\(\int \frac {A+B x}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx\) [97]

Optimal result

Integrand size = 31, antiderivative size = 106 \[ \int \frac {A+B x}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx=\frac {(B c-A d) \sqrt {c^2-d^2 x^2}}{2 c d^2 (c+d x)^{3/2}}-\frac {(3 B c+A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{2 \sqrt {2} c^{3/2} d^2} \] Output:

1/2*(-A*d+B*c)*(-d^2*x^2+c^2)^(1/2)/c/d^2/(d*x+c)^(3/2)-1/4*(A*d+3*B*c)*ar 
ctanh(2^(1/2)*c^(1/2)*(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2))*2^(1/2)/c^(3/2)/ 
d^2
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx=\frac {\frac {2 \sqrt {c} (B c-A d) \sqrt {c^2-d^2 x^2}}{(c+d x)^{3/2}}-\sqrt {2} (3 B c+A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{4 c^{3/2} d^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {671, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 471

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 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 = 0.36 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.60

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{2} x +3 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c d x +A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c d +3 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+2 A \sqrt {-d x +c}\, \sqrt {c}\, d -2 B \sqrt {-d x +c}\, c^{\frac {3}{2}}\right )}{4 c^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}} \sqrt {-d x +c}\, d^{2}}\)	\(170\)

Input:

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

Output:

-1/4*(-d^2*x^2+c^2)^(1/2)/c^(3/2)*(A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^ 
(1/2)/c^(1/2))*d^2*x+3*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2 
))*c*d*x+A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d+3*B*2^( 
1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2+2*A*(-d*x+c)^(1/2)*c^ 
(1/2)*d-2*B*(-d*x+c)^(1/2)*c^(3/2))/(d*x+c)^(3/2)/(-d*x+c)^(1/2)/d^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 353, normalized size of antiderivative = 3.33 \[ \int \frac {A+B x}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx=\left [\frac {\sqrt {2} {\left (3 \, B c^{3} + A c^{2} d + {\left (3 \, B c d^{2} + A d^{3}\right )} x^{2} + 2 \, {\left (3 \, B c^{2} d + A c d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 4 \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (B c^{2} - A c d\right )} \sqrt {d x + c}}{8 \, {\left (c^{2} d^{4} x^{2} + 2 \, c^{3} d^{3} x + c^{4} d^{2}\right )}}, \frac {\sqrt {2} {\left (3 \, B c^{3} + A c^{2} d + {\left (3 \, B c d^{2} + A d^{3}\right )} x^{2} + 2 \, {\left (3 \, B c^{2} d + A c d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + 2 \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (B c^{2} - A c d\right )} \sqrt {d x + c}}{4 \, {\left (c^{2} d^{4} x^{2} + 2 \, c^{3} d^{3} x + c^{4} d^{2}\right )}}\right ] \] Input:

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

Output:

[1/8*(sqrt(2)*(3*B*c^3 + A*c^2*d + (3*B*c*d^2 + A*d^3)*x^2 + 2*(3*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*sqrt(-d 
^2*x^2 + c^2)*(B*c^2 - A*c*d)*sqrt(d*x + c))/(c^2*d^4*x^2 + 2*c^3*d^3*x + 
c^4*d^2), 1/4*(sqrt(2)*(3*B*c^3 + A*c^2*d + (3*B*c*d^2 + A*d^3)*x^2 + 2*(3 
*B*c^2*d + A*c*d^2)*x)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sq 
rt(d*x + c)*sqrt(-c)/(c*d*x + c^2)) + 2*sqrt(-d^2*x^2 + c^2)*(B*c^2 - A*c* 
d)*sqrt(d*x + c))/(c^2*d^4*x^2 + 2*c^3*d^3*x + c^4*d^2)]
 

Sympy [F]

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

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

Output:

Integral((A + B*x)/(sqrt(-(-c + d*x)*(c + d*x))*(c + d*x)**(3/2)), x)
 

Maxima [F]

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

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

Output:

integrate((B*x + A)/(sqrt(-d^2*x^2 + c^2)*(d*x + c)^(3/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx=\frac {\frac {\sqrt {2} {\left (3 \, B c + A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c} + \frac {2 \, {\left (\sqrt {-d x + c} B c - \sqrt {-d x + c} A d\right )}}{{\left (d x + c\right )} c}}{4 \, d^{2}} \] Input:

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

Output:

1/4*(sqrt(2)*(3*B*c + A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c))/(sq 
rt(-c)*c) + 2*(sqrt(-d*x + c)*B*c - sqrt(-d*x + c)*A*d)/((d*x + c)*c))/d^2
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.22 \[ \int \frac {A+B x}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx=\frac {-4 \sqrt {-d x +c}\, a c d +4 \sqrt {-d x +c}\, b \,c^{2}+\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a c d +\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,d^{2} x +3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b \,c^{2}+3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c d x -\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a c d -\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,d^{2} x -3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b \,c^{2}-3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c d x}{8 c^{2} d^{2} \left (d x +c \right )} \] Input:

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

Output:

( - 4*sqrt(c - d*x)*a*c*d + 4*sqrt(c - d*x)*b*c**2 + sqrt(c)*sqrt(2)*log(s 
qrt(c - d*x) - sqrt(c)*sqrt(2))*a*c*d + sqrt(c)*sqrt(2)*log(sqrt(c - d*x) 
- sqrt(c)*sqrt(2))*a*d**2*x + 3*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c 
)*sqrt(2))*b*c**2 + 3*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2)) 
*b*c*d*x - sqrt(c)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*c*d - sq 
rt(c)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*d**2*x - 3*sqrt(c)*sq 
rt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**2 - 3*sqrt(c)*sqrt(2)*log( 
sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c*d*x)/(8*c**2*d**2*(c + d*x))