^10 + 3344*a^2*b^9*c^6*d
^5 - 6380*a^3*b^8*c^5*d^6 + 7496*a^4*b^7*c^4*d^7 - 5511*a^5*b^6*c^3*d^8 + 2445*a^6*b^5*c^2*d^9)/(16*a^9) - (((
(c + d*x)^(1/2)*(281*a^6*b^3*d^8 + 512*b^9*c^6*d^2 - 2816*a*b^8*c^5*d^3 - 1836*a^5*b^4*c*d^7 + 6400*a^2*b^7*c^
4*d^4 - 7680*a^3*b^6*c^3*d^5 + 5140*a^4*b^5*c^2*d^6))/(32*a^6) - (((80*a^11*b^2*d^6 - 304*a^10*b^3*c*d^5 - 128
*a^8*b^5*c^3*d^3 + 352*a^9*b^4*c^2*d^4)/(32*a^9) - ((256*a^9*b^2*d^3 - 512*a^8*b^3*c*d^2)*(c + d*x)^(1/2)*(5*a
^3*d^3 - 16*b^3*c^3 + 40*a*b^2*c^2*d - 30*a^2*b*c*d^2))/(512*a^10*c^(1/2)))*(5*a^3*d^3 - 16*b^3*c^3 + 40*a*b^2
*c^2*d - 30*a^2*b*c*d^2))/(16*a^4*c^(1/2)))*(5*a^3*d^3 - 16*b^3*c^3 + 40*a*b^2*c^2*d - 30*a^2*b*c*d^2))/(16*a^
4*c^(1/2)) + ((((c + d*x)^(1/2)*(281*a^6*b^3*d^8 + 512*b^9*c^6*d^2 - 2816*a*b^8*c^5*d^3 - 1836*a^5*b^4*c*d^7 +
6400*a^2*b^7*c^4*d^4 - 7680*a^3*b^6*c^3*d^5 + 5140*a^4*b^5*c^2*d^6))/(32*a^6) + (((80*a^11*b^2*d^6 - 304*a^10
*b^3*c*d^5 - 128*a^8*b^5*c^3*d^3 + 352*a^9*b^4*c^2*d^4)/(32*a^9) + ((256*a^9*b^2*d^3 - 512*a^8*b^3*c*d^2)*(c +
d*x)^(1/2)*(5*a^3*d^3 - 16*b^3*c^3 + 40*a*b^2*c^2*d - 30*a^2*b*c*d^2))/(512*a^10*c^(1/2)))*(5*a^3*d^3 - 16*b^
3*c^3 + 40*a*b^2*c^2*d - 30*a^2*b*c*d^2))/(16*a^4*c^(1/2)))*(5*a^3*d^3 - 16*b^3*c^3 + 40*a*b^2*c^2*d - 30*a^2*
b*c*d^2))/(16*a^4*c^(1/2))))*(5*a^3*d^3 - 16*b^3*c^3 + 40*a*b^2*c^2*d - 30*a^2*b*c*d^2)*1i)/(8*a^4*c^(1/2))