\(\int \frac {(A+B x) (c^2-d^2 x^2)^{3/2}}{(c+d x)^{13/2}} \, dx\) [81]

Optimal result

Integrand size = 31, antiderivative size = 233 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{13/2}} \, dx=-\frac {(19 B c-3 A d) \sqrt {c^2-d^2 x^2}}{24 d^2 (c+d x)^{7/2}}+\frac {(115 B c-3 A d) \sqrt {c^2-d^2 x^2}}{192 c d^2 (c+d x)^{5/2}}-\frac {(13 B c+3 A d) \sqrt {c^2-d^2 x^2}}{256 c^2 d^2 (c+d x)^{3/2}}+\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{3/2}}{4 d^2 (c+d x)^{11/2}}-\frac {(13 B c+3 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{256 \sqrt {2} c^{5/2} d^2} \] Output:

-1/24*(-3*A*d+19*B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(7/2)+1/192*(-3*A*d 
+115*B*c)*(-d^2*x^2+c^2)^(1/2)/c/d^2/(d*x+c)^(5/2)-1/256*(3*A*d+13*B*c)*(- 
d^2*x^2+c^2)^(1/2)/c^2/d^2/(d*x+c)^(3/2)+1/4*(-A*d+B*c)*(-d^2*x^2+c^2)^(3/ 
2)/d^2/(d*x+c)^(11/2)-1/512*(3*A*d+13*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 = 2.35 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{13/2}} \, dx=\frac {\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (B c \left (5 c^3+3 c^2 d x+343 c d^2 x^2-39 d^3 x^3\right )-3 A d \left (39 c^3-79 c^2 d x+13 c d^2 x^2+3 d^3 x^3\right )\right )}{(c+d x)^{9/2}}-3 \sqrt {2} (13 B c+3 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{1536 c^{5/2} d^2} \] Input:

Integrate[((A + B*x)*(c^2 - d^2*x^2)^(3/2))/(c + d*x)^(13/2),x]
 

Output:

((2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(B*c*(5*c^3 + 3*c^2*d*x + 343*c*d^2*x^2 - 
39*d^3*x^3) - 3*A*d*(39*c^3 - 79*c^2*d*x + 13*c*d^2*x^2 + 3*d^3*x^3)))/(c 
+ d*x)^(9/2) - 3*Sqrt[2]*(13*B*c + 3*A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c 
+ d*x])/Sqrt[c^2 - d^2*x^2]])/(1536*c^(5/2)*d^2)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {671, 465, 465, 470, 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.

\(\displaystyle \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{13/2}} \, dx\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {(3 A d+13 B c) \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{11/2}}dx}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{5/2} (B c-A d)}{8 c d^2 (c+d x)^{13/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(3 A d+13 B c) \left (-\frac {1}{2} \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{7/2}}dx-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{9/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{5/2} (B c-A d)}{8 c d^2 (c+d x)^{13/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(3 A d+13 B c) \left (\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}dx+\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{9/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{5/2} (B c-A d)}{8 c d^2 (c+d x)^{13/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(3 A d+13 B c) \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{4 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )+\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{9/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{5/2} (B c-A d)}{8 c d^2 (c+d x)^{13/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(3 A d+13 B c) \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{2 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )+\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{9/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{5/2} (B c-A d)}{8 c d^2 (c+d x)^{13/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(3 A d+13 B c) \left (\frac {1}{2} \left (\frac {1}{4} \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{2 \sqrt {2} c^{3/2} d}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )+\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{9/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{5/2} (B c-A d)}{8 c d^2 (c+d x)^{13/2}}\)

Input:

Int[((A + B*x)*(c^2 - d^2*x^2)^(3/2))/(c + d*x)^(13/2),x]
 

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(5/2))/(8*c*d^2*(c + d*x)^(13/2)) + ((13*B*c 
+ 3*A*d)*(-1/3*(c^2 - d^2*x^2)^(3/2)/(d*(c + d*x)^(9/2)) + (Sqrt[c^2 - d^2 
*x^2]/(2*d*(c + d*x)^(5/2)) + (-1/2*Sqrt[c^2 - d^2*x^2]/(c*d*(c + d*x)^(3/ 
2)) - ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])]/(2*Sqrt 
[2]*c^(3/2)*d))/4)/2))/(16*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 465

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 + p + 1))), x] - Simp[b*(p/(d^2*(n + 
 p + 1)))   Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, 
b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LtQ[n, -2] || EqQ[n 
+ 2*p + 1, 0]) && NeQ[n + p + 1, 0] && 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. \(477\) vs. \(2(195)=390\).

Time = 0.37 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.05

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (9 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{5} x^{4}+39 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{4} x^{4}+36 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{4} x^{3}+156 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{3} x^{3}+54 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{3} x^{2}+18 A \,d^{4} x^{3} \sqrt {-d x +c}\, \sqrt {c}+234 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{2} x^{2}+78 B \,c^{\frac {3}{2}} d^{3} x^{3} \sqrt {-d x +c}+36 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{2} x +78 A \,c^{\frac {3}{2}} d^{3} x^{2} \sqrt {-d x +c}+156 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d x -686 B \,c^{\frac {5}{2}} d^{2} x^{2} \sqrt {-d x +c}+9 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d -474 A \,c^{\frac {5}{2}} d^{2} x \sqrt {-d x +c}+39 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5}-6 B \,c^{\frac {7}{2}} d x \sqrt {-d x +c}+234 A \,c^{\frac {7}{2}} \sqrt {-d x +c}\, d -10 B \,c^{\frac {9}{2}} \sqrt {-d x +c}\right )}{1536 c^{\frac {5}{2}} \left (d x +c \right )^{\frac {9}{2}} \sqrt {-d x +c}\, d^{2}}\)	\(478\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(13/2),x,method=_RETURNVERBOSE)
 

Output:

-1/1536*(-d^2*x^2+c^2)^(1/2)/c^(5/2)*(9*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/ 
2)*2^(1/2)/c^(1/2))*d^5*x^4+39*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2 
)/c^(1/2))*c*d^4*x^4+36*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/ 
2))*c*d^4*x^3+156*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^ 
2*d^3*x^3+54*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^3 
*x^2+18*A*d^4*x^3*(-d*x+c)^(1/2)*c^(1/2)+234*B*2^(1/2)*arctanh(1/2*(-d*x+c 
)^(1/2)*2^(1/2)/c^(1/2))*c^3*d^2*x^2+78*B*c^(3/2)*d^3*x^3*(-d*x+c)^(1/2)+3 
6*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d^2*x+78*A*c^( 
3/2)*d^3*x^2*(-d*x+c)^(1/2)+156*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/ 
2)/c^(1/2))*c^4*d*x-686*B*c^(5/2)*d^2*x^2*(-d*x+c)^(1/2)+9*A*2^(1/2)*arcta 
nh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4*d-474*A*c^(5/2)*d^2*x*(-d*x+c)^ 
(1/2)+39*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^5-6*B*c^( 
7/2)*d*x*(-d*x+c)^(1/2)+234*A*c^(7/2)*(-d*x+c)^(1/2)*d-10*B*c^(9/2)*(-d*x+ 
c)^(1/2))/(d*x+c)^(9/2)/(-d*x+c)^(1/2)/d^2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 705, normalized size of antiderivative = 3.03 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{13/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (13 \, B c^{6} + 3 \, A c^{5} d + {\left (13 \, B c d^{5} + 3 \, A d^{6}\right )} x^{5} + 5 \, {\left (13 \, B c^{2} d^{4} + 3 \, A c d^{5}\right )} x^{4} + 10 \, {\left (13 \, B c^{3} d^{3} + 3 \, A c^{2} d^{4}\right )} x^{3} + 10 \, {\left (13 \, B c^{4} d^{2} + 3 \, A c^{3} d^{3}\right )} x^{2} + 5 \, {\left (13 \, B c^{5} d + 3 \, 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 (5 \, B c^{5} - 117 \, A c^{4} d - 3 \, {\left (13 \, B c^{2} d^{3} + 3 \, A c d^{4}\right )} x^{3} + {\left (343 \, B c^{3} d^{2} - 39 \, A c^{2} d^{3}\right )} x^{2} + 3 \, {\left (B c^{4} d + 79 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{3072 \, {\left (c^{3} d^{7} x^{5} + 5 \, c^{4} d^{6} x^{4} + 10 \, c^{5} d^{5} x^{3} + 10 \, c^{6} d^{4} x^{2} + 5 \, c^{7} d^{3} x + c^{8} d^{2}\right )}}, \frac {3 \, \sqrt {2} {\left (13 \, B c^{6} + 3 \, A c^{5} d + {\left (13 \, B c d^{5} + 3 \, A d^{6}\right )} x^{5} + 5 \, {\left (13 \, B c^{2} d^{4} + 3 \, A c d^{5}\right )} x^{4} + 10 \, {\left (13 \, B c^{3} d^{3} + 3 \, A c^{2} d^{4}\right )} x^{3} + 10 \, {\left (13 \, B c^{4} d^{2} + 3 \, A c^{3} d^{3}\right )} x^{2} + 5 \, {\left (13 \, B c^{5} d + 3 \, 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 (5 \, B c^{5} - 117 \, A c^{4} d - 3 \, {\left (13 \, B c^{2} d^{3} + 3 \, A c d^{4}\right )} x^{3} + {\left (343 \, B c^{3} d^{2} - 39 \, A c^{2} d^{3}\right )} x^{2} + 3 \, {\left (B c^{4} d + 79 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{1536 \, {\left (c^{3} d^{7} x^{5} + 5 \, c^{4} d^{6} x^{4} + 10 \, c^{5} d^{5} x^{3} + 10 \, c^{6} d^{4} x^{2} + 5 \, c^{7} d^{3} x + c^{8} d^{2}\right )}}\right ] \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(13/2),x, algorithm="fricas 
")
 

Output:

[1/3072*(3*sqrt(2)*(13*B*c^6 + 3*A*c^5*d + (13*B*c*d^5 + 3*A*d^6)*x^5 + 5* 
(13*B*c^2*d^4 + 3*A*c*d^5)*x^4 + 10*(13*B*c^3*d^3 + 3*A*c^2*d^4)*x^3 + 10* 
(13*B*c^4*d^2 + 3*A*c^3*d^3)*x^2 + 5*(13*B*c^5*d + 3*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)*sq 
rt(c) - 3*c^2)/(d^2*x^2 + 2*c*d*x + c^2)) + 4*(5*B*c^5 - 117*A*c^4*d - 3*( 
13*B*c^2*d^3 + 3*A*c*d^4)*x^3 + (343*B*c^3*d^2 - 39*A*c^2*d^3)*x^2 + 3*(B* 
c^4*d + 79*A*c^3*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^3*d^7*x^5 
+ 5*c^4*d^6*x^4 + 10*c^5*d^5*x^3 + 10*c^6*d^4*x^2 + 5*c^7*d^3*x + c^8*d^2) 
, 1/1536*(3*sqrt(2)*(13*B*c^6 + 3*A*c^5*d + (13*B*c*d^5 + 3*A*d^6)*x^5 + 5 
*(13*B*c^2*d^4 + 3*A*c*d^5)*x^4 + 10*(13*B*c^3*d^3 + 3*A*c^2*d^4)*x^3 + 10 
*(13*B*c^4*d^2 + 3*A*c^3*d^3)*x^2 + 5*(13*B*c^5*d + 3*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*(5*B*c^5 - 117*A*c^4*d - 3*(13*B*c^2*d^3 + 3*A*c*d^4)*x^3 + (34 
3*B*c^3*d^2 - 39*A*c^2*d^3)*x^2 + 3*(B*c^4*d + 79*A*c^3*d^2)*x)*sqrt(-d^2* 
x^2 + c^2)*sqrt(d*x + c))/(c^3*d^7*x^5 + 5*c^4*d^6*x^4 + 10*c^5*d^5*x^3 + 
10*c^6*d^4*x^2 + 5*c^7*d^3*x + c^8*d^2)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{13/2}} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{\left (c + d x\right )^{\frac {13}{2}}}\, dx \] Input:

integrate((B*x+A)*(-d**2*x**2+c**2)**(3/2)/(d*x+c)**(13/2),x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(3/2)*(A + B*x)/(c + d*x)**(13/2), x)
 

Maxima [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{13/2}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {13}{2}}} \,d x } \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(13/2),x, algorithm="maxima 
")
 

Output:

integrate((-d^2*x^2 + c^2)^(3/2)*(B*x + A)/(d*x + c)^(13/2), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{13/2}} \, dx=\frac {\frac {3 \, \sqrt {2} {\left (13 \, B c + 3 \, A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {2 \, {\left (39 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} B c - 226 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B c^{2} + 572 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c^{3} - 312 \, \sqrt {-d x + c} B c^{4} + 9 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} A d + 66 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} A c d + 132 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c^{2} d - 72 \, \sqrt {-d x + c} A c^{3} d\right )}}{{\left (d x + c\right )}^{4} c^{2}}}{1536 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(13/2),x, algorithm="giac")
 

Output:

1/1536*(3*sqrt(2)*(13*B*c + 3*A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt( 
-c))/(sqrt(-c)*c^2) - 2*(39*(d*x - c)^3*sqrt(-d*x + c)*B*c - 226*(d*x - c) 
^2*sqrt(-d*x + c)*B*c^2 + 572*(-d*x + c)^(3/2)*B*c^3 - 312*sqrt(-d*x + c)* 
B*c^4 + 9*(d*x - c)^3*sqrt(-d*x + c)*A*d + 66*(d*x - c)^2*sqrt(-d*x + c)*A 
*c*d + 132*(-d*x + c)^(3/2)*A*c^2*d - 72*sqrt(-d*x + c)*A*c^3*d)/((d*x + c 
)^4*c^2))/d^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{13/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{3/2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^{13/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x))/(c + d*x)^(13/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x))/(c + d*x)^(13/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.22 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{13/2}} \, dx =\text {Too large to display} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(13/2),x)
 

Output:

( - 234*sqrt(c - d*x)*a*c**4*d + 474*sqrt(c - d*x)*a*c**3*d**2*x - 78*sqrt 
(c - d*x)*a*c**2*d**3*x**2 - 18*sqrt(c - d*x)*a*c*d**4*x**3 + 10*sqrt(c - 
d*x)*b*c**5 + 6*sqrt(c - d*x)*b*c**4*d*x + 686*sqrt(c - d*x)*b*c**3*d**2*x 
**2 - 78*sqrt(c - d*x)*b*c**2*d**3*x**3 + 9*sqrt(c)*sqrt(2)*log(tan(asin(s 
qrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**4*d + 36*sqrt(c)*sqrt(2)*log(tan( 
asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**3*d**2*x + 54*sqrt(c)*sqrt( 
2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**2*d**3*x**2 + 36 
*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c*d** 
4*x**3 + 9*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2 
))*a*d**5*x**4 + 39*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sq 
rt(2)))/2))*b*c**5 + 156*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt( 
c)*sqrt(2)))/2))*b*c**4*d*x + 234*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d* 
x)/(sqrt(c)*sqrt(2)))/2))*b*c**3*d**2*x**2 + 156*sqrt(c)*sqrt(2)*log(tan(a 
sin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**2*d**3*x**3 + 39*sqrt(c)*sqr 
t(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c*d**4*x**4)/(153 
6*c**3*d**2*(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x 
**4))