\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{(c+d x)^{13/2}} \, dx\) [89]

Optimal result

Integrand size = 31, antiderivative size = 217 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{13/2}} \, dx=\frac {(33 B c-5 A d) \sqrt {c^2-d^2 x^2}}{8 d^2 (c+d x)^{3/2}}+\frac {2 B \sqrt {c^2-d^2 x^2}}{d^2 \sqrt {c+d x}}-\frac {(17 B c-5 A d) \left (c^2-d^2 x^2\right )^{3/2}}{12 d^2 (c+d x)^{7/2}}+\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{5/2}}{3 d^2 (c+d x)^{11/2}}-\frac {5 (13 B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{8 \sqrt {2} \sqrt {c} d^2} \] Output:

1/8*(-5*A*d+33*B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(3/2)+2*B*(-d^2*x^2+c 
^2)^(1/2)/d^2/(d*x+c)^(1/2)-1/12*(-5*A*d+17*B*c)*(-d^2*x^2+c^2)^(3/2)/d^2/ 
(d*x+c)^(7/2)+1/3*(-A*d+B*c)*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(11/2)-5/16* 
(-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^(1/2)/d^2
 

Mathematica [A] (verified)

Time = 2.24 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{13/2}} \, dx=\frac {\frac {2 \sqrt {c^2-d^2 x^2} \left (-A d \left (13 c^2+14 c d x+33 d^2 x^2\right )+B \left (121 c^3+326 c^2 d x+285 c d^2 x^2+48 d^3 x^3\right )\right )}{(c+d x)^{7/2}}-\frac {15 \sqrt {2} (13 B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {c}}}{48 d^2} \] Input:

Integrate[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(13/2),x]
 

Output:

((2*Sqrt[c^2 - d^2*x^2]*(-(A*d*(13*c^2 + 14*c*d*x + 33*d^2*x^2)) + B*(121* 
c^3 + 326*c^2*d*x + 285*c*d^2*x^2 + 48*d^3*x^3)))/(c + d*x)^(7/2) - (15*Sq 
rt[2]*(13*B*c - A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^ 
2*x^2]])/Sqrt[c])/(48*d^2)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, 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, 466, 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.

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {(13 B c-A d) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{11/2}}dx}{12 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{6 c d^2 (c+d x)^{13/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(13 B c-A d) \left (-\frac {5}{4} \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{7/2}}dx-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{2 d (c+d x)^{9/2}}\right )}{12 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{6 c d^2 (c+d x)^{13/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(13 B c-A d) \left (-\frac {5}{4} \left (-\frac {3}{2} \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{3/2}}dx-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{2 d (c+d x)^{9/2}}\right )}{12 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{6 c d^2 (c+d x)^{13/2}}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {(13 B c-A d) \left (-\frac {5}{4} \left (-\frac {3}{2} \left (2 c \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{2 d (c+d x)^{9/2}}\right )}{12 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{6 c d^2 (c+d x)^{13/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(13 B c-A d) \left (-\frac {5}{4} \left (-\frac {3}{2} \left (4 c d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{2 d (c+d x)^{9/2}}\right )}{12 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{6 c d^2 (c+d x)^{13/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(13 B c-A d) \left (-\frac {5}{4} \left (-\frac {3}{2} \left (\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}-\frac {2 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{d}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{2 d (c+d x)^{9/2}}\right )}{12 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{6 c d^2 (c+d x)^{13/2}}\)

Input:

Int[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(13/2),x]
 

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(7/2))/(6*c*d^2*(c + d*x)^(13/2)) + ((13*B*c 
- A*d)*(-1/2*(c^2 - d^2*x^2)^(5/2)/(d*(c + d*x)^(9/2)) - (5*(-((c^2 - d^2* 
x^2)^(3/2)/(d*(c + d*x)^(5/2))) - (3*((2*Sqrt[c^2 - d^2*x^2])/(d*Sqrt[c + 
d*x]) - (2*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sq 
rt[c + d*x])])/d))/2))/4))/(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 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 466

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 + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 
2*(n + 2*p + 1)))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr 
eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 
] || EqQ[n + p + 1, 0]) && NeQ[n + 2*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 [A] (verified)

Time = 0.39 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.99

method	result	size
risch	\(\frac {2 B \sqrt {-d x +c}\, \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}-\frac {2 \left (\frac {\left (-\frac {11 A d}{16}+\frac {47 B c}{16}\right ) \left (-d x +c \right )^{\frac {5}{2}}+\left (\frac {5}{3} A c d -\frac {29}{3} B \,c^{2}\right ) \left (-d x +c \right )^{\frac {3}{2}}+\left (\frac {33}{4} B \,c^{3}-\frac {5}{4} A \,c^{2} d \right ) \sqrt {-d x +c}}{\left (-d x -c \right )^{3}}-\frac {5 \left (A d -13 B c \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}\right ) \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(215\)
default	\(\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (15 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{4} x^{3}-195 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{3} x^{3}+45 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{3} x^{2}-585 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x^{2}+96 B \,d^{3} x^{3} \sqrt {-d x +c}\, \sqrt {c}+45 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x -66 A \,d^{3} x^{2} \sqrt {-d x +c}\, \sqrt {c}-585 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d x +570 B \,c^{\frac {3}{2}} d^{2} x^{2} \sqrt {-d x +c}+15 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d -28 A \,c^{\frac {3}{2}} d^{2} x \sqrt {-d x +c}-195 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}+652 B \,c^{\frac {5}{2}} d x \sqrt {-d x +c}-26 A \sqrt {-d x +c}\, c^{\frac {5}{2}} d +242 B \sqrt {-d x +c}\, c^{\frac {7}{2}}\right )}{48 \left (d x +c \right )^{\frac {7}{2}} \sqrt {-d x +c}\, d^{2} \sqrt {c}}\)	\(394\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(13/2),x,method=_RETURNVERBOSE)
 

Output:

2*B*(-d*x+c)^(1/2)/d^2*((-d^2*x^2+c^2)/(d*x+c))^(1/2)/(-d^2*x^2+c^2)^(1/2) 
*(d*x+c)^(1/2)-2/d^2*(((-11/16*A*d+47/16*B*c)*(-d*x+c)^(5/2)+(5/3*A*c*d-29 
/3*B*c^2)*(-d*x+c)^(3/2)+(33/4*B*c^3-5/4*A*c^2*d)*(-d*x+c)^(1/2))/(-d*x-c) 
^3-5/32*(A*d-13*B*c)*2^(1/2)/c^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^ 
(1/2)))*((-d^2*x^2+c^2)/(d*x+c))^(1/2)/(-d^2*x^2+c^2)^(1/2)*(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 607, normalized size of antiderivative = 2.80 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{13/2}} \, dx=\left [-\frac {15 \, \sqrt {2} {\left (13 \, B c^{5} - A c^{4} d + {\left (13 \, B c d^{4} - A d^{5}\right )} x^{4} + 4 \, {\left (13 \, B c^{2} d^{3} - A c d^{4}\right )} x^{3} + 6 \, {\left (13 \, B c^{3} d^{2} - A c^{2} d^{3}\right )} x^{2} + 4 \, {\left (13 \, B c^{4} d - A c^{3} d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x - 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 4 \, {\left (48 \, B c d^{3} x^{3} + 121 \, B c^{4} - 13 \, A c^{3} d + 3 \, {\left (95 \, B c^{2} d^{2} - 11 \, A c d^{3}\right )} x^{2} + 2 \, {\left (163 \, B c^{3} d - 7 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{96 \, {\left (c d^{6} x^{4} + 4 \, c^{2} d^{5} x^{3} + 6 \, c^{3} d^{4} x^{2} + 4 \, c^{4} d^{3} x + c^{5} d^{2}\right )}}, \frac {15 \, \sqrt {2} {\left (13 \, B c^{5} - A c^{4} d + {\left (13 \, B c d^{4} - A d^{5}\right )} x^{4} + 4 \, {\left (13 \, B c^{2} d^{3} - A c d^{4}\right )} x^{3} + 6 \, {\left (13 \, B c^{3} d^{2} - A c^{2} d^{3}\right )} x^{2} + 4 \, {\left (13 \, B c^{4} d - A c^{3} d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + 2 \, {\left (48 \, B c d^{3} x^{3} + 121 \, B c^{4} - 13 \, A c^{3} d + 3 \, {\left (95 \, B c^{2} d^{2} - 11 \, A c d^{3}\right )} x^{2} + 2 \, {\left (163 \, B c^{3} d - 7 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{48 \, {\left (c d^{6} x^{4} + 4 \, c^{2} d^{5} x^{3} + 6 \, c^{3} d^{4} x^{2} + 4 \, c^{4} d^{3} x + c^{5} d^{2}\right )}}\right ] \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(13/2),x, algorithm="fricas 
")
 

Output:

[-1/96*(15*sqrt(2)*(13*B*c^5 - A*c^4*d + (13*B*c*d^4 - A*d^5)*x^4 + 4*(13* 
B*c^2*d^3 - A*c*d^4)*x^3 + 6*(13*B*c^3*d^2 - A*c^2*d^3)*x^2 + 4*(13*B*c^4* 
d - A*c^3*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*(48 
*B*c*d^3*x^3 + 121*B*c^4 - 13*A*c^3*d + 3*(95*B*c^2*d^2 - 11*A*c*d^3)*x^2 
+ 2*(163*B*c^3*d - 7*A*c^2*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c* 
d^6*x^4 + 4*c^2*d^5*x^3 + 6*c^3*d^4*x^2 + 4*c^4*d^3*x + c^5*d^2), 1/48*(15 
*sqrt(2)*(13*B*c^5 - A*c^4*d + (13*B*c*d^4 - A*d^5)*x^4 + 4*(13*B*c^2*d^3 
- A*c*d^4)*x^3 + 6*(13*B*c^3*d^2 - A*c^2*d^3)*x^2 + 4*(13*B*c^4*d - A*c^3* 
d^2)*x)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqr 
t(-c)/(c*d*x + c^2)) + 2*(48*B*c*d^3*x^3 + 121*B*c^4 - 13*A*c^3*d + 3*(95* 
B*c^2*d^2 - 11*A*c*d^3)*x^2 + 2*(163*B*c^3*d - 7*A*c^2*d^2)*x)*sqrt(-d^2*x 
^2 + c^2)*sqrt(d*x + c))/(c*d^6*x^4 + 4*c^2*d^5*x^3 + 6*c^3*d^4*x^2 + 4*c^ 
4*d^3*x + c^5*d^2)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{13/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x+A)*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**(13/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{13/2}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {13}{2}}} \,d x } \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(13/2),x, algorithm="maxima 
")
 

Output:

integrate((-d^2*x^2 + c^2)^(5/2)*(B*x + A)/(d*x + c)^(13/2), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.75 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{13/2}} \, dx=\frac {\frac {15 \, \sqrt {2} {\left (13 \, B c - A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} + 96 \, \sqrt {-d x + c} B + \frac {2 \, {\left (141 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B c - 464 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c^{2} + 396 \, \sqrt {-d x + c} B c^{3} - 33 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} A d + 80 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c d - 60 \, \sqrt {-d x + c} A c^{2} d\right )}}{{\left (d x + c\right )}^{3}}}{48 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(13/2),x, algorithm="giac")
 

Output:

1/48*(15*sqrt(2)*(13*B*c - A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c) 
)/sqrt(-c) + 96*sqrt(-d*x + c)*B + 2*(141*(d*x - c)^2*sqrt(-d*x + c)*B*c - 
 464*(-d*x + c)^(3/2)*B*c^2 + 396*sqrt(-d*x + c)*B*c^3 - 33*(d*x - c)^2*sq 
rt(-d*x + c)*A*d + 80*(-d*x + c)^(3/2)*A*c*d - 60*sqrt(-d*x + c)*A*c^2*d)/ 
(d*x + c)^3)/d^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{13/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^{13/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(13/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(13/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.39 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{13/2}} \, dx =\text {Too large to display} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(13/2),x)
 

Output:

( - 208*sqrt(c - d*x)*a*c**3*d - 224*sqrt(c - d*x)*a*c**2*d**2*x - 528*sqr 
t(c - d*x)*a*c*d**3*x**2 + 1936*sqrt(c - d*x)*b*c**4 + 5216*sqrt(c - d*x)* 
b*c**3*d*x + 4560*sqrt(c - d*x)*b*c**2*d**2*x**2 + 768*sqrt(c - d*x)*b*c*d 
**3*x**3 - 120*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2) 
))/2))*a*c**3*d - 360*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)* 
sqrt(2)))/2))*a*c**2*d**2*x - 360*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d* 
x)/(sqrt(c)*sqrt(2)))/2))*a*c*d**3*x**2 - 120*sqrt(c)*sqrt(2)*log(tan(asin 
(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*d**4*x**3 + 1560*sqrt(c)*sqrt(2)*l 
og(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**4 + 4680*sqrt(c)*sqr 
t(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**3*d*x + 4680*s 
qrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**2*d* 
*2*x**2 + 1560*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2) 
))/2))*b*c*d**3*x**3 + 45*sqrt(c)*sqrt(2)*a*c**3*d + 135*sqrt(c)*sqrt(2)*a 
*c**2*d**2*x + 135*sqrt(c)*sqrt(2)*a*c*d**3*x**2 + 45*sqrt(c)*sqrt(2)*a*d* 
*4*x**3 - 1005*sqrt(c)*sqrt(2)*b*c**4 - 3015*sqrt(c)*sqrt(2)*b*c**3*d*x - 
3015*sqrt(c)*sqrt(2)*b*c**2*d**2*x**2 - 1005*sqrt(c)*sqrt(2)*b*c*d**3*x**3 
)/(384*c*d**2*(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3))