\(\int \frac {(c+d x)^3 \sqrt {c^2-d^2 x^2}}{x^4} \, dx\) [111]

Optimal result

Integrand size = 27, antiderivative size = 152 \[ \int \frac {(c+d x)^3 \sqrt {c^2-d^2 x^2}}{x^4} \, dx=d^3 \sqrt {c^2-d^2 x^2}-\frac {3 c^2 d \sqrt {c^2-d^2 x^2}}{2 x^2}-\frac {3 c d^2 \sqrt {c^2-d^2 x^2}}{x}-\frac {c \left (c^2-d^2 x^2\right )^{3/2}}{3 x^3}-3 c d^3 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )+\frac {1}{2} c d^3 \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.91 \[ \int \frac {(c+d x)^3 \sqrt {c^2-d^2 x^2}}{x^4} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (-2 c^3-9 c^2 d x-16 c d^2 x^2+6 d^3 x^3\right )}{6 x^3}-c d^3 \text {arctanh}\left (\frac {\sqrt {-d^2} x}{c}-\frac {\sqrt {c^2-d^2 x^2}}{c}\right )+3 c \left (-d^2\right )^{3/2} \log \left (-\sqrt {-d^2} x+\sqrt {c^2-d^2 x^2}\right ) \] Input:

Integrate[((c + d*x)^3*Sqrt[c^2 - d^2*x^2])/x^4,x]
 

Output:

(Sqrt[c^2 - d^2*x^2]*(-2*c^3 - 9*c^2*d*x - 16*c*d^2*x^2 + 6*d^3*x^3))/(6*x 
^3) - c*d^3*ArcTanh[(Sqrt[-d^2]*x)/c - Sqrt[c^2 - d^2*x^2]/c] + 3*c*(-d^2) 
^(3/2)*Log[-(Sqrt[-d^2]*x) + Sqrt[c^2 - d^2*x^2]]
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.95, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {540, 27, 2338, 25, 27, 536, 538, 224, 216, 243, 73, 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.

\(\displaystyle \int \frac {(c+d x)^3 \sqrt {c^2-d^2 x^2}}{x^4} \, dx\)

\(\Big \downarrow \) 540

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2338

\(\displaystyle \frac {-\frac {\int -\frac {c^4 d^2 (6 c-d x) \sqrt {c^2-d^2 x^2}}{x^2}dx}{2 c^2}-\frac {3 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}{2 x^2}}{c^2}-\frac {c \left (c^2-d^2 x^2\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {c^4 d^2 (6 c-d x) \sqrt {c^2-d^2 x^2}}{x^2}dx}{2 c^2}-\frac {3 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}{2 x^2}}{c^2}-\frac {c \left (c^2-d^2 x^2\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{2} c^2 d^2 \int \frac {(6 c-d x) \sqrt {c^2-d^2 x^2}}{x^2}dx-\frac {3 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}{2 x^2}}{c^2}-\frac {c \left (c^2-d^2 x^2\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 536

\(\displaystyle \frac {\frac {1}{2} c^2 d^2 \left (\int \frac {-d c^2-6 d^2 x c}{x \sqrt {c^2-d^2 x^2}}dx-\frac {(6 c+d x) \sqrt {c^2-d^2 x^2}}{x}\right )-\frac {3 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}{2 x^2}}{c^2}-\frac {c \left (c^2-d^2 x^2\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 538

\(\displaystyle \frac {\frac {1}{2} c^2 d^2 \left (c^2 (-d) \int \frac {1}{x \sqrt {c^2-d^2 x^2}}dx-6 c d^2 \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx-\frac {(6 c+d x) \sqrt {c^2-d^2 x^2}}{x}\right )-\frac {3 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}{2 x^2}}{c^2}-\frac {c \left (c^2-d^2 x^2\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{2} c^2 d^2 \left (c^2 (-d) \int \frac {1}{x \sqrt {c^2-d^2 x^2}}dx-6 c d^2 \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}-\frac {(6 c+d x) \sqrt {c^2-d^2 x^2}}{x}\right )-\frac {3 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}{2 x^2}}{c^2}-\frac {c \left (c^2-d^2 x^2\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{2} c^2 d^2 \left (c^2 (-d) \int \frac {1}{x \sqrt {c^2-d^2 x^2}}dx-6 c d \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )-\frac {(6 c+d x) \sqrt {c^2-d^2 x^2}}{x}\right )-\frac {3 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}{2 x^2}}{c^2}-\frac {c \left (c^2-d^2 x^2\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {1}{2} c^2 d^2 \left (-\frac {1}{2} c^2 d \int \frac {1}{x^2 \sqrt {c^2-d^2 x^2}}dx^2-6 c d \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )-\frac {(6 c+d x) \sqrt {c^2-d^2 x^2}}{x}\right )-\frac {3 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}{2 x^2}}{c^2}-\frac {c \left (c^2-d^2 x^2\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {1}{2} c^2 d^2 \left (\frac {c^2 \int \frac {1}{\frac {c^2}{d^2}-\frac {x^4}{d^2}}d\sqrt {c^2-d^2 x^2}}{d}-6 c d \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )-\frac {(6 c+d x) \sqrt {c^2-d^2 x^2}}{x}\right )-\frac {3 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}{2 x^2}}{c^2}-\frac {c \left (c^2-d^2 x^2\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {1}{2} c^2 d^2 \left (-6 c d \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )+c d \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right )-\frac {\sqrt {c^2-d^2 x^2} (6 c+d x)}{x}\right )-\frac {3 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}{2 x^2}}{c^2}-\frac {c \left (c^2-d^2 x^2\right )^{3/2}}{3 x^3}\)

Input:

Int[((c + d*x)^3*Sqrt[c^2 - d^2*x^2])/x^4,x]
 

Output:

-1/3*(c*(c^2 - d^2*x^2)^(3/2))/x^3 + ((-3*c^2*d*(c^2 - d^2*x^2)^(3/2))/(2* 
x^2) + (c^2*d^2*(-(((6*c + d*x)*Sqrt[c^2 - d^2*x^2])/x) - 6*c*d*ArcTan[(d* 
x)/Sqrt[c^2 - d^2*x^2]] + c*d*ArcTanh[Sqrt[c^2 - d^2*x^2]/c]))/2)/c^2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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]
 

rule 224

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 536

Int[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_)^2, x_Symbol] :> S 
imp[(-(2*c*p - d*x))*((a + b*x^2)^p/(2*p*x)), x] + Int[(a*d + 2*b*c*p*x)*(( 
a + b*x^2)^(p - 1)/x), x] /; FreeQ[{a, b, c, d}, x] && GtQ[p, 0] && Integer 
Q[2*p]
 

rule 538

Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp 
[c   Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d   Int[1/Sqrt[a + b*x^2], x] 
, x] /; FreeQ[{a, b, c, d}, x]
 

rule 540

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain 
der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) 
, x] + Simp[1/(a*(m + 1))   Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 
1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG 
tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.90

method	result	size
risch	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, c \left (16 d^{2} x^{2}+9 c d x +2 c^{2}\right )}{6 x^{3}}+\frac {d^{3} c^{2} \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{2 \sqrt {c^{2}}}-\frac {3 d^{4} c \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}+d^{3} \sqrt {-d^{2} x^{2}+c^{2}}\)	\(137\)
default	\(-\frac {c \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3 x^{3}}+d^{3} \left (\sqrt {-d^{2} x^{2}+c^{2}}-\frac {c^{2} \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{\sqrt {c^{2}}}\right )+3 c \,d^{2} \left (-\frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{c^{2} x}-\frac {2 d^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{c^{2}}\right )+3 c^{2} d \left (-\frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{2 c^{2} x^{2}}-\frac {d^{2} \left (\sqrt {-d^{2} x^{2}+c^{2}}-\frac {c^{2} \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{\sqrt {c^{2}}}\right )}{2 c^{2}}\right )\)	\(264\)

Input:

int((d*x+c)^3*(-d^2*x^2+c^2)^(1/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/6*(-d^2*x^2+c^2)^(1/2)*c*(16*d^2*x^2+9*c*d*x+2*c^2)/x^3+1/2*d^3*c^2/(c^ 
2)^(1/2)*ln((2*c^2+2*(c^2)^(1/2)*(-d^2*x^2+c^2)^(1/2))/x)-3*d^4*c/(d^2)^(1 
/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))+d^3*(-d^2*x^2+c^2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.85 \[ \int \frac {(c+d x)^3 \sqrt {c^2-d^2 x^2}}{x^4} \, dx=\frac {36 \, c d^{3} x^{3} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - 3 \, c d^{3} x^{3} \log \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{x}\right ) + 6 \, c d^{3} x^{3} + {\left (6 \, d^{3} x^{3} - 16 \, c d^{2} x^{2} - 9 \, c^{2} d x - 2 \, c^{3}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{6 \, x^{3}} \] Input:

integrate((d*x+c)^3*(-d^2*x^2+c^2)^(1/2)/x^4,x, algorithm="fricas")
 

Output:

1/6*(36*c*d^3*x^3*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) - 3*c*d^3*x^3* 
log(-(c - sqrt(-d^2*x^2 + c^2))/x) + 6*c*d^3*x^3 + (6*d^3*x^3 - 16*c*d^2*x 
^2 - 9*c^2*d*x - 2*c^3)*sqrt(-d^2*x^2 + c^2))/x^3
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.49 (sec) , antiderivative size = 461, normalized size of antiderivative = 3.03 \[ \int \frac {(c+d x)^3 \sqrt {c^2-d^2 x^2}}{x^4} \, dx=c^{3} \left (\begin {cases} - \frac {d \sqrt {\frac {c^{2}}{d^{2} x^{2}} - 1}}{3 x^{2}} + \frac {d^{3} \sqrt {\frac {c^{2}}{d^{2} x^{2}} - 1}}{3 c^{2}} & \text {for}\: \left |{\frac {c^{2}}{d^{2} x^{2}}}\right | > 1 \\- \frac {i d \sqrt {- \frac {c^{2}}{d^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i d^{3} \sqrt {- \frac {c^{2}}{d^{2} x^{2}} + 1}}{3 c^{2}} & \text {otherwise} \end {cases}\right ) + 3 c^{2} d \left (\begin {cases} - \frac {d \sqrt {\frac {c^{2}}{d^{2} x^{2}} - 1}}{2 x} + \frac {d^{2} \operatorname {acosh}{\left (\frac {c}{d x} \right )}}{2 c} & \text {for}\: \left |{\frac {c^{2}}{d^{2} x^{2}}}\right | > 1 \\\frac {i c^{2}}{2 d x^{3} \sqrt {- \frac {c^{2}}{d^{2} x^{2}} + 1}} - \frac {i d}{2 x \sqrt {- \frac {c^{2}}{d^{2} x^{2}} + 1}} - \frac {i d^{2} \operatorname {asin}{\left (\frac {c}{d x} \right )}}{2 c} & \text {otherwise} \end {cases}\right ) + 3 c d^{2} \left (\begin {cases} \frac {i c}{x \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}}} + i d \operatorname {acosh}{\left (\frac {d x}{c} \right )} - \frac {i d^{2} x}{c \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}}} & \text {for}\: \left |{\frac {d^{2} x^{2}}{c^{2}}}\right | > 1 \\- \frac {c}{x \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}}} - d \operatorname {asin}{\left (\frac {d x}{c} \right )} + \frac {d^{2} x}{c \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}}} & \text {otherwise} \end {cases}\right ) + d^{3} \left (\begin {cases} \frac {c^{2}}{d x \sqrt {\frac {c^{2}}{d^{2} x^{2}} - 1}} - c \operatorname {acosh}{\left (\frac {c}{d x} \right )} - \frac {d x}{\sqrt {\frac {c^{2}}{d^{2} x^{2}} - 1}} & \text {for}\: \left |{\frac {c^{2}}{d^{2} x^{2}}}\right | > 1 \\- \frac {i c^{2}}{d x \sqrt {- \frac {c^{2}}{d^{2} x^{2}} + 1}} + i c \operatorname {asin}{\left (\frac {c}{d x} \right )} + \frac {i d x}{\sqrt {- \frac {c^{2}}{d^{2} x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((d*x+c)**3*(-d**2*x**2+c**2)**(1/2)/x**4,x)
 

Output:

c**3*Piecewise((-d*sqrt(c**2/(d**2*x**2) - 1)/(3*x**2) + d**3*sqrt(c**2/(d 
**2*x**2) - 1)/(3*c**2), Abs(c**2/(d**2*x**2)) > 1), (-I*d*sqrt(-c**2/(d** 
2*x**2) + 1)/(3*x**2) + I*d**3*sqrt(-c**2/(d**2*x**2) + 1)/(3*c**2), True) 
) + 3*c**2*d*Piecewise((-d*sqrt(c**2/(d**2*x**2) - 1)/(2*x) + d**2*acosh(c 
/(d*x))/(2*c), Abs(c**2/(d**2*x**2)) > 1), (I*c**2/(2*d*x**3*sqrt(-c**2/(d 
**2*x**2) + 1)) - I*d/(2*x*sqrt(-c**2/(d**2*x**2) + 1)) - I*d**2*asin(c/(d 
*x))/(2*c), True)) + 3*c*d**2*Piecewise((I*c/(x*sqrt(-1 + d**2*x**2/c**2)) 
 + I*d*acosh(d*x/c) - I*d**2*x/(c*sqrt(-1 + d**2*x**2/c**2)), Abs(d**2*x** 
2/c**2) > 1), (-c/(x*sqrt(1 - d**2*x**2/c**2)) - d*asin(d*x/c) + d**2*x/(c 
*sqrt(1 - d**2*x**2/c**2)), True)) + d**3*Piecewise((c**2/(d*x*sqrt(c**2/( 
d**2*x**2) - 1)) - c*acosh(c/(d*x)) - d*x/sqrt(c**2/(d**2*x**2) - 1), Abs( 
c**2/(d**2*x**2)) > 1), (-I*c**2/(d*x*sqrt(-c**2/(d**2*x**2) + 1)) + I*c*a 
sin(c/(d*x)) + I*d*x/sqrt(-c**2/(d**2*x**2) + 1), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.88 \[ \int \frac {(c+d x)^3 \sqrt {c^2-d^2 x^2}}{x^4} \, dx=-3 \, c d^{3} \arcsin \left (\frac {d x}{c}\right ) + \frac {1}{2} \, c d^{3} \log \left (\frac {2 \, c^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} c}{{\left | x \right |}}\right ) - \frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} d^{3} - \frac {3 \, \sqrt {-d^{2} x^{2} + c^{2}} c d^{2}}{x} - \frac {3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d}{2 \, x^{2}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c}{3 \, x^{3}} \] Input:

integrate((d*x+c)^3*(-d^2*x^2+c^2)^(1/2)/x^4,x, algorithm="maxima")
 

Output:

-3*c*d^3*arcsin(d*x/c) + 1/2*c*d^3*log(2*c^2/abs(x) + 2*sqrt(-d^2*x^2 + c^ 
2)*c/abs(x)) - 1/2*sqrt(-d^2*x^2 + c^2)*d^3 - 3*sqrt(-d^2*x^2 + c^2)*c*d^2 
/x - 3/2*(-d^2*x^2 + c^2)^(3/2)*d/x^2 - 1/3*(-d^2*x^2 + c^2)^(3/2)*c/x^3
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (134) = 268\).

Time = 0.14 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.87 \[ \int \frac {(c+d x)^3 \sqrt {c^2-d^2 x^2}}{x^4} \, dx=-\frac {3 \, c d^{4} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{{\left | d \right |}} + \frac {{\left (c d^{4} + \frac {9 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} c d^{2}}{x} + \frac {33 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} c}{x^{2}}\right )} d^{6} x^{3}}{24 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} {\left | d \right |}} + \frac {c d^{4} \log \left (\frac {{\left | -2 \, c d - 2 \, \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |} \right |}}{2 \, d^{2} {\left | x \right |}}\right )}{2 \, {\left | d \right |}} + \sqrt {-d^{2} x^{2} + c^{2}} d^{3} - \frac {\frac {33 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} c d^{4}}{x} + \frac {9 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} c d^{2}}{x^{2}} + \frac {{\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} c}{x^{3}}}{24 \, d^{2} {\left | d \right |}} \] Input:

integrate((d*x+c)^3*(-d^2*x^2+c^2)^(1/2)/x^4,x, algorithm="giac")
 

Output:

-3*c*d^4*arcsin(d*x/c)*sgn(c)*sgn(d)/abs(d) + 1/24*(c*d^4 + 9*(c*d + sqrt( 
-d^2*x^2 + c^2)*abs(d))*c*d^2/x + 33*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2 
*c/x^2)*d^6*x^3/((c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*abs(d)) + 1/2*c*d^4 
*log(1/2*abs(-2*c*d - 2*sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*abs(x)))/abs(d) 
+ sqrt(-d^2*x^2 + c^2)*d^3 - 1/24*(33*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))* 
c*d^4/x + 9*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*c*d^2/x^2 + (c*d + sqrt( 
-d^2*x^2 + c^2)*abs(d))^3*c/x^3)/(d^2*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^3 \sqrt {c^2-d^2 x^2}}{x^4} \, dx=\int \frac {\sqrt {c^2-d^2\,x^2}\,{\left (c+d\,x\right )}^3}{x^4} \,d x \] Input:

int(((c^2 - d^2*x^2)^(1/2)*(c + d*x)^3)/x^4,x)
 

Output:

int(((c^2 - d^2*x^2)^(1/2)*(c + d*x)^3)/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.07 \[ \int \frac {(c+d x)^3 \sqrt {c^2-d^2 x^2}}{x^4} \, dx=\frac {-36 \mathit {asin} \left (\frac {d x}{c}\right ) c \,d^{3} x^{3}-4 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3}-18 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x -32 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{2} x^{2}+12 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{3} x^{3}-3 \,\mathrm {log}\left (\frac {\sqrt {-d^{2} x^{2}+c^{2}}-c}{c}\right ) c \,d^{3} x^{3}+3 \,\mathrm {log}\left (\frac {\sqrt {-d^{2} x^{2}+c^{2}}+c}{c}\right ) c \,d^{3} x^{3}}{12 x^{3}} \] Input:

int((d*x+c)^3*(-d^2*x^2+c^2)^(1/2)/x^4,x)
 

Output:

( - 36*asin((d*x)/c)*c*d**3*x**3 - 4*sqrt(c**2 - d**2*x**2)*c**3 - 18*sqrt 
(c**2 - d**2*x**2)*c**2*d*x - 32*sqrt(c**2 - d**2*x**2)*c*d**2*x**2 + 12*s 
qrt(c**2 - d**2*x**2)*d**3*x**3 - 3*log((sqrt(c**2 - d**2*x**2) - c)/c)*c* 
d**3*x**3 + 3*log((sqrt(c**2 - d**2*x**2) + c)/c)*c*d**3*x**3)/(12*x**3)