\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{(c+d x)^{17/2}} \, dx\) [91]

Optimal result

Integrand size = 31, antiderivative size = 274 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{17/2}} \, dx=\frac {(47 B c-3 A d) \sqrt {c^2-d^2 x^2}}{48 d^2 (c+d x)^{7/2}}-\frac {(239 B c-3 A d) \sqrt {c^2-d^2 x^2}}{384 c d^2 (c+d x)^{5/2}}+\frac {(17 B c+3 A d) \sqrt {c^2-d^2 x^2}}{512 c^2 d^2 (c+d x)^{3/2}}-\frac {(5 B c-A d) \left (c^2-d^2 x^2\right )^{3/2}}{8 d^2 (c+d x)^{11/2}}+\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{5/2}}{5 d^2 (c+d x)^{15/2}}+\frac {(17 B c+3 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{512 \sqrt {2} c^{5/2} d^2} \] Output:

1/48*(-3*A*d+47*B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(7/2)-1/384*(-3*A*d+ 
239*B*c)*(-d^2*x^2+c^2)^(1/2)/c/d^2/(d*x+c)^(5/2)+1/512*(3*A*d+17*B*c)*(-d 
^2*x^2+c^2)^(1/2)/c^2/d^2/(d*x+c)^(3/2)-1/8*(-A*d+5*B*c)*(-d^2*x^2+c^2)^(3 
/2)/d^2/(d*x+c)^(11/2)+1/5*(-A*d+B*c)*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(15 
/2)+1/1024*(3*A*d+17*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 = 4.23 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{17/2}} \, dx=\frac {-\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (B c \left (269 c^4+1352 c^3 d x-1046 c^2 d^2 x^2+3760 c d^3 x^3-255 d^4 x^4\right )-3 A d \left (-317 c^4+824 c^3 d x-842 c^2 d^2 x^2+80 c d^3 x^3+15 d^4 x^4\right )\right )}{(c+d x)^{11/2}}+15 \sqrt {2} (17 B c+3 A d) \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{15360 c^{5/2} d^2} \] Input:

Integrate[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(17/2),x]
 

Output:

((-2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(B*c*(269*c^4 + 1352*c^3*d*x - 1046*c^2*d 
^2*x^2 + 3760*c*d^3*x^3 - 255*d^4*x^4) - 3*A*d*(-317*c^4 + 824*c^3*d*x - 8 
42*c^2*d^2*x^2 + 80*c*d^3*x^3 + 15*d^4*x^4)))/(c + d*x)^(11/2) + 15*Sqrt[2 
]*(17*B*c + 3*A*d)*ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d 
*x])])/(15360*c^(5/2)*d^2)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {671, 465, 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.

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {(3 A d+17 B c) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{15/2}}dx}{20 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{10 c d^2 (c+d x)^{17/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(3 A d+17 B c) \left (-\frac {5}{8} \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{11/2}}dx-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{4 d (c+d x)^{13/2}}\right )}{20 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{10 c d^2 (c+d x)^{17/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(3 A d+17 B c) \left (-\frac {5}{8} \left (-\frac {1}{2} \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{7/2}}dx-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{9/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{4 d (c+d x)^{13/2}}\right )}{20 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{10 c d^2 (c+d x)^{17/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(3 A d+17 B c) \left (-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}dx+\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{9/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{4 d (c+d x)^{13/2}}\right )}{20 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{10 c d^2 (c+d x)^{17/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(3 A d+17 B c) \left (-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{4 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )+\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{9/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{4 d (c+d x)^{13/2}}\right )}{20 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{10 c d^2 (c+d x)^{17/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(3 A d+17 B c) \left (-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{2 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )+\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{9/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{4 d (c+d x)^{13/2}}\right )}{20 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{10 c d^2 (c+d x)^{17/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(3 A d+17 B c) \left (-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{2 \sqrt {2} c^{3/2} d}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )+\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{9/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{4 d (c+d x)^{13/2}}\right )}{20 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{10 c d^2 (c+d x)^{17/2}}\)

Input:

Int[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(17/2),x]
 

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(7/2))/(10*c*d^2*(c + d*x)^(17/2)) + ((17*B*c 
 + 3*A*d)*(-1/4*(c^2 - d^2*x^2)^(5/2)/(d*(c + d*x)^(13/2)) - (5*(-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))/8))/(20*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. \(581\) vs. \(2(230)=460\).

Time = 0.37 (sec) , antiderivative size = 582, normalized size of antiderivative = 2.12

method	result	size
default	\(\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (45 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{6} x^{5}+255 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{5} x^{5}+225 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{5} x^{4}+1275 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{4} x^{4}+450 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{4} x^{3}+90 A \,d^{5} x^{4} \sqrt {-d x +c}\, \sqrt {c}+2550 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{3} x^{3}+510 B \,c^{\frac {3}{2}} d^{4} x^{4} \sqrt {-d x +c}+450 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{3} x^{2}+480 A \,c^{\frac {3}{2}} d^{4} x^{3} \sqrt {-d x +c}+2550 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d^{2} x^{2}-7520 B \,c^{\frac {5}{2}} d^{3} x^{3} \sqrt {-d x +c}+225 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d^{2} x -5052 A \,c^{\frac {5}{2}} d^{3} x^{2} \sqrt {-d x +c}+1275 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5} d x +2092 B \,c^{\frac {7}{2}} d^{2} x^{2} \sqrt {-d x +c}+45 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5} d +4944 A \,c^{\frac {7}{2}} d^{2} x \sqrt {-d x +c}+255 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{6}-2704 B \,c^{\frac {9}{2}} d x \sqrt {-d x +c}-1902 A \sqrt {-d x +c}\, c^{\frac {9}{2}} d -538 B \sqrt {-d x +c}\, c^{\frac {11}{2}}\right )}{15360 c^{\frac {5}{2}} \left (d x +c \right )^{\frac {11}{2}} \sqrt {-d x +c}\, d^{2}}\)	\(582\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(17/2),x,method=_RETURNVERBOSE)
 

Output:

1/15360*(-d^2*x^2+c^2)^(1/2)/c^(5/2)*(45*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1 
/2)*2^(1/2)/c^(1/2))*d^6*x^5+255*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1 
/2)/c^(1/2))*c*d^5*x^5+225*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^ 
(1/2))*c*d^5*x^4+1275*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2) 
)*c^2*d^4*x^4+450*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^ 
2*d^4*x^3+90*A*d^5*x^4*(-d*x+c)^(1/2)*c^(1/2)+2550*B*2^(1/2)*arctanh(1/2*( 
-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d^3*x^3+510*B*c^(3/2)*d^4*x^4*(-d*x+c)^ 
(1/2)+450*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d^3*x^ 
2+480*A*c^(3/2)*d^4*x^3*(-d*x+c)^(1/2)+2550*B*2^(1/2)*arctanh(1/2*(-d*x+c) 
^(1/2)*2^(1/2)/c^(1/2))*c^4*d^2*x^2-7520*B*c^(5/2)*d^3*x^3*(-d*x+c)^(1/2)+ 
225*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4*d^2*x-5052*A 
*c^(5/2)*d^3*x^2*(-d*x+c)^(1/2)+1275*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)* 
2^(1/2)/c^(1/2))*c^5*d*x+2092*B*c^(7/2)*d^2*x^2*(-d*x+c)^(1/2)+45*A*2^(1/2 
)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^5*d+4944*A*c^(7/2)*d^2*x*( 
-d*x+c)^(1/2)+255*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^ 
6-2704*B*c^(9/2)*d*x*(-d*x+c)^(1/2)-1902*A*(-d*x+c)^(1/2)*c^(9/2)*d-538*B* 
(-d*x+c)^(1/2)*c^(11/2))/(d*x+c)^(11/2)/(-d*x+c)^(1/2)/d^2
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 827, normalized size of antiderivative = 3.02 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{17/2}} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(17/2),x, algorithm="fricas 
")
 

Output:

[1/30720*(15*sqrt(2)*(17*B*c^7 + 3*A*c^6*d + (17*B*c*d^6 + 3*A*d^7)*x^6 + 
6*(17*B*c^2*d^5 + 3*A*c*d^6)*x^5 + 15*(17*B*c^3*d^4 + 3*A*c^2*d^5)*x^4 + 2 
0*(17*B*c^4*d^3 + 3*A*c^3*d^4)*x^3 + 15*(17*B*c^5*d^2 + 3*A*c^4*d^3)*x^2 + 
 6*(17*B*c^6*d + 3*A*c^5*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*(269*B*c^6 + 951*A*c^5*d - 15*(17*B*c^2*d^4 + 3*A*c*d^5)*x^4 + 
 80*(47*B*c^3*d^3 - 3*A*c^2*d^4)*x^3 - 2*(523*B*c^4*d^2 - 1263*A*c^3*d^3)* 
x^2 + 8*(169*B*c^5*d - 309*A*c^4*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c 
))/(c^3*d^8*x^6 + 6*c^4*d^7*x^5 + 15*c^5*d^6*x^4 + 20*c^6*d^5*x^3 + 15*c^7 
*d^4*x^2 + 6*c^8*d^3*x + c^9*d^2), -1/15360*(15*sqrt(2)*(17*B*c^7 + 3*A*c^ 
6*d + (17*B*c*d^6 + 3*A*d^7)*x^6 + 6*(17*B*c^2*d^5 + 3*A*c*d^6)*x^5 + 15*( 
17*B*c^3*d^4 + 3*A*c^2*d^5)*x^4 + 20*(17*B*c^4*d^3 + 3*A*c^3*d^4)*x^3 + 15 
*(17*B*c^5*d^2 + 3*A*c^4*d^3)*x^2 + 6*(17*B*c^6*d + 3*A*c^5*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*(269*B*c^6 + 951*A*c^5*d - 15*(17*B*c^2*d^4 + 3*A*c*d^5)*x^4 + 
80*(47*B*c^3*d^3 - 3*A*c^2*d^4)*x^3 - 2*(523*B*c^4*d^2 - 1263*A*c^3*d^3)*x 
^2 + 8*(169*B*c^5*d - 309*A*c^4*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c) 
)/(c^3*d^8*x^6 + 6*c^4*d^7*x^5 + 15*c^5*d^6*x^4 + 20*c^6*d^5*x^3 + 15*c^7* 
d^4*x^2 + 6*c^8*d^3*x + c^9*d^2)]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{17/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x+A)*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**(17/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{17/2}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {17}{2}}} \,d x } \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(17/2),x, algorithm="maxima 
")
 

Output:

integrate((-d^2*x^2 + c^2)^(5/2)*(B*x + A)/(d*x + c)^(17/2), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{17/2}} \, dx=-\frac {\frac {15 \, \sqrt {2} {\left (17 \, B c + 3 \, A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {2 \, {\left (255 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} B c - 2740 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} B c^{2} - 8704 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B c^{3} + 9520 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c^{4} - 4080 \, \sqrt {-d x + c} B c^{5} + 45 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} A d + 420 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} A c d - 1536 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} A c^{2} d + 1680 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c^{3} d - 720 \, \sqrt {-d x + c} A c^{4} d\right )}}{{\left (d x + c\right )}^{5} c^{2}}}{15360 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(17/2),x, algorithm="giac")
 

Output:

-1/15360*(15*sqrt(2)*(17*B*c + 3*A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sq 
rt(-c))/(sqrt(-c)*c^2) - 2*(255*(d*x - c)^4*sqrt(-d*x + c)*B*c - 2740*(d*x 
 - c)^3*sqrt(-d*x + c)*B*c^2 - 8704*(d*x - c)^2*sqrt(-d*x + c)*B*c^3 + 952 
0*(-d*x + c)^(3/2)*B*c^4 - 4080*sqrt(-d*x + c)*B*c^5 + 45*(d*x - c)^4*sqrt 
(-d*x + c)*A*d + 420*(d*x - c)^3*sqrt(-d*x + c)*A*c*d - 1536*(d*x - c)^2*s 
qrt(-d*x + c)*A*c^2*d + 1680*(-d*x + c)^(3/2)*A*c^3*d - 720*sqrt(-d*x + c) 
*A*c^4*d)/((d*x + c)^5*c^2))/d^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{17/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^{17/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(17/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(17/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 639, normalized size of antiderivative = 2.33 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{17/2}} \, dx =\text {Too large to display} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(17/2),x)
 

Output:

( - 1902*sqrt(c - d*x)*a*c**5*d + 4944*sqrt(c - d*x)*a*c**4*d**2*x - 5052* 
sqrt(c - d*x)*a*c**3*d**3*x**2 + 480*sqrt(c - d*x)*a*c**2*d**4*x**3 + 90*s 
qrt(c - d*x)*a*c*d**5*x**4 - 538*sqrt(c - d*x)*b*c**6 - 2704*sqrt(c - d*x) 
*b*c**5*d*x + 2092*sqrt(c - d*x)*b*c**4*d**2*x**2 - 7520*sqrt(c - d*x)*b*c 
**3*d**3*x**3 + 510*sqrt(c - d*x)*b*c**2*d**4*x**4 - 45*sqrt(c)*sqrt(2)*lo 
g(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**5*d - 225*sqrt(c)*sqr 
t(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**4*d**2*x - 450 
*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**3* 
d**3*x**2 - 450*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2 
)))/2))*a*c**2*d**4*x**3 - 225*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/ 
(sqrt(c)*sqrt(2)))/2))*a*c*d**5*x**4 - 45*sqrt(c)*sqrt(2)*log(tan(asin(sqr 
t(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*d**6*x**5 - 255*sqrt(c)*sqrt(2)*log(ta 
n(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**6 - 1275*sqrt(c)*sqrt(2)* 
log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**5*d*x - 2550*sqrt(c 
)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**4*d**2*x* 
*2 - 2550*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2) 
)*b*c**3*d**3*x**3 - 1275*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt 
(c)*sqrt(2)))/2))*b*c**2*d**4*x**4 - 255*sqrt(c)*sqrt(2)*log(tan(asin(sqrt 
(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c*d**5*x**5)/(15360*c**3*d**2*(c**5 + 5 
*c**4*d*x + 10*c**3*d**2*x**2 + 10*c**2*d**3*x**3 + 5*c*d**4*x**4 + d**...