\(\int \frac {A+B x}{\sqrt {c+d x} (c^2-d^2 x^2)^{3/2}} \, dx\) [104]

Optimal result

Integrand size = 31, antiderivative size = 148 \[ \int \frac {A+B x}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {(B c+A d) \sqrt {c+d x}}{2 c^2 d^2 \sqrt {c^2-d^2 x^2}}+\frac {(B c-A d) \sqrt {c^2-d^2 x^2}}{4 c^2 d^2 (c+d x)^{3/2}}-\frac {(B c+3 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{4 \sqrt {2} c^{5/2} d^2} \] Output:

1/2*(A*d+B*c)*(d*x+c)^(1/2)/c^2/d^2/(-d^2*x^2+c^2)^(1/2)+1/4*(-A*d+B*c)*(- 
d^2*x^2+c^2)^(1/2)/c^2/d^2/(d*x+c)^(3/2)-1/8*(3*A*d+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^(5/2)/d^2
 

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.84 \[ \int \frac {A+B x}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} (B c (3 c+d x)+A d (c+3 d x))}{(c-d x) (c+d x)^{3/2}}-\sqrt {2} (B c+3 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{8 c^{5/2} d^2} \] Input:

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

Output:

((2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(B*c*(3*c + d*x) + A*d*(c + 3*d*x)))/((c - 
 d*x)*(c + d*x)^(3/2)) - Sqrt[2]*(B*c + 3*A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sq 
rt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/(8*c^(5/2)*d^2)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {671, 467, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 467

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

\(\Big \downarrow \) 471

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(B*c - A*d)/(2*c*d^2*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2]) + ((B*c + 3*A*d)*( 
Sqrt[c + d*x]/(c*d*Sqrt[c^2 - d^2*x^2]) - ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqr 
t[2]*Sqrt[c]*Sqrt[c + d*x])]/(Sqrt[2]*c^(3/2)*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 467

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-c)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*a*d*(p + 1))), x] + Simp[c*((n + 2 
*p + 2)/(2*a*(p + 1)))   Int[(c + d*x)^(n - 1)*(a + b*x^2)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[p, -1] && LtQ[0, 
n, 1] && 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 = 0.36 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.38

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (3 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{2} x +B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c d x +3 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c d +B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-6 B \,c^{\frac {5}{2}}-2 B \,c^{\frac {3}{2}} d x -2 A \,c^{\frac {3}{2}} d -6 A \sqrt {c}\, d^{2} x \right )}{8 \left (d x +c \right )^{\frac {3}{2}} \left (-d x +c \right ) d^{2} c^{\frac {5}{2}}}\)	\(204\)

Input:

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

Output:

-1/8/(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2)*(3*A*(-d*x+c)^(1/2)*2^(1/2)*arctan 
h(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*d^2*x+B*(-d*x+c)^(1/2)*2^(1/2)*arcta 
nh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d*x+3*A*(-d*x+c)^(1/2)*2^(1/2)*ar 
ctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d+B*(-d*x+c)^(1/2)*2^(1/2)*arc 
tanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2-6*B*c^(5/2)-2*B*c^(3/2)*d*x-2 
*A*c^(3/2)*d-6*A*c^(1/2)*d^2*x)/(-d*x+c)/d^2/c^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 457, normalized size of antiderivative = 3.09 \[ \int \frac {A+B x}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\left [-\frac {\sqrt {2} {\left (B c^{4} + 3 \, A c^{3} d - {\left (B c d^{3} + 3 \, A d^{4}\right )} x^{3} - {\left (B c^{2} d^{2} + 3 \, A c d^{3}\right )} x^{2} + {\left (B c^{3} d + 3 \, A c^{2} d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 4 \, {\left (3 \, B c^{3} + A c^{2} d + {\left (B c^{2} d + 3 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{16 \, {\left (c^{3} d^{5} x^{3} + c^{4} d^{4} x^{2} - c^{5} d^{3} x - c^{6} d^{2}\right )}}, -\frac {\sqrt {2} {\left (B c^{4} + 3 \, A c^{3} d - {\left (B c d^{3} + 3 \, A d^{4}\right )} x^{3} - {\left (B c^{2} d^{2} + 3 \, A c d^{3}\right )} x^{2} + {\left (B c^{3} d + 3 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + 2 \, {\left (3 \, B c^{3} + A c^{2} d + {\left (B c^{2} d + 3 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{8 \, {\left (c^{3} d^{5} x^{3} + c^{4} d^{4} x^{2} - c^{5} d^{3} x - c^{6} d^{2}\right )}}\right ] \] Input:

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

Output:

[-1/16*(sqrt(2)*(B*c^4 + 3*A*c^3*d - (B*c*d^3 + 3*A*d^4)*x^3 - (B*c^2*d^2 
+ 3*A*c*d^3)*x^2 + (B*c^3*d + 3*A*c^2*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*(3*B*c^3 + A*c^2*d + (B*c^2*d + 3*A*c*d^2)*x)*sqr 
t(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^3*d^5*x^3 + c^4*d^4*x^2 - c^5*d^3*x - 
c^6*d^2), -1/8*(sqrt(2)*(B*c^4 + 3*A*c^3*d - (B*c*d^3 + 3*A*d^4)*x^3 - (B* 
c^2*d^2 + 3*A*c*d^3)*x^2 + (B*c^3*d + 3*A*c^2*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*(3* 
B*c^3 + A*c^2*d + (B*c^2*d + 3*A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + 
 c))/(c^3*d^5*x^3 + c^4*d^4*x^2 - c^5*d^3*x - c^6*d^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.75 \[ \int \frac {A+B x}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {2} {\left (B c + 3 \, A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{2} d} - \frac {2 \, {\left ({\left (d x - c\right )} B c + 4 \, B c^{2} + 3 \, {\left (d x - c\right )} A d + 4 \, A c d\right )}}{{\left ({\left (-d x + c\right )}^{\frac {3}{2}} - 2 \, \sqrt {-d x + c} c\right )} c^{2} d}}{8 \, d} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.05 \[ \int \frac {A+B x}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a c d +3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,d^{2} x +\sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b \,c^{2}+\sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c d x -3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a c d -3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,d^{2} x -\sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b \,c^{2}-\sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c d x +4 a \,c^{2} d +12 a c \,d^{2} x +12 b \,c^{3}+4 b \,c^{2} d x}{16 \sqrt {-d x +c}\, c^{3} d^{2} \left (d x +c \right )} \] Input:

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

Output:

(3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*a*c* 
d + 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*a 
*d**2*x + sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2 
))*b*c**2 + sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt 
(2))*b*c*d*x - 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c) 
*sqrt(2))*a*c*d - 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt 
(c)*sqrt(2))*a*d**2*x - sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + 
sqrt(c)*sqrt(2))*b*c**2 - sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) 
+ sqrt(c)*sqrt(2))*b*c*d*x + 4*a*c**2*d + 12*a*c*d**2*x + 12*b*c**3 + 4*b* 
c**2*d*x)/(16*sqrt(c - d*x)*c**3*d**2*(c + d*x))