\(\int \frac {A+B x}{\sqrt {c+d x} (c^2-d^2 x^2)^{5/2}} \, dx\) [112]

Optimal result

Integrand size = 31, antiderivative size = 194 \[ \int \frac {A+B x}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2 B}{7 d^2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}+\frac {(B c+7 A d) (c+7 d x)}{168 c^2 d^2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}+\frac {5 (B c+7 A d) (c+3 d x)}{192 c^4 d^2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}-\frac {5 (B c+7 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{64 \sqrt {2} c^{9/2} d^2} \] Output:

2/7*B/d^2/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(3/2)+1/168*(7*A*d+B*c)*(7*d*x+c)/c 
^2/d^2/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(3/2)+5/192*(7*A*d+B*c)*(3*d*x+c)/c^4/ 
d^2/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2)-5/128*(7*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^(9/2)/d^2
 

Mathematica [A] (verified)

Time = 1.84 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (A d \left (43 c^3+161 c^2 d x-35 c d^2 x^2-105 d^3 x^3\right )+B c \left (61 c^3+23 c^2 d x-5 c d^2 x^2-15 d^3 x^3\right )\right )}{(c-d x)^2 (c+d x)^{5/2}}-15 \sqrt {2} (B c+7 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{384 c^{9/2} d^2} \] Input:

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

Output:

((2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(A*d*(43*c^3 + 161*c^2*d*x - 35*c*d^2*x^2 
- 105*d^3*x^3) + B*c*(61*c^3 + 23*c^2*d*x - 5*c*d^2*x^2 - 15*d^3*x^3)))/(( 
c - d*x)^2*(c + d*x)^(5/2)) - 15*Sqrt[2]*(B*c + 7*A*d)*ArcTanh[(Sqrt[2]*Sq 
rt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/(384*c^(9/2)*d^2)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.20, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {671, 467, 470, 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.

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {(7 A d+B c) \int \frac {\sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{5/2}}dx}{8 c d}+\frac {B c-A d}{4 c d^2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 467

\(\displaystyle \frac {(7 A d+B c) \left (\frac {5 \int \frac {1}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}dx}{6 c}+\frac {\sqrt {c+d x}}{3 c d \left (c^2-d^2 x^2\right )^{3/2}}\right )}{8 c d}+\frac {B c-A d}{4 c d^2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(7 A d+B c) \left (\frac {5 \left (\frac {3 \int \frac {\sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c}+\frac {\sqrt {c+d x}}{3 c d \left (c^2-d^2 x^2\right )^{3/2}}\right )}{8 c d}+\frac {B c-A d}{4 c d^2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 467

\(\displaystyle \frac {(7 A d+B c) \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{2 c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c}+\frac {\sqrt {c+d x}}{3 c d \left (c^2-d^2 x^2\right )^{3/2}}\right )}{8 c d}+\frac {B c-A d}{4 c d^2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(7 A d+B c) \left (\frac {5 \left (\frac {3 \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c}+\frac {\sqrt {c+d x}}{3 c d \left (c^2-d^2 x^2\right )^{3/2}}\right )}{8 c d}+\frac {B c-A d}{4 c d^2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(7 A d+B c) \left (\frac {5 \left (\frac {3 \left (\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {2} c^{3/2} d}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c}+\frac {\sqrt {c+d x}}{3 c d \left (c^2-d^2 x^2\right )^{3/2}}\right )}{8 c d}+\frac {B c-A d}{4 c d^2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\)

Input:

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

Output:

(B*c - A*d)/(4*c*d^2*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(3/2)) + ((B*c + 7*A*d) 
*(Sqrt[c + d*x]/(3*c*d*(c^2 - d^2*x^2)^(3/2)) + (5*(-1/2*1/(c*d*Sqrt[c + d 
*x]*Sqrt[c^2 - d^2*x^2]) + (3*(Sqrt[c + d*x]/(c*d*Sqrt[c^2 - d^2*x^2]) - A 
rcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])]/(Sqrt[2]*c^(3/ 
2)*d)))/(4*c)))/(6*c)))/(8*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 470

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 
2*p + 2)/(2*c*(n + p + 1))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; 
 FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + 
 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(413\) vs. \(2(162)=324\).

Time = 0.36 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.13

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (-105 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{4} x^{3}-15 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{3} x^{3}-105 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{3} x^{2}-15 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x^{2}+105 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x +210 A \sqrt {c}\, d^{4} x^{3}+15 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d x +30 B \,c^{\frac {3}{2}} d^{3} x^{3}+105 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d +70 A \,c^{\frac {3}{2}} d^{3} x^{2}+15 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}+10 B \,c^{\frac {5}{2}} d^{2} x^{2}-322 A \,c^{\frac {5}{2}} d^{2} x -46 B \,c^{\frac {7}{2}} d x -86 A \,c^{\frac {7}{2}} d -122 B \,c^{\frac {9}{2}}\right )}{384 \left (d x +c \right )^{\frac {5}{2}} \left (-d x +c \right )^{2} d^{2} c^{\frac {9}{2}}}\)	\(414\)

Input:

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

Output:

-1/384*(-d^2*x^2+c^2)^(1/2)*(-105*A*(-d*x+c)^(1/2)*2^(1/2)*arctanh(1/2*(-d 
*x+c)^(1/2)*2^(1/2)/c^(1/2))*d^4*x^3-15*B*(-d*x+c)^(1/2)*2^(1/2)*arctanh(1 
/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^3*x^3-105*A*(-d*x+c)^(1/2)*2^(1/2)* 
arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^3*x^2-15*B*(-d*x+c)^(1/2)* 
2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^2*x^2+105*A*(-d* 
x+c)^(1/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^2*x+2 
10*A*c^(1/2)*d^4*x^3+15*B*(-d*x+c)^(1/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2 
)*2^(1/2)/c^(1/2))*c^3*d*x+30*B*c^(3/2)*d^3*x^3+105*A*(-d*x+c)^(1/2)*2^(1/ 
2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d+70*A*c^(3/2)*d^3*x^2+ 
15*B*(-d*x+c)^(1/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^ 
4+10*B*c^(5/2)*d^2*x^2-322*A*c^(5/2)*d^2*x-46*B*c^(7/2)*d*x-86*A*c^(7/2)*d 
-122*B*c^(9/2))/(d*x+c)^(5/2)/(-d*x+c)^2/d^2/c^(9/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 329 vs. \(2 (162) = 324\).

Time = 0.10 (sec) , antiderivative size = 683, normalized size of antiderivative = 3.52 \[ \int \frac {A+B x}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\left [\frac {15 \, \sqrt {2} {\left (B c^{6} + 7 \, A c^{5} d + {\left (B c d^{5} + 7 \, A d^{6}\right )} x^{5} + {\left (B c^{2} d^{4} + 7 \, A c d^{5}\right )} x^{4} - 2 \, {\left (B c^{3} d^{3} + 7 \, A c^{2} d^{4}\right )} x^{3} - 2 \, {\left (B c^{4} d^{2} + 7 \, A c^{3} d^{3}\right )} x^{2} + {\left (B c^{5} d + 7 \, A c^{4} d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 4 \, {\left (61 \, B c^{5} + 43 \, A c^{4} d - 15 \, {\left (B c^{2} d^{3} + 7 \, A c d^{4}\right )} x^{3} - 5 \, {\left (B c^{3} d^{2} + 7 \, A c^{2} d^{3}\right )} x^{2} + 23 \, {\left (B c^{4} d + 7 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{768 \, {\left (c^{5} d^{7} x^{5} + c^{6} d^{6} x^{4} - 2 \, c^{7} d^{5} x^{3} - 2 \, c^{8} d^{4} x^{2} + c^{9} d^{3} x + c^{10} d^{2}\right )}}, \frac {15 \, \sqrt {2} {\left (B c^{6} + 7 \, A c^{5} d + {\left (B c d^{5} + 7 \, A d^{6}\right )} x^{5} + {\left (B c^{2} d^{4} + 7 \, A c d^{5}\right )} x^{4} - 2 \, {\left (B c^{3} d^{3} + 7 \, A c^{2} d^{4}\right )} x^{3} - 2 \, {\left (B c^{4} d^{2} + 7 \, A c^{3} d^{3}\right )} x^{2} + {\left (B c^{5} d + 7 \, A c^{4} d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + 2 \, {\left (61 \, B c^{5} + 43 \, A c^{4} d - 15 \, {\left (B c^{2} d^{3} + 7 \, A c d^{4}\right )} x^{3} - 5 \, {\left (B c^{3} d^{2} + 7 \, A c^{2} d^{3}\right )} x^{2} + 23 \, {\left (B c^{4} d + 7 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{384 \, {\left (c^{5} d^{7} x^{5} + c^{6} d^{6} x^{4} - 2 \, c^{7} d^{5} x^{3} - 2 \, c^{8} d^{4} x^{2} + c^{9} d^{3} x + c^{10} d^{2}\right )}}\right ] \] Input:

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

Output:

[1/768*(15*sqrt(2)*(B*c^6 + 7*A*c^5*d + (B*c*d^5 + 7*A*d^6)*x^5 + (B*c^2*d 
^4 + 7*A*c*d^5)*x^4 - 2*(B*c^3*d^3 + 7*A*c^2*d^4)*x^3 - 2*(B*c^4*d^2 + 7*A 
*c^3*d^3)*x^2 + (B*c^5*d + 7*A*c^4*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*(61*B*c^5 + 43*A*c^4*d - 15*(B*c^2*d^3 + 7*A*c*d^4)* 
x^3 - 5*(B*c^3*d^2 + 7*A*c^2*d^3)*x^2 + 23*(B*c^4*d + 7*A*c^3*d^2)*x)*sqrt 
(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^5*d^7*x^5 + c^6*d^6*x^4 - 2*c^7*d^5*x^3 
 - 2*c^8*d^4*x^2 + c^9*d^3*x + c^10*d^2), 1/384*(15*sqrt(2)*(B*c^6 + 7*A*c 
^5*d + (B*c*d^5 + 7*A*d^6)*x^5 + (B*c^2*d^4 + 7*A*c*d^5)*x^4 - 2*(B*c^3*d^ 
3 + 7*A*c^2*d^4)*x^3 - 2*(B*c^4*d^2 + 7*A*c^3*d^3)*x^2 + (B*c^5*d + 7*A*c^ 
4*d^2)*x)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*s 
qrt(-c)/(c*d*x + c^2)) + 2*(61*B*c^5 + 43*A*c^4*d - 15*(B*c^2*d^3 + 7*A*c* 
d^4)*x^3 - 5*(B*c^3*d^2 + 7*A*c^2*d^3)*x^2 + 23*(B*c^4*d + 7*A*c^3*d^2)*x) 
*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^5*d^7*x^5 + c^6*d^6*x^4 - 2*c^7*d^ 
5*x^3 - 2*c^8*d^4*x^2 + c^9*d^3*x + c^10*d^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\frac {15 \, \sqrt {2} {\left (B c + 7 \, A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{4} d} + \frac {16 \, {\left (3 \, {\left (d x - c\right )} B c - 2 \, B c^{2} + 9 \, {\left (d x - c\right )} A d - 2 \, A c d\right )}}{{\left (d x - c\right )} \sqrt {-d x + c} c^{4} d} - \frac {6 \, {\left (3 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c - 10 \, \sqrt {-d x + c} B c^{2} - 11 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d + 26 \, \sqrt {-d x + c} A c d\right )}}{{\left (d x + c\right )}^{2} c^{4} d}}{384 \, d} \] Input:

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

Output:

1/384*(15*sqrt(2)*(B*c + 7*A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c) 
)/(sqrt(-c)*c^4*d) + 16*(3*(d*x - c)*B*c - 2*B*c^2 + 9*(d*x - c)*A*d - 2*A 
*c*d)/((d*x - c)*sqrt(-d*x + c)*c^4*d) - 6*(3*(-d*x + c)^(3/2)*B*c - 10*sq 
rt(-d*x + c)*B*c^2 - 11*(-d*x + c)^(3/2)*A*d + 26*sqrt(-d*x + c)*A*c*d)/(( 
d*x + c)^2*c^4*d))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 676, normalized size of antiderivative = 3.48 \[ \int \frac {A+B x}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(105*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*a* 
c**3*d + 105*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqr 
t(2))*a*c**2*d**2*x - 105*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) 
- sqrt(c)*sqrt(2))*a*c*d**3*x**2 - 105*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(s 
qrt(c - d*x) - sqrt(c)*sqrt(2))*a*d**4*x**3 + 15*sqrt(c)*sqrt(c - d*x)*sqr 
t(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c**4 + 15*sqrt(c)*sqrt(c - d*x 
)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c**3*d*x - 15*sqrt(c)*sqr 
t(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c**2*d**2*x**2 - 
 15*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c 
*d**3*x**3 - 105*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c) 
*sqrt(2))*a*c**3*d - 105*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + 
 sqrt(c)*sqrt(2))*a*c**2*d**2*x + 105*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sq 
rt(c - d*x) + sqrt(c)*sqrt(2))*a*c*d**3*x**2 + 105*sqrt(c)*sqrt(c - d*x)*s 
qrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*d**4*x**3 - 15*sqrt(c)*sqrt( 
c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**4 - 15*sqrt(c)* 
sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**3*d*x + 15 
*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**2 
*d**2*x**2 + 15*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)* 
sqrt(2))*b*c*d**3*x**3 + 172*a*c**4*d + 644*a*c**3*d**2*x - 140*a*c**2*d** 
3*x**2 - 420*a*c*d**4*x**3 + 244*b*c**5 + 92*b*c**4*d*x - 20*b*c**3*d**...