\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{(c+d x)^{19/2}} \, dx\) [92]

Optimal result

Integrand size = 31, antiderivative size = 318 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{19/2}} \, dx=\frac {(109 B c-5 A d) \sqrt {c^2-d^2 x^2}}{160 d^2 (c+d x)^{9/2}}-\frac {(749 B c-5 A d) \sqrt {c^2-d^2 x^2}}{1920 c d^2 (c+d x)^{7/2}}+\frac {(19 B c+5 A d) \sqrt {c^2-d^2 x^2}}{3072 c^2 d^2 (c+d x)^{5/2}}+\frac {(19 B c+5 A d) \sqrt {c^2-d^2 x^2}}{4096 c^3 d^2 (c+d x)^{3/2}}-\frac {(29 B c-5 A d) \left (c^2-d^2 x^2\right )^{3/2}}{60 d^2 (c+d x)^{13/2}}+\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{5/2}}{6 d^2 (c+d x)^{17/2}}+\frac {(19 B c+5 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{4096 \sqrt {2} c^{7/2} d^2} \] Output:

1/160*(-5*A*d+109*B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(9/2)-1/1920*(-5*A 
*d+749*B*c)*(-d^2*x^2+c^2)^(1/2)/c/d^2/(d*x+c)^(7/2)+1/3072*(5*A*d+19*B*c) 
*(-d^2*x^2+c^2)^(1/2)/c^2/d^2/(d*x+c)^(5/2)+1/4096*(5*A*d+19*B*c)*(-d^2*x^ 
2+c^2)^(1/2)/c^3/d^2/(d*x+c)^(3/2)-1/60*(-5*A*d+29*B*c)*(-d^2*x^2+c^2)^(3/ 
2)/d^2/(d*x+c)^(13/2)+1/6*(-A*d+B*c)*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(17/ 
2)+1/8192*(5*A*d+19*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^(7/2)/d^2
 

Mathematica [A] (verified)

Time = 4.30 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.66 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{19/2}} \, dx=\frac {-\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (B c \left (903 c^5+5727 c^4 d x-15018 c^3 d^2 x^2+19598 c^2 d^3 x^3-1805 c d^4 x^4-285 d^5 x^5\right )-5 A d \left (-1341 c^5+3579 c^4 d x-3090 c^3 d^2 x^2+262 c^2 d^3 x^3+95 c d^4 x^4+15 d^5 x^5\right )\right )}{(c+d x)^{13/2}}+15 \sqrt {2} (19 B c+5 A d) \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{122880 c^{7/2} d^2} \] Input:

Integrate[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(19/2),x]
 

Output:

((-2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(B*c*(903*c^5 + 5727*c^4*d*x - 15018*c^3* 
d^2*x^2 + 19598*c^2*d^3*x^3 - 1805*c*d^4*x^4 - 285*d^5*x^5) - 5*A*d*(-1341 
*c^5 + 3579*c^4*d*x - 3090*c^3*d^2*x^2 + 262*c^2*d^3*x^3 + 95*c*d^4*x^4 + 
15*d^5*x^5)))/(c + d*x)^(13/2) + 15*Sqrt[2]*(19*B*c + 5*A*d)*ArcTanh[Sqrt[ 
c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])])/(122880*c^(7/2)*d^2)
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 307, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {671, 465, 465, 465, 470, 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.

\(\displaystyle \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{19/2}} \, dx\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {(5 A d+19 B c) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{17/2}}dx}{24 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{12 c d^2 (c+d x)^{19/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(5 A d+19 B c) \left (-\frac {1}{2} \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{13/2}}dx-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^{15/2}}\right )}{24 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{12 c d^2 (c+d x)^{19/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(5 A d+19 B c) \left (\frac {1}{2} \left (\frac {3}{8} \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}}dx+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^{15/2}}\right )}{24 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{12 c d^2 (c+d x)^{19/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(5 A d+19 B c) \left (\frac {1}{2} \left (\frac {3}{8} \left (-\frac {1}{6} \int \frac {1}{(c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}dx-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^{15/2}}\right )}{24 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{12 c d^2 (c+d x)^{19/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(5 A d+19 B c) \left (\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \int \frac {1}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}dx}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^{15/2}}\right )}{24 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{12 c d^2 (c+d x)^{19/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(5 A d+19 B c) \left (\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{4 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^{15/2}}\right )}{24 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{12 c d^2 (c+d x)^{19/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(5 A d+19 B c) \left (\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{2 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^{15/2}}\right )}{24 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{12 c d^2 (c+d x)^{19/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(5 A d+19 B c) \left (\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{2 \sqrt {2} c^{3/2} d}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^{15/2}}\right )}{24 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{12 c d^2 (c+d x)^{19/2}}\)

Input:

Int[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(19/2),x]
 

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(7/2))/(12*c*d^2*(c + d*x)^(19/2)) + ((19*B*c 
 + 5*A*d)*(-1/5*(c^2 - d^2*x^2)^(5/2)/(d*(c + d*x)^(15/2)) + ((c^2 - d^2*x 
^2)^(3/2)/(4*d*(c + d*x)^(11/2)) + (3*(-1/3*Sqrt[c^2 - d^2*x^2]/(d*(c + d* 
x)^(7/2)) + (Sqrt[c^2 - d^2*x^2]/(4*c*d*(c + d*x)^(5/2)) - (3*(-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)))/(8*c))/6))/8)/2))/(24*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. \(685\) vs. \(2(268)=536\).

Time = 0.38 (sec) , antiderivative size = 686, normalized size of antiderivative = 2.16

method	result	size
default	\(\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (-1806 B \sqrt {-d x +c}\, c^{\frac {13}{2}}+285 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{7}-39196 B \,c^{\frac {7}{2}} d^{3} x^{3} \sqrt {-d x +c}+4275 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5} d^{2} x^{2}+450 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5} d^{2} x +1710 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{6} d x +1500 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{4} x^{3}+5700 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d^{3} x^{3}+1125 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d^{3} x^{2}+4275 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{4} x^{4}+1125 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{5} x^{4}+570 B \,c^{\frac {3}{2}} d^{5} x^{5} \sqrt {-d x +c}+950 A \,c^{\frac {3}{2}} d^{5} x^{4} \sqrt {-d x +c}+150 A \,d^{6} x^{5} \sqrt {-d x +c}\, \sqrt {c}+75 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{7} x^{6}-11454 B \,c^{\frac {11}{2}} d x \sqrt {-d x +c}-30900 A \,c^{\frac {7}{2}} d^{3} x^{2} \sqrt {-d x +c}+30036 B \,c^{\frac {9}{2}} d^{2} x^{2} \sqrt {-d x +c}+35790 A \,c^{\frac {9}{2}} d^{2} x \sqrt {-d x +c}+3610 B \,c^{\frac {5}{2}} d^{4} x^{4} \sqrt {-d x +c}+2620 A \,c^{\frac {5}{2}} d^{4} x^{3} \sqrt {-d x +c}-13410 A \sqrt {-d x +c}\, c^{\frac {11}{2}} d +75 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{6} d +285 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{6} x^{6}+450 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{6} x^{5}+1710 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{5} x^{5}\right )}{122880 c^{\frac {7}{2}} \left (d x +c \right )^{\frac {13}{2}} \sqrt {-d x +c}\, d^{2}}\)	\(686\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(19/2),x,method=_RETURNVERBOSE)
 

Output:

1/122880*(-d^2*x^2+c^2)^(1/2)/c^(7/2)*(-1806*B*(-d*x+c)^(1/2)*c^(13/2)+285 
*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^7-39196*B*c^(7/2) 
*d^3*x^3*(-d*x+c)^(1/2)+4275*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/ 
c^(1/2))*c^5*d^2*x^2+450*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1 
/2))*c^5*d^2*x+1710*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))* 
c^6*d*x+1500*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d^4 
*x^3+5700*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4*d^3*x^ 
3+1125*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4*d^3*x^2+4 
275*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d^4*x^4+1125 
*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^5*x^4+570*B*c 
^(3/2)*d^5*x^5*(-d*x+c)^(1/2)+950*A*c^(3/2)*d^5*x^4*(-d*x+c)^(1/2)+150*A*d 
^6*x^5*(-d*x+c)^(1/2)*c^(1/2)+75*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1 
/2)/c^(1/2))*d^7*x^6-11454*B*c^(11/2)*d*x*(-d*x+c)^(1/2)-30900*A*c^(7/2)*d 
^3*x^2*(-d*x+c)^(1/2)+30036*B*c^(9/2)*d^2*x^2*(-d*x+c)^(1/2)+35790*A*c^(9/ 
2)*d^2*x*(-d*x+c)^(1/2)+3610*B*c^(5/2)*d^4*x^4*(-d*x+c)^(1/2)+2620*A*c^(5/ 
2)*d^4*x^3*(-d*x+c)^(1/2)-13410*A*(-d*x+c)^(1/2)*c^(11/2)*d+75*A*2^(1/2)*a 
rctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^6*d+285*B*2^(1/2)*arctanh(1/2 
*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^6*x^6+450*A*2^(1/2)*arctanh(1/2*(-d*x 
+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^6*x^5+1710*B*2^(1/2)*arctanh(1/2*(-d*x+c)^( 
1/2)*2^(1/2)/c^(1/2))*c^2*d^5*x^5)/(d*x+c)^(13/2)/(-d*x+c)^(1/2)/d^2
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 945, normalized size of antiderivative = 2.97 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{19/2}} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(19/2),x, algorithm="fricas 
")
 

Output:

[1/245760*(15*sqrt(2)*(19*B*c^8 + 5*A*c^7*d + (19*B*c*d^7 + 5*A*d^8)*x^7 + 
 7*(19*B*c^2*d^6 + 5*A*c*d^7)*x^6 + 21*(19*B*c^3*d^5 + 5*A*c^2*d^6)*x^5 + 
35*(19*B*c^4*d^4 + 5*A*c^3*d^5)*x^4 + 35*(19*B*c^5*d^3 + 5*A*c^4*d^4)*x^3 
+ 21*(19*B*c^6*d^2 + 5*A*c^5*d^3)*x^2 + 7*(19*B*c^7*d + 5*A*c^6*d^2)*x)*sq 
rt(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*(903*B*c^7 + 6705*A*c^6 
*d - 15*(19*B*c^2*d^5 + 5*A*c*d^6)*x^5 - 95*(19*B*c^3*d^4 + 5*A*c^2*d^5)*x 
^4 + 2*(9799*B*c^4*d^3 - 655*A*c^3*d^4)*x^3 - 6*(2503*B*c^5*d^2 - 2575*A*c 
^4*d^3)*x^2 + 3*(1909*B*c^6*d - 5965*A*c^5*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sq 
rt(d*x + c))/(c^4*d^9*x^7 + 7*c^5*d^8*x^6 + 21*c^6*d^7*x^5 + 35*c^7*d^6*x^ 
4 + 35*c^8*d^5*x^3 + 21*c^9*d^4*x^2 + 7*c^10*d^3*x + c^11*d^2), -1/122880* 
(15*sqrt(2)*(19*B*c^8 + 5*A*c^7*d + (19*B*c*d^7 + 5*A*d^8)*x^7 + 7*(19*B*c 
^2*d^6 + 5*A*c*d^7)*x^6 + 21*(19*B*c^3*d^5 + 5*A*c^2*d^6)*x^5 + 35*(19*B*c 
^4*d^4 + 5*A*c^3*d^5)*x^4 + 35*(19*B*c^5*d^3 + 5*A*c^4*d^4)*x^3 + 21*(19*B 
*c^6*d^2 + 5*A*c^5*d^3)*x^2 + 7*(19*B*c^7*d + 5*A*c^6*d^2)*x)*sqrt(-c)*arc 
tan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + c^2)) 
 + 2*(903*B*c^7 + 6705*A*c^6*d - 15*(19*B*c^2*d^5 + 5*A*c*d^6)*x^5 - 95*(1 
9*B*c^3*d^4 + 5*A*c^2*d^5)*x^4 + 2*(9799*B*c^4*d^3 - 655*A*c^3*d^4)*x^3 - 
6*(2503*B*c^5*d^2 - 2575*A*c^4*d^3)*x^2 + 3*(1909*B*c^6*d - 5965*A*c^5*d^2 
)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^4*d^9*x^7 + 7*c^5*d^8*x^6 +...
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{19/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x+A)*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**(19/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{19/2}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {19}{2}}} \,d x } \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(19/2),x, algorithm="maxima 
")
 

Output:

integrate((-d^2*x^2 + c^2)^(5/2)*(B*x + A)/(d*x + c)^(19/2), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 299, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{19/2}} \, dx=-\frac {\frac {15 \, \sqrt {2} {\left (19 \, B c + 5 \, A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{3}} - \frac {2 \, {\left (285 \, {\left (d x - c\right )}^{5} \sqrt {-d x + c} B c + 3230 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} B c^{2} - 9528 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} B c^{3} - 30096 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B c^{4} + 25840 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c^{5} - 9120 \, \sqrt {-d x + c} B c^{6} + 75 \, {\left (d x - c\right )}^{5} \sqrt {-d x + c} A d + 850 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} A c d + 3960 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} A c^{2} d - 7920 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} A c^{3} d + 6800 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c^{4} d - 2400 \, \sqrt {-d x + c} A c^{5} d\right )}}{{\left (d x + c\right )}^{6} c^{3}}}{122880 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(19/2),x, algorithm="giac")
 

Output:

-1/122880*(15*sqrt(2)*(19*B*c + 5*A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/s 
qrt(-c))/(sqrt(-c)*c^3) - 2*(285*(d*x - c)^5*sqrt(-d*x + c)*B*c + 3230*(d* 
x - c)^4*sqrt(-d*x + c)*B*c^2 - 9528*(d*x - c)^3*sqrt(-d*x + c)*B*c^3 - 30 
096*(d*x - c)^2*sqrt(-d*x + c)*B*c^4 + 25840*(-d*x + c)^(3/2)*B*c^5 - 9120 
*sqrt(-d*x + c)*B*c^6 + 75*(d*x - c)^5*sqrt(-d*x + c)*A*d + 850*(d*x - c)^ 
4*sqrt(-d*x + c)*A*c*d + 3960*(d*x - c)^3*sqrt(-d*x + c)*A*c^2*d - 7920*(d 
*x - c)^2*sqrt(-d*x + c)*A*c^3*d + 6800*(-d*x + c)^(3/2)*A*c^4*d - 2400*sq 
rt(-d*x + c)*A*c^5*d)/((d*x + c)^6*c^3))/d^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{19/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^{19/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(19/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(19/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 760, normalized size of antiderivative = 2.39 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{19/2}} \, dx =\text {Too large to display} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(19/2),x)
 

Output:

( - 13410*sqrt(c - d*x)*a*c**6*d + 35790*sqrt(c - d*x)*a*c**5*d**2*x - 309 
00*sqrt(c - d*x)*a*c**4*d**3*x**2 + 2620*sqrt(c - d*x)*a*c**3*d**4*x**3 + 
950*sqrt(c - d*x)*a*c**2*d**5*x**4 + 150*sqrt(c - d*x)*a*c*d**6*x**5 - 180 
6*sqrt(c - d*x)*b*c**7 - 11454*sqrt(c - d*x)*b*c**6*d*x + 30036*sqrt(c - d 
*x)*b*c**5*d**2*x**2 - 39196*sqrt(c - d*x)*b*c**4*d**3*x**3 + 3610*sqrt(c 
- d*x)*b*c**3*d**4*x**4 + 570*sqrt(c - d*x)*b*c**2*d**5*x**5 - 75*sqrt(c)* 
sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**6*d - 450*s 
qrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**5*d* 
*2*x - 1125*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/ 
2))*a*c**4*d**3*x**2 - 1500*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sq 
rt(c)*sqrt(2)))/2))*a*c**3*d**4*x**3 - 1125*sqrt(c)*sqrt(2)*log(tan(asin(s 
qrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**2*d**5*x**4 - 450*sqrt(c)*sqrt(2) 
*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c*d**6*x**5 - 75*sqrt 
(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*d**7*x**6 
- 285*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b* 
c**7 - 1710*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/ 
2))*b*c**6*d*x - 4275*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)* 
sqrt(2)))/2))*b*c**5*d**2*x**2 - 5700*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c 
+ d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**4*d**3*x**3 - 4275*sqrt(c)*sqrt(2)*log( 
tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**3*d**4*x**4 - 1710*s...