\(\int \frac {(c+d x)^{5/2}}{x^3 \sqrt {a+b x}} \, dx\) [337]

Optimal result

Integrand size = 22, antiderivative size = 177 \[ \int \frac {(c+d x)^{5/2}}{x^3 \sqrt {a+b x}} \, dx=\frac {c (3 b c-7 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a^2 x}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}-\frac {\sqrt {c} \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2}}+\frac {2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}} \] Output:

1/4*c*(-7*a*d+3*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2/x-1/2*c*(b*x+a)^(1/2) 
*(d*x+c)^(3/2)/a/x^2-1/4*c^(1/2)*(15*a^2*d^2-10*a*b*c*d+3*b^2*c^2)*arctanh 
(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(5/2)+2*d^(5/2)*arctanh(d^ 
(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(1/2)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x)^{5/2}}{x^3 \sqrt {a+b x}} \, dx=\frac {1}{4} \left (\frac {c \sqrt {a+b x} \sqrt {c+d x} (-2 a c+3 b c x-9 a d x)}{a^2 x^2}-\frac {\sqrt {c} \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{5/2}}+\frac {8 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}\right ) \] Input:

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

Output:

((c*Sqrt[a + b*x]*Sqrt[c + d*x]*(-2*a*c + 3*b*c*x - 9*a*d*x))/(a^2*x^2) - 
(Sqrt[c]*(3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b 
*x])/(Sqrt[a]*Sqrt[c + d*x])])/a^(5/2) + (8*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[ 
a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b])/4
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {109, 27, 166, 27, 175, 66, 104, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(c+d x)^{5/2}}{x^3 \sqrt {a+b x}} \, dx\)

\(\Big \downarrow \) 109

\(\displaystyle -\frac {\int \frac {\sqrt {c+d x} \left (c (3 b c-7 a d)-4 a d^2 x\right )}{2 x^2 \sqrt {a+b x}}dx}{2 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\sqrt {c+d x} \left (c (3 b c-7 a d)-4 a d^2 x\right )}{x^2 \sqrt {a+b x}}dx}{4 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}\)

\(\Big \downarrow \) 166

\(\displaystyle -\frac {\frac {\int -\frac {8 a^2 x d^3+c \left (3 b^2 c^2-10 a b d c+15 a^2 d^2\right )}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{a}-\frac {c \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{a x}}{4 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\int \frac {8 a^2 x d^3+c \left (3 b^2 c^2-10 a b d c+15 a^2 d^2\right )}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 a}-\frac {c \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{a x}}{4 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}\)

\(\Big \downarrow \) 175

\(\displaystyle -\frac {-\frac {c \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+8 a^2 d^3 \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 a}-\frac {c \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{a x}}{4 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}\)

\(\Big \downarrow \) 66

\(\displaystyle -\frac {-\frac {c \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx+16 a^2 d^3 \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 a}-\frac {c \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{a x}}{4 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {-\frac {2 c \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+16 a^2 d^3 \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 a}-\frac {c \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{a x}}{4 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {-\frac {\frac {16 a^2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}-\frac {2 \sqrt {c} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}}{2 a}-\frac {c \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{a x}}{4 a}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}\)

Input:

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

Output:

-1/2*(c*Sqrt[a + b*x]*(c + d*x)^(3/2))/(a*x^2) - (-((c*(3*b*c - 7*a*d)*Sqr 
t[a + b*x]*Sqrt[c + d*x])/(a*x)) - ((-2*Sqrt[c]*(3*b^2*c^2 - 10*a*b*c*d + 
15*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt 
[a] + (16*a^2*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d* 
x])])/Sqrt[b])/(2*a))/(4*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> 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] + Simp[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 166

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

rule 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(b*g - a*h)/b   Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] 
 /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
 

rule 221

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(353\) vs. \(2(139)=278\).

Time = 0.22 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.00

method	result	size
default	\(\frac {\sqrt {x d +c}\, \sqrt {b x +a}\, \left (8 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} d^{3} x^{2} \sqrt {a c}-15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{2} c \,d^{2} x^{2} \sqrt {d b}+10 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a b \,c^{2} d \,x^{2} \sqrt {d b}-3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) b^{2} c^{3} x^{2} \sqrt {d b}-18 a c d x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}+6 b \,c^{2} x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}-4 a \,c^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\right )}{8 a^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, x^{2} \sqrt {d b}\, \sqrt {a c}}\)	\(354\)

Input:

int((d*x+c)^(5/2)/x^3/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/8*(d*x+c)^(1/2)*(b*x+a)^(1/2)/a^2*(8*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c)) 
^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*d^3*x^2*(a*c)^(1/2)-15*ln((a* 
d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a^2*c*d^2*x^2*(d 
*b)^(1/2)+10*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/ 
x)*a*b*c^2*d*x^2*(d*b)^(1/2)-3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x 
+c))^(1/2)+2*a*c)/x)*b^2*c^3*x^2*(d*b)^(1/2)-18*a*c*d*x*((b*x+a)*(d*x+c))^ 
(1/2)*(d*b)^(1/2)*(a*c)^(1/2)+6*b*c^2*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2 
)*(a*c)^(1/2)-4*a*c^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2))/((b 
*x+a)*(d*x+c))^(1/2)/x^2/(d*b)^(1/2)/(a*c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.92 (sec) , antiderivative size = 1031, normalized size of antiderivative = 5.82 \[ \int \frac {(c+d x)^{5/2}}{x^3 \sqrt {a+b x}} \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)^(5/2)/x^3/(b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

[1/16*(8*a^2*d^2*x^2*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a 
^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/ 
b) + 8*(b^2*c*d + a*b*d^2)*x) + (3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*x^2* 
sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2* 
c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 
+ a^2*c*d)*x)/x^2) - 4*(2*a*c^2 - 3*(b*c^2 - 3*a*c*d)*x)*sqrt(b*x + a)*sqr 
t(d*x + c))/(a^2*x^2), -1/16*(16*a^2*d^2*x^2*sqrt(-d/b)*arctan(1/2*(2*b*d* 
x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + 
 (b*c*d + a*d^2)*x)) - (3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*x^2*sqrt(c/a) 
*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b* 
c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d 
)*x)/x^2) + 4*(2*a*c^2 - 3*(b*c^2 - 3*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c 
))/(a^2*x^2), 1/8*(4*a^2*d^2*x^2*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6 
*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x 
+ c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + (3*b^2*c^2 - 10*a*b*c*d + 15*a 
^2*d^2)*x^2*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sq 
rt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) - 2*(2*a*c 
^2 - 3*(b*c^2 - 3*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*x^2), -1/8*( 
8*a^2*d^2*x^2*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sq 
rt(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) - (3*b^...
 

Sympy [F]

\[ \int \frac {(c+d x)^{5/2}}{x^3 \sqrt {a+b x}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x^{3} \sqrt {a + b x}}\, dx \] Input:

integrate((d*x+c)**(5/2)/x**3/(b*x+a)**(1/2),x)
 

Output:

Integral((c + d*x)**(5/2)/(x**3*sqrt(a + b*x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^3 \sqrt {a+b x}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x+c)^(5/2)/x^3/(b*x+a)^(1/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1149 vs. \(2 (139) = 278\).

Time = 0.58 (sec) , antiderivative size = 1149, normalized size of antiderivative = 6.49 \[ \int \frac {(c+d x)^{5/2}}{x^3 \sqrt {a+b x}} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^(5/2)/x^3/(b*x+a)^(1/2),x, algorithm="giac")
 

Output:

-1/4*(4*sqrt(b*d)*d^2*abs(b)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + ( 
b*x + a)*b*d - a*b*d))^2)/b + (3*sqrt(b*d)*b^3*c^3*abs(b) - 10*sqrt(b*d)*a 
*b^2*c^2*d*abs(b) + 15*sqrt(b*d)*a^2*b*c*d^2*abs(b))*arctan(-1/2*(b^2*c + 
a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2) 
/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*a^2*b) - 2*(3*sqrt(b*d)*b^9*c^6*abs(b 
) - 21*sqrt(b*d)*a*b^8*c^5*d*abs(b) + 54*sqrt(b*d)*a^2*b^7*c^4*d^2*abs(b) 
- 66*sqrt(b*d)*a^3*b^6*c^3*d^3*abs(b) + 39*sqrt(b*d)*a^4*b^5*c^2*d^4*abs(b 
) - 9*sqrt(b*d)*a^5*b^4*c*d^5*abs(b) - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a 
) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^7*c^5*abs(b) + 32*sqrt(b*d)*( 
sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^6*c^4 
*d*abs(b) - 10*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a) 
*b*d - a*b*d))^2*a^2*b^5*c^3*d^2*abs(b) - 40*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x 
 + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^4*c^2*d^3*abs(b) + 27 
*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) 
^2*a^4*b^3*c*d^4*abs(b) + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2* 
c + (b*x + a)*b*d - a*b*d))^4*b^5*c^4*abs(b) - 21*sqrt(b*d)*(sqrt(b*d)*sqr 
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^4*c^3*d*abs(b) - 9 
*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) 
^4*a^2*b^3*c^2*d^2*abs(b) - 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b 
^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^2*c*d^3*abs(b) - 3*sqrt(b*d)*(sq...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/2}}{x^3 \sqrt {a+b x}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/2}}{x^3\,\sqrt {a+b\,x}} \,d x \] Input:

int((c + d*x)^(5/2)/(x^3*(a + b*x)^(1/2)),x)
                                                                                    
                                                                                    
 

Output:

int((c + d*x)^(5/2)/(x^3*(a + b*x)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 937, normalized size of antiderivative = 5.29 \[ \int \frac {(c+d x)^{5/2}}{x^3 \sqrt {a+b x}} \, dx =\text {Too large to display} \] Input:

int((d*x+c)^(5/2)/x^3/(b*x+a)^(1/2),x)
 

Output:

( - 4*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*c**2*d - 18*sqrt(c + d*x)*sqrt(a 
+ b*x)*a**3*b*c*d**2*x - 4*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**2*c**3 - 12 
*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**2*c**2*d*x + 6*sqrt(c + d*x)*sqrt(a + 
 b*x)*a*b**3*c**3*x + 15*sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqr 
t(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x)) 
*a**3*b*d**3*x**2 + 5*sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b 
)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a* 
*2*b**2*c*d**2*x**2 - 7*sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt 
(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))* 
a*b**3*c**2*d*x**2 + 3*sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt( 
b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*b 
**4*c**3*x**2 + 15*sqrt(c)*sqrt(a)*log(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt 
(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**3*b*d 
**3*x**2 + 5*sqrt(c)*sqrt(a)*log(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + 
a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**2*b**2*c*d* 
*2*x**2 - 7*sqrt(c)*sqrt(a)*log(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a 
*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a*b**3*c**2*d*x 
**2 + 3*sqrt(c)*sqrt(a)*log(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + 
 b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*b**4*c**3*x**2 - 15 
*sqrt(c)*sqrt(a)*log(2*sqrt(d)*sqrt(b)*sqrt(c + d*x)*sqrt(a + b*x) + 2*...