\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{(c+d x)^4} \, dx\) [32]

Optimal result

Integrand size = 29, antiderivative size = 170 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^4} \, dx=\frac {5 (4 B c-3 A d) x \sqrt {c^2-d^2 x^2}}{2 d}-\frac {B \left (c^2-d^2 x^2\right )^{3/2}}{3 d^2}+\frac {2 (7 B c-5 A d) \left (c^2-d^2 x^2\right )^{3/2}}{d^2 (c+d x)}+\frac {2 (B c-A d) \left (c^2-d^2 x^2\right )^{5/2}}{d^2 (c+d x)^3}+\frac {5 c^2 (4 B c-3 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^2} \] Output:

5/2*(-3*A*d+4*B*c)*x*(-d^2*x^2+c^2)^(1/2)/d-1/3*B*(-d^2*x^2+c^2)^(3/2)/d^2 
+2*(-5*A*d+7*B*c)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)+2*(-A*d+B*c)*(-d^2*x^2+ 
c^2)^(5/2)/d^2/(d*x+c)^3+5/2*c^2*(-3*A*d+4*B*c)*arctan(d*x/(-d^2*x^2+c^2)^ 
(1/2))/d^2
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^4} \, dx=\frac {\frac {d^3 \sqrt {c^2-d^2 x^2} \left (3 A d \left (-24 c^2-7 c d x+d^2 x^2\right )+2 B \left (47 c^3+17 c^2 d x-5 c d^2 x^2+d^3 x^3\right )\right )}{c+d x}+15 c^2 \left (-d^2\right )^{3/2} (-4 B c+3 A d) \log \left (-\sqrt {-d^2} x+\sqrt {c^2-d^2 x^2}\right )}{6 d^5} \] Input:

Integrate[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^4,x]
 

Output:

((d^3*Sqrt[c^2 - d^2*x^2]*(3*A*d*(-24*c^2 - 7*c*d*x + d^2*x^2) + 2*B*(47*c 
^3 + 17*c^2*d*x - 5*c*d^2*x^2 + d^3*x^3)))/(c + d*x) + 15*c^2*(-d^2)^(3/2) 
*(-4*B*c + 3*A*d)*Log[-(Sqrt[-d^2]*x) + Sqrt[c^2 - d^2*x^2]])/(6*d^5)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {671, 465, 466, 211, 224, 216}

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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 465

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

Int[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^4,x]
 

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(7/2))/(c*d^2*(c + d*x)^4) + ((4*B*c - 3*A*d) 
*((2*(c^2 - d^2*x^2)^(5/2))/(d*(c + d*x)^2) + 5*((c^2 - d^2*x^2)^(3/2)/(3* 
d) + c*((x*Sqrt[c^2 - d^2*x^2])/2 + (c^2*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]] 
)/(2*d)))))/(c*d)
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

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 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 = 0.49 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.04

method	result	size
risch	\(-\frac {\left (-2 B \,d^{2} x^{2}-3 A \,d^{2} x +12 B c d x +24 A c d -46 B \,c^{2}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{6 d^{2}}-\frac {c^{2} \left (\frac {15 A d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}-\frac {20 B c \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}+\frac {16 \left (A d -B c \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{d^{2} \left (x +\frac {c}{d}\right )}\right )}{2 d}\)	\(176\)
default	\(\frac {B \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{c d \left (x +\frac {c}{d}\right )^{3}}+\frac {4 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{3 c d \left (x +\frac {c}{d}\right )^{2}}+\frac {5 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{5}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{8 d^{2}}+\frac {3 c^{2} \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )\right )}{3 c}\right )}{c}\right )}{d^{4}}+\frac {\left (A d -B c \right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{c d \left (x +\frac {c}{d}\right )^{4}}-\frac {3 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{c d \left (x +\frac {c}{d}\right )^{3}}+\frac {4 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{3 c d \left (x +\frac {c}{d}\right )^{2}}+\frac {5 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{5}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{8 d^{2}}+\frac {3 c^{2} \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )\right )}{3 c}\right )}{c}\right )}{c}\right )}{d^{5}}\)	\(651\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^4,x,method=_RETURNVERBOSE)
 

Output:

-1/6*(-2*B*d^2*x^2-3*A*d^2*x+12*B*c*d*x+24*A*c*d-46*B*c^2)/d^2*(-d^2*x^2+c 
^2)^(1/2)-1/2*c^2/d*(15*A*d/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2 
)^(1/2))-20*B*c/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))+16* 
(A*d-B*c)/d^2/(x+c/d)*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^4} \, dx=\frac {94 \, B c^{4} - 72 \, A c^{3} d + 2 \, {\left (47 \, B c^{3} d - 36 \, A c^{2} d^{2}\right )} x - 30 \, {\left (4 \, B c^{4} - 3 \, A c^{3} d + {\left (4 \, B c^{3} d - 3 \, A c^{2} d^{2}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (2 \, B d^{3} x^{3} + 94 \, B c^{3} - 72 \, A c^{2} d - {\left (10 \, B c d^{2} - 3 \, A d^{3}\right )} x^{2} + {\left (34 \, B c^{2} d - 21 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{6 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^4,x, algorithm="fricas")
 

Output:

1/6*(94*B*c^4 - 72*A*c^3*d + 2*(47*B*c^3*d - 36*A*c^2*d^2)*x - 30*(4*B*c^4 
 - 3*A*c^3*d + (4*B*c^3*d - 3*A*c^2*d^2)*x)*arctan(-(c - sqrt(-d^2*x^2 + c 
^2))/(d*x)) + (2*B*d^3*x^3 + 94*B*c^3 - 72*A*c^2*d - (10*B*c*d^2 - 3*A*d^3 
)*x^2 + (34*B*c^2*d - 21*A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2))/(d^3*x + c*d^2)
 

Sympy [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^4} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{\left (c + d x\right )^{4}}\, dx \] Input:

integrate((B*x+A)*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**4,x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(5/2)*(A + B*x)/(c + d*x)**4, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (154) = 308\).

Time = 0.13 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.23 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^4} \, dx=-\frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B c}{2 \, {\left (d^{5} x^{3} + 3 \, c d^{4} x^{2} + 3 \, c^{2} d^{3} x + c^{3} d^{2}\right )}} - \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c^{2}}{2 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} + \frac {15 \, \sqrt {-d^{2} x^{2} + c^{2}} B c^{3}}{d^{3} x + c d^{2}} + \frac {10 \, B c^{3} \arcsin \left (\frac {d x}{c}\right )}{d^{2}} - \frac {15 \, A c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} A}{2 \, {\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )}} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B}{3 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} + \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A c}{2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} + \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c}{6 \, {\left (d^{3} x + c d^{2}\right )}} - \frac {15 \, \sqrt {-d^{2} x^{2} + c^{2}} A c^{2}}{d^{2} x + c d} + \frac {5 \, \sqrt {-d^{2} x^{2} + c^{2}} B c^{2}}{2 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^4,x, algorithm="maxima")
 

Output:

-1/2*(-d^2*x^2 + c^2)^(5/2)*B*c/(d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3 
*d^2) - 5/2*(-d^2*x^2 + c^2)^(3/2)*B*c^2/(d^4*x^2 + 2*c*d^3*x + c^2*d^2) + 
 15*sqrt(-d^2*x^2 + c^2)*B*c^3/(d^3*x + c*d^2) + 10*B*c^3*arcsin(d*x/c)/d^ 
2 - 15/2*A*c^2*arcsin(d*x/c)/d + 1/2*(-d^2*x^2 + c^2)^(5/2)*A/(d^4*x^3 + 3 
*c*d^3*x^2 + 3*c^2*d^2*x + c^3*d) + 1/3*(-d^2*x^2 + c^2)^(5/2)*B/(d^4*x^2 
+ 2*c*d^3*x + c^2*d^2) + 5/2*(-d^2*x^2 + c^2)^(3/2)*A*c/(d^3*x^2 + 2*c*d^2 
*x + c^2*d) + 5/6*(-d^2*x^2 + c^2)^(3/2)*B*c/(d^3*x + c*d^2) - 15*sqrt(-d^ 
2*x^2 + c^2)*A*c^2/(d^2*x + c*d) + 5/2*sqrt(-d^2*x^2 + c^2)*B*c^2/d^2
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^4} \, dx=\frac {5 \, {\left (4 \, B c^{3} - 3 \, A c^{2} d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{2 \, d {\left | d \right |}} + \frac {1}{6} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, B x - \frac {3 \, {\left (4 \, B c d^{3} - A d^{4}\right )}}{d^{4}}\right )} x + \frac {2 \, {\left (23 \, B c^{2} d^{2} - 12 \, A c d^{3}\right )}}{d^{4}}\right )} - \frac {16 \, {\left (B c^{3} - A c^{2} d\right )}}{d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )} {\left | d \right |}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^4,x, algorithm="giac")
 

Output:

5/2*(4*B*c^3 - 3*A*c^2*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) + 1/6*sqr 
t(-d^2*x^2 + c^2)*((2*B*x - 3*(4*B*c*d^3 - A*d^4)/d^4)*x + 2*(23*B*c^2*d^2 
 - 12*A*c*d^3)/d^4) - 16*(B*c^3 - A*c^2*d)/(d*((c*d + sqrt(-d^2*x^2 + c^2) 
*abs(d))/(d^2*x) + 1)*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^4} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^4} \,d x \] Input:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^4,x)
 

Output:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.18 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^4} \, dx=\frac {-45 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d +60 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3}+45 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{3} d +45 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d^{2} x -60 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{4}-60 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3} d x +102 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d +21 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{2} x -3 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x^{2}-120 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3}-34 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d x +10 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x^{2}-2 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{3} x^{3}-102 a \,c^{3} d +21 a \,c^{2} d^{2} x +24 a c \,d^{3} x^{2}-3 a \,d^{4} x^{3}+120 b \,c^{4}-34 b \,c^{3} d x -44 b \,c^{2} d^{2} x^{2}+12 b c \,d^{3} x^{3}-2 b \,d^{4} x^{4}}{6 d^{2} \left (\sqrt {-d^{2} x^{2}+c^{2}}-c -d x \right )} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^4,x)
 

Output:

( - 45*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*a*c**2*d + 60*sqrt(c**2 - d**2 
*x**2)*asin((d*x)/c)*b*c**3 + 45*asin((d*x)/c)*a*c**3*d + 45*asin((d*x)/c) 
*a*c**2*d**2*x - 60*asin((d*x)/c)*b*c**4 - 60*asin((d*x)/c)*b*c**3*d*x + 1 
02*sqrt(c**2 - d**2*x**2)*a*c**2*d + 21*sqrt(c**2 - d**2*x**2)*a*c*d**2*x 
- 3*sqrt(c**2 - d**2*x**2)*a*d**3*x**2 - 120*sqrt(c**2 - d**2*x**2)*b*c**3 
 - 34*sqrt(c**2 - d**2*x**2)*b*c**2*d*x + 10*sqrt(c**2 - d**2*x**2)*b*c*d* 
*2*x**2 - 2*sqrt(c**2 - d**2*x**2)*b*d**3*x**3 - 102*a*c**3*d + 21*a*c**2* 
d**2*x + 24*a*c*d**3*x**2 - 3*a*d**4*x**3 + 120*b*c**4 - 34*b*c**3*d*x - 4 
4*b*c**2*d**2*x**2 + 12*b*c*d**3*x**3 - 2*b*d**4*x**4)/(6*d**2*(sqrt(c**2 
- d**2*x**2) - c - d*x))