\(\int \frac {A+B x}{(c+d x)^{5/2} (c^2-d^2 x^2)^{3/2}} \, dx\) [106]

Optimal result

Integrand size = 31, antiderivative size = 201 \[ \int \frac {A+B x}{(c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {B c-A d}{6 c d^2 (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}-\frac {5 B c+7 A d}{48 c^2 d^2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}+\frac {5 (5 B c+7 A d) (c+3 d x)}{384 c^4 d^2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}-\frac {5 (5 B c+7 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{128 \sqrt {2} c^{9/2} d^2} \] Output:

1/6*(-A*d+B*c)/c/d^2/(d*x+c)^(5/2)/(-d^2*x^2+c^2)^(1/2)-1/48*(7*A*d+5*B*c) 
/c^2/d^2/(d*x+c)^(3/2)/(-d^2*x^2+c^2)^(1/2)+5/384*(7*A*d+5*B*c)*(3*d*x+c)/ 
c^4/d^2/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2)-5/256*(7*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^(9/2)/d^2
 

Mathematica [A] (verified)

Time = 1.92 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{(c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (B c \left (49 c^3+85 c^2 d x+175 c d^2 x^2+75 d^3 x^3\right )+A d \left (-85 c^3+119 c^2 d x+245 c d^2 x^2+105 d^3 x^3\right )\right )}{(c-d x) (c+d x)^{7/2}}-15 \sqrt {2} (5 B c+7 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{768 c^{9/2} d^2} \] Input:

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

Output:

((2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(B*c*(49*c^3 + 85*c^2*d*x + 175*c*d^2*x^2 
+ 75*d^3*x^3) + A*d*(-85*c^3 + 119*c^2*d*x + 245*c*d^2*x^2 + 105*d^3*x^3)) 
)/((c - d*x)*(c + d*x)^(7/2)) - 15*Sqrt[2]*(5*B*c + 7*A*d)*ArcTanh[(Sqrt[2 
]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/(768*c^(9/2)*d^2)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {671, 470, 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.

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {(7 A d+5 B c) \int \frac {1}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}dx}{12 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(7 A d+5 B c) \left (\frac {5 \int \frac {1}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}dx}{8 c}-\frac {1}{4 c d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\right )}{12 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(7 A d+5 B c) \left (\frac {5 \left (\frac {3 \int \frac {\sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{8 c}-\frac {1}{4 c d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\right )}{12 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 467

\(\displaystyle \frac {(7 A d+5 B c) \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{2 c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{8 c}-\frac {1}{4 c d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\right )}{12 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(7 A d+5 B c) \left (\frac {5 \left (\frac {3 \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{8 c}-\frac {1}{4 c d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\right )}{12 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(7 A d+5 B c) \left (\frac {5 \left (\frac {3 \left (\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {2} c^{3/2} d}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{8 c}-\frac {1}{4 c d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\right )}{12 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}\)

Input:

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

Output:

(B*c - A*d)/(6*c*d^2*(c + d*x)^(5/2)*Sqrt[c^2 - d^2*x^2]) + ((5*B*c + 7*A* 
d)*(-1/4*1/(c*d*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2]) + (5*(-1/2*1/(c*d*Sqr 
t[c + d*x]*Sqrt[c^2 - d^2*x^2]) + (3*(Sqrt[c + d*x]/(c*d*Sqrt[c^2 - d^2*x^ 
2]) - ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])]/(Sqrt[2 
]*c^(3/2)*d)))/(4*c)))/(8*c)))/(12*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. \(2(169)=338\).

Time = 0.37 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.06

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (-490 A \,c^{\frac {3}{2}} d^{3} x^{2}+75 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}-350 B \,c^{\frac {5}{2}} d^{2} x^{2}+75 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{3} x^{3}+315 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{3} x^{2}+225 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x^{2}+315 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x +225 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d x -150 B \,c^{\frac {3}{2}} d^{3} x^{3}+105 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d -210 A \sqrt {c}\, d^{4} x^{3}+105 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{4} x^{3}-238 A \,c^{\frac {5}{2}} d^{2} x -170 B \,c^{\frac {7}{2}} d x +170 A \,c^{\frac {7}{2}} d -98 B \,c^{\frac {9}{2}}\right )}{768 \left (d x +c \right )^{\frac {7}{2}} \left (-d x +c \right ) d^{2} c^{\frac {9}{2}}}\)	\(414\)

Input:

int((B*x+A)/(d*x+c)^(5/2)/(-d^2*x^2+c^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/768/(d*x+c)^(7/2)*(-d^2*x^2+c^2)^(1/2)*(-490*A*c^(3/2)*d^3*x^2+75*B*(-d 
*x+c)^(1/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4-350*B* 
c^(5/2)*d^2*x^2+75*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+315*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+225*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+315*A*(-d*x+c)^(1/2)*2^(1/2)*a 
rctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^2*x+225*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-150*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-210*A*c^(1/2)*d^4*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))*d^4*x^3-238*A*c^(5/2)*d^2*x-170*B*c^(7/2 
)*d*x+170*A*c^(7/2)*d-98*B*c^(9/2))/(-d*x+c)/d^2/c^(9/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (169) = 338\).

Time = 0.11 (sec) , antiderivative size = 713, normalized size of antiderivative = 3.55 \[ \int \frac {A+B x}{(c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\left [-\frac {15 \, \sqrt {2} {\left (5 \, B c^{6} + 7 \, A c^{5} d - {\left (5 \, B c d^{5} + 7 \, A d^{6}\right )} x^{5} - 3 \, {\left (5 \, B c^{2} d^{4} + 7 \, A c d^{5}\right )} x^{4} - 2 \, {\left (5 \, B c^{3} d^{3} + 7 \, A c^{2} d^{4}\right )} x^{3} + 2 \, {\left (5 \, B c^{4} d^{2} + 7 \, A c^{3} d^{3}\right )} x^{2} + 3 \, {\left (5 \, 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 (49 \, B c^{5} - 85 \, A c^{4} d + 15 \, {\left (5 \, B c^{2} d^{3} + 7 \, A c d^{4}\right )} x^{3} + 35 \, {\left (5 \, B c^{3} d^{2} + 7 \, A c^{2} d^{3}\right )} x^{2} + 17 \, {\left (5 \, B c^{4} d + 7 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{1536 \, {\left (c^{5} d^{7} x^{5} + 3 \, c^{6} d^{6} x^{4} + 2 \, c^{7} d^{5} x^{3} - 2 \, c^{8} d^{4} x^{2} - 3 \, c^{9} d^{3} x - c^{10} d^{2}\right )}}, -\frac {15 \, \sqrt {2} {\left (5 \, B c^{6} + 7 \, A c^{5} d - {\left (5 \, B c d^{5} + 7 \, A d^{6}\right )} x^{5} - 3 \, {\left (5 \, B c^{2} d^{4} + 7 \, A c d^{5}\right )} x^{4} - 2 \, {\left (5 \, B c^{3} d^{3} + 7 \, A c^{2} d^{4}\right )} x^{3} + 2 \, {\left (5 \, B c^{4} d^{2} + 7 \, A c^{3} d^{3}\right )} x^{2} + 3 \, {\left (5 \, 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 (49 \, B c^{5} - 85 \, A c^{4} d + 15 \, {\left (5 \, B c^{2} d^{3} + 7 \, A c d^{4}\right )} x^{3} + 35 \, {\left (5 \, B c^{3} d^{2} + 7 \, A c^{2} d^{3}\right )} x^{2} + 17 \, {\left (5 \, 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} + 3 \, c^{6} d^{6} x^{4} + 2 \, c^{7} d^{5} x^{3} - 2 \, c^{8} d^{4} x^{2} - 3 \, c^{9} d^{3} x - c^{10} d^{2}\right )}}\right ] \] Input:

integrate((B*x+A)/(d*x+c)^(5/2)/(-d^2*x^2+c^2)^(3/2),x, algorithm="fricas" 
)
 

Output:

[-1/1536*(15*sqrt(2)*(5*B*c^6 + 7*A*c^5*d - (5*B*c*d^5 + 7*A*d^6)*x^5 - 3* 
(5*B*c^2*d^4 + 7*A*c*d^5)*x^4 - 2*(5*B*c^3*d^3 + 7*A*c^2*d^4)*x^3 + 2*(5*B 
*c^4*d^2 + 7*A*c^3*d^3)*x^2 + 3*(5*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*(49*B*c^5 - 85*A*c^4*d + 15*(5*B*c 
^2*d^3 + 7*A*c*d^4)*x^3 + 35*(5*B*c^3*d^2 + 7*A*c^2*d^3)*x^2 + 17*(5*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 + 3* 
c^6*d^6*x^4 + 2*c^7*d^5*x^3 - 2*c^8*d^4*x^2 - 3*c^9*d^3*x - c^10*d^2), -1/ 
768*(15*sqrt(2)*(5*B*c^6 + 7*A*c^5*d - (5*B*c*d^5 + 7*A*d^6)*x^5 - 3*(5*B* 
c^2*d^4 + 7*A*c*d^5)*x^4 - 2*(5*B*c^3*d^3 + 7*A*c^2*d^4)*x^3 + 2*(5*B*c^4* 
d^2 + 7*A*c^3*d^3)*x^2 + 3*(5*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)*sqrt(-c)/(c*d*x + c^2)) + 2*( 
49*B*c^5 - 85*A*c^4*d + 15*(5*B*c^2*d^3 + 7*A*c*d^4)*x^3 + 35*(5*B*c^3*d^2 
 + 7*A*c^2*d^3)*x^2 + 17*(5*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 + 3*c^6*d^6*x^4 + 2*c^7*d^5*x^3 - 2*c^8*d^4*x 
^2 - 3*c^9*d^3*x - c^10*d^2)]
 

Sympy [F]

\[ \int \frac {A+B x}{(c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\int \frac {A + B x}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((B*x+A)/(d*x+c)**(5/2)/(-d**2*x**2+c**2)**(3/2),x)
 

Output:

Integral((A + B*x)/((-(-c + d*x)*(c + d*x))**(3/2)*(c + d*x)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {A+B x}{(c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\int { \frac {B x + A}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((B*x+A)/(d*x+c)^(5/2)/(-d^2*x^2+c^2)^(3/2),x, algorithm="maxima" 
)
 

Output:

integrate((B*x + A)/((-d^2*x^2 + c^2)^(3/2)*(d*x + c)^(5/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x}{(c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\frac {15 \, \sqrt {2} {\left (5 \, B c + 7 \, A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{4} d} + \frac {96 \, {\left (B c + A d\right )}}{\sqrt {-d x + c} c^{4} d} - \frac {2 \, {\left (27 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B c - 112 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c^{2} + 84 \, \sqrt {-d x + c} B c^{3} + 57 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} A d - 272 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c d + 348 \, \sqrt {-d x + c} A c^{2} d\right )}}{{\left (d x + c\right )}^{3} c^{4} d}}{768 \, d} \] Input:

integrate((B*x+A)/(d*x+c)^(5/2)/(-d^2*x^2+c^2)^(3/2),x, algorithm="giac")
 

Output:

1/768*(15*sqrt(2)*(5*B*c + 7*A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(- 
c))/(sqrt(-c)*c^4*d) + 96*(B*c + A*d)/(sqrt(-d*x + c)*c^4*d) - 2*(27*(d*x 
- c)^2*sqrt(-d*x + c)*B*c - 112*(-d*x + c)^(3/2)*B*c^2 + 84*sqrt(-d*x + c) 
*B*c^3 + 57*(d*x - c)^2*sqrt(-d*x + c)*A*d - 272*(-d*x + c)^(3/2)*A*c*d + 
348*sqrt(-d*x + c)*A*c^2*d)/((d*x + c)^3*c^4*d))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{(c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x}{{\left (c^2-d^2\,x^2\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int((A + B*x)/((c^2 - d^2*x^2)^(3/2)*(c + d*x)^(5/2)),x)
 

Output:

int((A + B*x)/((c^2 - d^2*x^2)^(3/2)*(c + d*x)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 676, normalized size of antiderivative = 3.36 \[ \int \frac {A+B x}{(c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int((B*x+A)/(d*x+c)^(5/2)/(-d^2*x^2+c^2)^(3/2),x)
 

Output:

(105*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*a* 
c**3*d + 315*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqr 
t(2))*a*c**2*d**2*x + 315*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 + 75*sqrt(c)*sqrt(c - d*x)*sqr 
t(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c**4 + 225*sqrt(c)*sqrt(c - d* 
x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c**3*d*x + 225*sqrt(c)*s 
qrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c**2*d**2*x**2 
 + 75*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 - 315*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) 
 + sqrt(c)*sqrt(2))*a*c**2*d**2*x - 315*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(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*d**4*x**3 - 75*sqrt(c)*sqr 
t(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**4 - 225*sqrt( 
c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**3*d*x - 
 225*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b* 
c**2*d**2*x**2 - 75*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt 
(c)*sqrt(2))*b*c*d**3*x**3 - 340*a*c**4*d + 476*a*c**3*d**2*x + 980*a*c**2 
*d**3*x**2 + 420*a*c*d**4*x**3 + 196*b*c**5 + 340*b*c**4*d*x + 700*b*c*...