\(\int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{5/2}} \, dx\) [69]

Optimal result

Integrand size = 31, antiderivative size = 130 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{5/2}} \, dx=\frac {(B c-A d) \sqrt {c^2-d^2 x^2}}{d^2 (c+d x)^{3/2}}+\frac {2 B \sqrt {c^2-d^2 x^2}}{d^2 \sqrt {c+d x}}-\frac {(5 B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {2} \sqrt {c} d^2} \] Output:

(-A*d+B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(3/2)+2*B*(-d^2*x^2+c^2)^(1/2) 
/d^2/(d*x+c)^(1/2)-1/2*(-A*d+5*B*c)*arctanh(2^(1/2)*c^(1/2)*(d*x+c)^(1/2)/ 
(-d^2*x^2+c^2)^(1/2))*2^(1/2)/c^(1/2)/d^2
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.82 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{5/2}} \, dx=\frac {\frac {2 (3 B c-A d+2 B d x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{3/2}}-\frac {\sqrt {2} (5 B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {c}}}{2 d^2} \] Input:

Integrate[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x)^(5/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {671, 466, 471, 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.

\(\displaystyle \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {(5 B c-A d) \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{3/2}}dx}{4 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{2 c d^2 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {(5 B c-A d) \left (2 c \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )}{4 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{2 c d^2 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(5 B c-A d) \left (4 c d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )}{4 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{2 c d^2 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(5 B c-A d) \left (\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}-\frac {2 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{d}\right )}{4 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{2 c d^2 (c+d x)^{5/2}}\)

Input:

Int[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x)^(5/2),x]
 

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(3/2))/(2*c*d^2*(c + d*x)^(5/2)) + ((5*B*c - 
A*d)*((2*Sqrt[c^2 - d^2*x^2])/(d*Sqrt[c + d*x]) - (2*Sqrt[2]*Sqrt[c]*ArcTa 
nh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])])/d))/(4*c*d)
 

Defintions of rubi rules used

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 466

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 
2*(n + 2*p + 1)))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr 
eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 
] || EqQ[n + p + 1, 0]) && NeQ[n + 2*p + 1, 0] && IntegerQ[2*p]
 

rule 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m 
 + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m 
+ p + 1))   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, 
e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p 
 + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p 
 + 1, 0]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.30

method	result	size
risch	\(\frac {2 B \sqrt {-d x +c}\, \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}-\frac {2 \left (\frac {\left (-\frac {A d}{2}+\frac {B c}{2}\right ) \sqrt {-d x +c}}{-d x -c}-\frac {\left (A d -5 B c \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\right ) \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(169\)
default	\(\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{2} x -5 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c d x +A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c d -5 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+4 B \sqrt {-d x +c}\, \sqrt {c}\, d x -2 A \sqrt {-d x +c}\, \sqrt {c}\, d +6 B \sqrt {-d x +c}\, c^{\frac {3}{2}}\right )}{2 \left (d x +c \right )^{\frac {3}{2}} \sqrt {-d x +c}\, d^{2} \sqrt {c}}\)	\(186\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2*B*(-d*x+c)^(1/2)/d^2*((-d^2*x^2+c^2)/(d*x+c))^(1/2)/(-d^2*x^2+c^2)^(1/2) 
*(d*x+c)^(1/2)-2/d^2*((-1/2*A*d+1/2*B*c)*(-d*x+c)^(1/2)/(-d*x-c)-1/4*(A*d- 
5*B*c)*2^(1/2)/c^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2)))*((-d^2 
*x^2+c^2)/(d*x+c))^(1/2)/(-d^2*x^2+c^2)^(1/2)*(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.84 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{5/2}} \, dx=\left [-\frac {\sqrt {2} {\left (5 \, B c^{3} - A c^{2} d + {\left (5 \, B c d^{2} - A d^{3}\right )} x^{2} + 2 \, {\left (5 \, B c^{2} d - A c d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x - 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 4 \, {\left (2 \, B c d x + 3 \, B c^{2} - A c d\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{4 \, {\left (c d^{4} x^{2} + 2 \, c^{2} d^{3} x + c^{3} d^{2}\right )}}, \frac {\sqrt {2} {\left (5 \, B c^{3} - A c^{2} d + {\left (5 \, B c d^{2} - A d^{3}\right )} x^{2} + 2 \, {\left (5 \, B c^{2} d - A c d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + 2 \, {\left (2 \, B c d x + 3 \, B c^{2} - A c d\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{2 \, {\left (c d^{4} x^{2} + 2 \, c^{2} d^{3} x + c^{3} d^{2}\right )}}\right ] \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(5/2),x, algorithm="fricas" 
)
 

Output:

[-1/4*(sqrt(2)*(5*B*c^3 - A*c^2*d + (5*B*c*d^2 - A*d^3)*x^2 + 2*(5*B*c^2*d 
 - A*c*d^2)*x)*sqrt(c)*log(-(d^2*x^2 - 2*c*d*x - 2*sqrt(2)*sqrt(-d^2*x^2 + 
 c^2)*sqrt(d*x + c)*sqrt(c) - 3*c^2)/(d^2*x^2 + 2*c*d*x + c^2)) - 4*(2*B*c 
*d*x + 3*B*c^2 - A*c*d)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c*d^4*x^2 + 2 
*c^2*d^3*x + c^3*d^2), 1/2*(sqrt(2)*(5*B*c^3 - A*c^2*d + (5*B*c*d^2 - A*d^ 
3)*x^2 + 2*(5*B*c^2*d - A*c*d^2)*x)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-d^2* 
x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + c^2)) + 2*(2*B*c*d*x + 3*B*c^2 
- A*c*d)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c*d^4*x^2 + 2*c^2*d^3*x + c^ 
3*d^2)]
 

Sympy [F]

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

integrate((B*x+A)*(-d**2*x**2+c**2)**(1/2)/(d*x+c)**(5/2),x)
 

Output:

Integral(sqrt(-(-c + d*x)*(c + d*x))*(A + B*x)/(c + d*x)**(5/2), x)
 

Maxima [F]

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

integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(5/2),x, algorithm="maxima" 
)
 

Output:

integrate(sqrt(-d^2*x^2 + c^2)*(B*x + A)/(d*x + c)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.67 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{5/2}} \, dx=\frac {\frac {\sqrt {2} {\left (5 \, B c - A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} + 4 \, \sqrt {-d x + c} B + \frac {2 \, {\left (\sqrt {-d x + c} B c - \sqrt {-d x + c} A d\right )}}{d x + c}}{2 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

1/2*(sqrt(2)*(5*B*c - A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c))/sqr 
t(-c) + 4*sqrt(-d*x + c)*B + 2*(sqrt(-d*x + c)*B*c - sqrt(-d*x + c)*A*d)/( 
d*x + c))/d^2
 

Mupad [F(-1)]

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

int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x)^(5/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.63 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{5/2}} \, dx=\frac {-8 \sqrt {-d x +c}\, a c d +24 \sqrt {-d x +c}\, b \,c^{2}+16 \sqrt {-d x +c}\, b c d x -4 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a c d -4 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a \,d^{2} x +20 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{2}+20 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b c d x +\sqrt {c}\, \sqrt {2}\, a c d +\sqrt {c}\, \sqrt {2}\, a \,d^{2} x -17 \sqrt {c}\, \sqrt {2}\, b \,c^{2}-17 \sqrt {c}\, \sqrt {2}\, b c d x}{8 c \,d^{2} \left (d x +c \right )} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(5/2),x)
 

Output:

( - 8*sqrt(c - d*x)*a*c*d + 24*sqrt(c - d*x)*b*c**2 + 16*sqrt(c - d*x)*b*c 
*d*x - 4*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2)) 
*a*c*d - 4*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2 
))*a*d**2*x + 20*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt( 
2)))/2))*b*c**2 + 20*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*s 
qrt(2)))/2))*b*c*d*x + sqrt(c)*sqrt(2)*a*c*d + sqrt(c)*sqrt(2)*a*d**2*x - 
17*sqrt(c)*sqrt(2)*b*c**2 - 17*sqrt(c)*sqrt(2)*b*c*d*x)/(8*c*d**2*(c + d*x 
))