\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{(c+d x)^{9/2}} \, dx\) [87]

Optimal result

Integrand size = 31, antiderivative size = 213 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{9/2}} \, dx=\frac {2 c (9 B c-5 A d) \sqrt {c^2-d^2 x^2}}{d^2 \sqrt {c+d x}}+\frac {(9 B c-5 A d) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^2 (c+d x)^{3/2}}+\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{5/2}}{d^2 (c+d x)^{7/2}}+\frac {2 B \left (c^2-d^2 x^2\right )^{5/2}}{5 d^2 (c+d x)^{5/2}}-\frac {2 \sqrt {2} c^{3/2} (9 B c-5 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{d^2} \] Output:

2*c*(-5*A*d+9*B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(1/2)+1/3*(-5*A*d+9*B* 
c)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)^(3/2)+(-A*d+B*c)*(-d^2*x^2+c^2)^(5/2)/ 
d^2/(d*x+c)^(7/2)+2/5*B*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(5/2)-2*2^(1/2)*c 
^(3/2)*(-5*A*d+9*B*c)*arctanh(2^(1/2)*c^(1/2)*(d*x+c)^(1/2)/(-d^2*x^2+c^2) 
^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{9/2}} \, dx=\frac {2 \left (\frac {\sqrt {c^2-d^2 x^2} \left (5 A d \left (-19 c^2-12 c d x+d^2 x^2\right )+3 B \left (56 c^3+39 c^2 d x-6 c d^2 x^2+d^3 x^3\right )\right )}{(c+d x)^{3/2}}-15 \sqrt {2} c^{3/2} (9 B c-5 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )\right )}{15 d^2} \] Input:

Integrate[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(9/2),x]
 

Output:

(2*((Sqrt[c^2 - d^2*x^2]*(5*A*d*(-19*c^2 - 12*c*d*x + d^2*x^2) + 3*B*(56*c 
^3 + 39*c^2*d*x - 6*c*d^2*x^2 + d^3*x^3)))/(c + d*x)^(3/2) - 15*Sqrt[2]*c^ 
(3/2)*(9*B*c - 5*A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d 
^2*x^2]]))/(15*d^2)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {671, 466, 466, 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.

\(\displaystyle \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{9/2}} \, dx\)

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 471

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

\(\Big \downarrow \) 221

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

Input:

Int[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(9/2),x]
 

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(7/2))/(2*c*d^2*(c + d*x)^(9/2)) + ((9*B*c - 
5*A*d)*((2*(c^2 - d^2*x^2)^(5/2))/(5*d*(c + d*x)^(5/2)) + 2*c*((2*(c^2 - d 
^2*x^2)^(3/2))/(3*d*(c + d*x)^(3/2)) + 2*c*((2*Sqrt[c^2 - d^2*x^2])/(d*Sqr 
t[c + d*x]) - (2*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt 
[c]*Sqrt[c + d*x])])/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 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 = 206, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {2 \left (-3 B \,d^{2} x^{2}-5 A \,d^{2} x +21 B c d x +65 A c d -138 B \,c^{2}\right ) \sqrt {-d x +c}\, \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{15 d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}-\frac {8 c^{2} \left (\frac {\left (-\frac {A d}{2}+\frac {B c}{2}\right ) \sqrt {-d x +c}}{-d x -c}-\frac {\left (5 A d -9 B c \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\right ) \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(206\)
default	\(\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (3 B \,d^{3} x^{3} \sqrt {-d x +c}\, \sqrt {c}+75 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x +5 A \,d^{3} x^{2} \sqrt {-d x +c}\, \sqrt {c}-135 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d x -18 B \,c^{\frac {3}{2}} d^{2} x^{2} \sqrt {-d x +c}+75 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d -60 A \,c^{\frac {3}{2}} d^{2} x \sqrt {-d x +c}-135 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}+117 B \,c^{\frac {5}{2}} d x \sqrt {-d x +c}-95 A \sqrt {-d x +c}\, c^{\frac {5}{2}} d +168 B \sqrt {-d x +c}\, c^{\frac {7}{2}}\right )}{15 \left (d x +c \right )^{\frac {3}{2}} \sqrt {-d x +c}\, d^{2} \sqrt {c}}\)	\(273\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(9/2),x,method=_RETURNVERBOSE)
 

Output:

-2/15*(-3*B*d^2*x^2-5*A*d^2*x+21*B*c*d*x+65*A*c*d-138*B*c^2)*(-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)-8* 
c^2/d^2*((-1/2*A*d+1/2*B*c)*(-d*x+c)^(1/2)/(-d*x-c)-1/4*(5*A*d-9*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 = 0.11 (sec) , antiderivative size = 458, normalized size of antiderivative = 2.15 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{9/2}} \, dx=\left [-\frac {15 \, \sqrt {2} {\left (9 \, B c^{4} - 5 \, A c^{3} d + {\left (9 \, B c^{2} d^{2} - 5 \, A c d^{3}\right )} x^{2} + 2 \, {\left (9 \, B c^{3} d - 5 \, A c^{2} d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x - 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 2 \, {\left (3 \, B d^{3} x^{3} + 168 \, B c^{3} - 95 \, A c^{2} d - {\left (18 \, B c d^{2} - 5 \, A d^{3}\right )} x^{2} + 3 \, {\left (39 \, B c^{2} d - 20 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{15 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}}, \frac {2 \, {\left (15 \, \sqrt {2} {\left (9 \, B c^{4} - 5 \, A c^{3} d + {\left (9 \, B c^{2} d^{2} - 5 \, A c d^{3}\right )} x^{2} + 2 \, {\left (9 \, B c^{3} d - 5 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + {\left (3 \, B d^{3} x^{3} + 168 \, B c^{3} - 95 \, A c^{2} d - {\left (18 \, B c d^{2} - 5 \, A d^{3}\right )} x^{2} + 3 \, {\left (39 \, B c^{2} d - 20 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}\right )}}{15 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}}\right ] \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(9/2),x, algorithm="fricas" 
)
 

Output:

[-1/15*(15*sqrt(2)*(9*B*c^4 - 5*A*c^3*d + (9*B*c^2*d^2 - 5*A*c*d^3)*x^2 + 
2*(9*B*c^3*d - 5*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)) - 2*(3*B*d^3*x^3 + 168*B*c^3 - 95*A*c^2*d - (18*B*c*d^2 - 5*A*d^3)*x 
^2 + 3*(39*B*c^2*d - 20*A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(d 
^4*x^2 + 2*c*d^3*x + c^2*d^2), 2/15*(15*sqrt(2)*(9*B*c^4 - 5*A*c^3*d + (9* 
B*c^2*d^2 - 5*A*c*d^3)*x^2 + 2*(9*B*c^3*d - 5*A*c^2*d^2)*x)*sqrt(-c)*arcta 
n(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + c^2)) + 
 (3*B*d^3*x^3 + 168*B*c^3 - 95*A*c^2*d - (18*B*c*d^2 - 5*A*d^3)*x^2 + 3*(3 
9*B*c^2*d - 20*A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(d^4*x^2 + 
2*c*d^3*x + c^2*d^2)]
 

Sympy [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{9/2}} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{\left (c + d x\right )^{\frac {9}{2}}}\, dx \] Input:

integrate((B*x+A)*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**(9/2),x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(5/2)*(A + B*x)/(c + d*x)**(9/2), x)
 

Maxima [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{9/2}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {9}{2}}} \,d x } \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(9/2),x, algorithm="maxima" 
)
 

Output:

integrate((-d^2*x^2 + c^2)^(5/2)*(B*x + A)/(d*x + c)^(9/2), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.74 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{9/2}} \, dx=\frac {2 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B + 15 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c + 120 \, \sqrt {-d x + c} B c^{2} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d - 60 \, \sqrt {-d x + c} A c d + \frac {15 \, \sqrt {2} {\left (9 \, B c^{3} - 5 \, A c^{2} d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} + \frac {30 \, {\left (\sqrt {-d x + c} B c^{3} - \sqrt {-d x + c} A c^{2} d\right )}}{d x + c}\right )}}{15 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(9/2),x, algorithm="giac")
 

Output:

2/15*(3*(d*x - c)^2*sqrt(-d*x + c)*B + 15*(-d*x + c)^(3/2)*B*c + 120*sqrt( 
-d*x + c)*B*c^2 - 5*(-d*x + c)^(3/2)*A*d - 60*sqrt(-d*x + c)*A*c*d + 15*sq 
rt(2)*(9*B*c^3 - 5*A*c^2*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c))/sq 
rt(-c) + 30*(sqrt(-d*x + c)*B*c^3 - sqrt(-d*x + c)*A*c^2*d)/(d*x + c))/d^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{9/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^{9/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(9/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(9/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.36 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{9/2}} \, dx=\frac {-380 \sqrt {-d x +c}\, a \,c^{2} d -240 \sqrt {-d x +c}\, a c \,d^{2} x +20 \sqrt {-d x +c}\, a \,d^{3} x^{2}+672 \sqrt {-d x +c}\, b \,c^{3}+468 \sqrt {-d x +c}\, b \,c^{2} d x -72 \sqrt {-d x +c}\, b c \,d^{2} x^{2}+12 \sqrt {-d x +c}\, b \,d^{3} x^{3}-300 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a \,c^{2} d -300 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a c \,d^{2} x +540 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{3}+540 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{2} d x +355 \sqrt {c}\, \sqrt {2}\, a \,c^{2} d +355 \sqrt {c}\, \sqrt {2}\, a c \,d^{2} x -723 \sqrt {c}\, \sqrt {2}\, b \,c^{3}-723 \sqrt {c}\, \sqrt {2}\, b \,c^{2} d x}{30 d^{2} \left (d x +c \right )} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(9/2),x)
 

Output:

( - 380*sqrt(c - d*x)*a*c**2*d - 240*sqrt(c - d*x)*a*c*d**2*x + 20*sqrt(c 
- d*x)*a*d**3*x**2 + 672*sqrt(c - d*x)*b*c**3 + 468*sqrt(c - d*x)*b*c**2*d 
*x - 72*sqrt(c - d*x)*b*c*d**2*x**2 + 12*sqrt(c - d*x)*b*d**3*x**3 - 300*s 
qrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**2*d 
- 300*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a* 
c*d**2*x + 540*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2) 
))/2))*b*c**3 + 540*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sq 
rt(2)))/2))*b*c**2*d*x + 355*sqrt(c)*sqrt(2)*a*c**2*d + 355*sqrt(c)*sqrt(2 
)*a*c*d**2*x - 723*sqrt(c)*sqrt(2)*b*c**3 - 723*sqrt(c)*sqrt(2)*b*c**2*d*x 
)/(30*d**2*(c + d*x))