\(\int e^{3 \text {arctanh}(a x)} (c-a^2 c x^2)^4 \, dx\) [1171]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 152 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {55}{128} c^4 x \sqrt {1-a^2 x^2}+\frac {55}{192} c^4 x \left (1-a^2 x^2\right )^{3/2}+\frac {11}{48} c^4 x \left (1-a^2 x^2\right )^{5/2}-\frac {29 c^4 \left (1-a^2 x^2\right )^{7/2}}{63 a}-\frac {3}{8} c^4 x \left (1-a^2 x^2\right )^{7/2}-\frac {1}{9} a c^4 x^2 \left (1-a^2 x^2\right )^{7/2}+\frac {55 c^4 \arcsin (a x)}{128 a} \] Output:

55/128*c^4*x*(-a^2*x^2+1)^(1/2)+55/192*c^4*x*(-a^2*x^2+1)^(3/2)+11/48*c^4* 
x*(-a^2*x^2+1)^(5/2)-29/63*c^4*(-a^2*x^2+1)^(7/2)/a-3/8*c^4*x*(-a^2*x^2+1) 
^(7/2)-1/9*a*c^4*x^2*(-a^2*x^2+1)^(7/2)+55/128*c^4*arcsin(a*x)/a
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.70 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {c^4 \left (\sqrt {1-a^2 x^2} \left (-3712+4599 a x+10240 a^2 x^2+3066 a^3 x^3-8448 a^4 x^4-7224 a^5 x^5+1024 a^6 x^6+3024 a^7 x^7+896 a^8 x^8\right )-6930 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{8064 a} \] Input:

Integrate[E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^4,x]
 

Output:

(c^4*(Sqrt[1 - a^2*x^2]*(-3712 + 4599*a*x + 10240*a^2*x^2 + 3066*a^3*x^3 - 
 8448*a^4*x^4 - 7224*a^5*x^5 + 1024*a^6*x^6 + 3024*a^7*x^7 + 896*a^8*x^8) 
- 6930*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(8064*a)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6688, 469, 469, 455, 211, 211, 211, 223}

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.

\(\displaystyle \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx\)

\(\Big \downarrow \) 6688

\(\displaystyle c^4 \int (a x+1)^3 \left (1-a^2 x^2\right )^{5/2}dx\)

\(\Big \downarrow \) 469

\(\displaystyle c^4 \left (\frac {11}{9} \int (a x+1)^2 \left (1-a^2 x^2\right )^{5/2}dx-\frac {(a x+1)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 469

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \int (a x+1) \left (1-a^2 x^2\right )^{5/2}dx-\frac {(a x+1) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )-\frac {(a x+1)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 455

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \left (\int \left (1-a^2 x^2\right )^{5/2}dx-\frac {\left (1-a^2 x^2\right )^{7/2}}{7 a}\right )-\frac {(a x+1) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )-\frac {(a x+1)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \left (\frac {5}{6} \int \left (1-a^2 x^2\right )^{3/2}dx-\frac {\left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2}\right )-\frac {(a x+1) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )-\frac {(a x+1)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \left (\frac {5}{6} \left (\frac {3}{4} \int \sqrt {1-a^2 x^2}dx+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2}\right )-\frac {\left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2}\right )-\frac {(a x+1) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )-\frac {(a x+1)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {1}{2} x \sqrt {1-a^2 x^2}\right )+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2}\right )-\frac {\left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2}\right )-\frac {(a x+1) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )-\frac {(a x+1)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} x \sqrt {1-a^2 x^2}+\frac {\arcsin (a x)}{2 a}\right )+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2}\right )-\frac {\left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2}\right )-\frac {(a x+1) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )-\frac {(a x+1)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

Input:

Int[E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^4,x]
 

Output:

c^4*(-1/9*((1 + a*x)^2*(1 - a^2*x^2)^(7/2))/a + (11*(-1/8*((1 + a*x)*(1 - 
a^2*x^2)^(7/2))/a + (9*((x*(1 - a^2*x^2)^(5/2))/6 - (1 - a^2*x^2)^(7/2)/(7 
*a) + (5*((x*(1 - a^2*x^2)^(3/2))/4 + (3*((x*Sqrt[1 - a^2*x^2])/2 + ArcSin 
[a*x]/(2*a)))/4))/6))/8))/9)
 

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 223
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 455
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 469
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* 
((n + p)/(n + 2*p + 1))   Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; Fr 
eeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[n, 0] && NeQ[n + 2* 
p + 1, 0] && IntegerQ[2*p]
 

rule 6688
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> 
 Simp[c^p   Int[(1 - a^2*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, 
 d, p}, x] && EqQ[a^2*c + d, 0] && IntegerQ[p] && IGtQ[(n + 1)/2, 0] &&  !I 
ntegerQ[p - n/2]
 
Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.81

method result size
risch \(-\frac {\left (896 a^{8} x^{8}+3024 a^{7} x^{7}+1024 x^{6} a^{6}-7224 a^{5} x^{5}-8448 a^{4} x^{4}+3066 a^{3} x^{3}+10240 a^{2} x^{2}+4599 a x -3712\right ) \left (a^{2} x^{2}-1\right ) c^{4}}{8064 a \sqrt {-a^{2} x^{2}+1}}+\frac {55 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{4}}{128 \sqrt {a^{2}}}\) \(123\)
meijerg \(-\frac {11 c^{4} \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {9}{2}} \left (-24 x^{6} a^{6}-42 a^{4} x^{4}-105 a^{2} x^{2}+315\right )}{144 a^{8} \sqrt {-a^{2} x^{2}+1}}-\frac {35 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {9}{2}} \arcsin \left (a x \right )}{16 a^{9}}\right )}{\sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {14 c^{4} \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {7}{2}} \left (-14 a^{4} x^{4}-35 a^{2} x^{2}+105\right )}{56 a^{6} \sqrt {-a^{2} x^{2}+1}}-\frac {15 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {7}{2}} \arcsin \left (a x \right )}{8 a^{7}}\right )}{\sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {6 c^{4} \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (-5 a^{2} x^{2}+15\right )}{10 a^{4} \sqrt {-a^{2} x^{2}+1}}-\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{2 a^{5}}\right )}{\sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {c^{4} \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{\sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {c^{4} x}{\sqrt {-a^{2} x^{2}+1}}+\frac {c^{4} \left (-\frac {256 \sqrt {\pi }}{63}+\frac {\sqrt {\pi }\, \left (-28 a^{10} x^{10}-40 a^{8} x^{8}-64 x^{6} a^{6}-128 a^{4} x^{4}-512 a^{2} x^{2}+1024\right )}{252 \sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}+\frac {c^{4} \left (\frac {128 \sqrt {\pi }}{35}-\frac {\sqrt {\pi }\, \left (-10 a^{8} x^{8}-16 x^{6} a^{6}-32 a^{4} x^{4}-128 a^{2} x^{2}+256\right )}{70 \sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}-\frac {6 c^{4} \left (-\frac {16 \sqrt {\pi }}{5}+\frac {\sqrt {\pi }\, \left (-8 x^{6} a^{6}-16 a^{4} x^{4}-64 a^{2} x^{2}+128\right )}{40 \sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}-\frac {14 c^{4} \left (\frac {8 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-2 a^{4} x^{4}-8 a^{2} x^{2}+16\right )}{6 \sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}-\frac {11 c^{4} \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (-4 a^{2} x^{2}+8\right )}{4 \sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}-\frac {3 c^{4} \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {11}{2}} \left (-176 a^{8} x^{8}-264 x^{6} a^{6}-462 a^{4} x^{4}-1155 a^{2} x^{2}+3465\right )}{1408 a^{10} \sqrt {-a^{2} x^{2}+1}}-\frac {315 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {11}{2}} \arcsin \left (a x \right )}{128 a^{11}}\right )}{\sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {3 c^{4} \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}\) \(750\)
default \(\text {Expression too large to display}\) \(1036\)

Input:

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^4,x,method=_RETURNVERBOSE)
 

Output:

-1/8064*(896*a^8*x^8+3024*a^7*x^7+1024*a^6*x^6-7224*a^5*x^5-8448*a^4*x^4+3 
066*a^3*x^3+10240*a^2*x^2+4599*a*x-3712)*(a^2*x^2-1)/a/(-a^2*x^2+1)^(1/2)* 
c^4+55/128/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))*c^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.90 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=-\frac {6930 \, c^{4} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (896 \, a^{8} c^{4} x^{8} + 3024 \, a^{7} c^{4} x^{7} + 1024 \, a^{6} c^{4} x^{6} - 7224 \, a^{5} c^{4} x^{5} - 8448 \, a^{4} c^{4} x^{4} + 3066 \, a^{3} c^{4} x^{3} + 10240 \, a^{2} c^{4} x^{2} + 4599 \, a c^{4} x - 3712 \, c^{4}\right )} \sqrt {-a^{2} x^{2} + 1}}{8064 \, a} \] Input:

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^4,x, algorithm="fric 
as")
 

Output:

-1/8064*(6930*c^4*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (896*a^8*c^4*x^ 
8 + 3024*a^7*c^4*x^7 + 1024*a^6*c^4*x^6 - 7224*a^5*c^4*x^5 - 8448*a^4*c^4* 
x^4 + 3066*a^3*c^4*x^3 + 10240*a^2*c^4*x^2 + 4599*a*c^4*x - 3712*c^4)*sqrt 
(-a^2*x^2 + 1))/a
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 11.99 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.69 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\begin {cases} \frac {55 c^{4} \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{128 \sqrt {- a^{2}}} + \sqrt {- a^{2} x^{2} + 1} \left (\frac {a^{7} c^{4} x^{8}}{9} + \frac {3 a^{6} c^{4} x^{7}}{8} + \frac {8 a^{5} c^{4} x^{6}}{63} - \frac {43 a^{4} c^{4} x^{5}}{48} - \frac {22 a^{3} c^{4} x^{4}}{21} + \frac {73 a^{2} c^{4} x^{3}}{192} + \frac {80 a c^{4} x^{2}}{63} + \frac {73 c^{4} x}{128} - \frac {29 c^{4}}{63 a}\right ) & \text {for}\: a^{2} \neq 0 \\- \frac {a^{9} c^{4} x^{10}}{10} - \frac {a^{8} c^{4} x^{9}}{3} + \frac {8 a^{6} c^{4} x^{7}}{7} + a^{5} c^{4} x^{6} - \frac {6 a^{4} c^{4} x^{5}}{5} - 2 a^{3} c^{4} x^{4} + \frac {3 a c^{4} x^{2}}{2} + c^{4} x & \text {otherwise} \end {cases} \] Input:

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*(-a**2*c*x**2+c)**4,x)
 

Output:

Piecewise((55*c**4*log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/(12 
8*sqrt(-a**2)) + sqrt(-a**2*x**2 + 1)*(a**7*c**4*x**8/9 + 3*a**6*c**4*x**7 
/8 + 8*a**5*c**4*x**6/63 - 43*a**4*c**4*x**5/48 - 22*a**3*c**4*x**4/21 + 7 
3*a**2*c**4*x**3/192 + 80*a*c**4*x**2/63 + 73*c**4*x/128 - 29*c**4/(63*a)) 
, Ne(a**2, 0)), (-a**9*c**4*x**10/10 - a**8*c**4*x**9/3 + 8*a**6*c**4*x**7 
/7 + a**5*c**4*x**6 - 6*a**4*c**4*x**5/5 - 2*a**3*c**4*x**4 + 3*a*c**4*x** 
2/2 + c**4*x, True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (126) = 252\).

Time = 0.12 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.68 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=-\frac {a^{9} c^{4} x^{10}}{9 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {3 \, a^{8} c^{4} x^{9}}{8 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {a^{7} c^{4} x^{8}}{63 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {61 \, a^{6} c^{4} x^{7}}{48 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {74 \, a^{5} c^{4} x^{6}}{63 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {245 \, a^{4} c^{4} x^{5}}{192 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {146 \, a^{3} c^{4} x^{4}}{63 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {73 \, a^{2} c^{4} x^{3}}{384 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {109 \, a c^{4} x^{2}}{63 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {73 \, c^{4} x}{128 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {55 \, c^{4} \arcsin \left (a x\right )}{128 \, a} - \frac {29 \, c^{4}}{63 \, \sqrt {-a^{2} x^{2} + 1} a} \] Input:

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^4,x, algorithm="maxi 
ma")
 

Output:

-1/9*a^9*c^4*x^10/sqrt(-a^2*x^2 + 1) - 3/8*a^8*c^4*x^9/sqrt(-a^2*x^2 + 1) 
- 1/63*a^7*c^4*x^8/sqrt(-a^2*x^2 + 1) + 61/48*a^6*c^4*x^7/sqrt(-a^2*x^2 + 
1) + 74/63*a^5*c^4*x^6/sqrt(-a^2*x^2 + 1) - 245/192*a^4*c^4*x^5/sqrt(-a^2* 
x^2 + 1) - 146/63*a^3*c^4*x^4/sqrt(-a^2*x^2 + 1) - 73/384*a^2*c^4*x^3/sqrt 
(-a^2*x^2 + 1) + 109/63*a*c^4*x^2/sqrt(-a^2*x^2 + 1) + 73/128*c^4*x/sqrt(- 
a^2*x^2 + 1) + 55/128*c^4*arcsin(a*x)/a - 29/63*c^4/(sqrt(-a^2*x^2 + 1)*a)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.82 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {55 \, c^{4} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{128 \, {\left | a \right |}} - \frac {1}{8064} \, \sqrt {-a^{2} x^{2} + 1} {\left (\frac {3712 \, c^{4}}{a} - {\left (4599 \, c^{4} + 2 \, {\left (5120 \, a c^{4} + {\left (1533 \, a^{2} c^{4} - 4 \, {\left (1056 \, a^{3} c^{4} + {\left (903 \, a^{4} c^{4} - 2 \, {\left (64 \, a^{5} c^{4} + 7 \, {\left (8 \, a^{7} c^{4} x + 27 \, a^{6} c^{4}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \] Input:

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^4,x, algorithm="giac 
")
 

Output:

55/128*c^4*arcsin(a*x)*sgn(a)/abs(a) - 1/8064*sqrt(-a^2*x^2 + 1)*(3712*c^4 
/a - (4599*c^4 + 2*(5120*a*c^4 + (1533*a^2*c^4 - 4*(1056*a^3*c^4 + (903*a^ 
4*c^4 - 2*(64*a^5*c^4 + 7*(8*a^7*c^4*x + 27*a^6*c^4)*x)*x)*x)*x)*x)*x)*x)
 

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.45 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {73\,c^4\,x\,\sqrt {1-a^2\,x^2}}{128}+\frac {55\,c^4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{128\,\sqrt {-a^2}}-\frac {29\,c^4\,\sqrt {1-a^2\,x^2}}{63\,a}+\frac {80\,a\,c^4\,x^2\,\sqrt {1-a^2\,x^2}}{63}+\frac {73\,a^2\,c^4\,x^3\,\sqrt {1-a^2\,x^2}}{192}-\frac {22\,a^3\,c^4\,x^4\,\sqrt {1-a^2\,x^2}}{21}-\frac {43\,a^4\,c^4\,x^5\,\sqrt {1-a^2\,x^2}}{48}+\frac {8\,a^5\,c^4\,x^6\,\sqrt {1-a^2\,x^2}}{63}+\frac {3\,a^6\,c^4\,x^7\,\sqrt {1-a^2\,x^2}}{8}+\frac {a^7\,c^4\,x^8\,\sqrt {1-a^2\,x^2}}{9} \] Input:

int(((c - a^2*c*x^2)^4*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)
 

Output:

(73*c^4*x*(1 - a^2*x^2)^(1/2))/128 + (55*c^4*asinh(x*(-a^2)^(1/2)))/(128*( 
-a^2)^(1/2)) - (29*c^4*(1 - a^2*x^2)^(1/2))/(63*a) + (80*a*c^4*x^2*(1 - a^ 
2*x^2)^(1/2))/63 + (73*a^2*c^4*x^3*(1 - a^2*x^2)^(1/2))/192 - (22*a^3*c^4* 
x^4*(1 - a^2*x^2)^(1/2))/21 - (43*a^4*c^4*x^5*(1 - a^2*x^2)^(1/2))/48 + (8 
*a^5*c^4*x^6*(1 - a^2*x^2)^(1/2))/63 + (3*a^6*c^4*x^7*(1 - a^2*x^2)^(1/2)) 
/8 + (a^7*c^4*x^8*(1 - a^2*x^2)^(1/2))/9
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.16 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {c^{4} \left (3465 \mathit {asin} \left (a x \right )+896 \sqrt {-a^{2} x^{2}+1}\, a^{8} x^{8}+3024 \sqrt {-a^{2} x^{2}+1}\, a^{7} x^{7}+1024 \sqrt {-a^{2} x^{2}+1}\, a^{6} x^{6}-7224 \sqrt {-a^{2} x^{2}+1}\, a^{5} x^{5}-8448 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}+3066 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}+10240 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+4599 \sqrt {-a^{2} x^{2}+1}\, a x -3712 \sqrt {-a^{2} x^{2}+1}+3712\right )}{8064 a} \] Input:

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^4,x)
 

Output:

(c**4*(3465*asin(a*x) + 896*sqrt( - a**2*x**2 + 1)*a**8*x**8 + 3024*sqrt( 
- a**2*x**2 + 1)*a**7*x**7 + 1024*sqrt( - a**2*x**2 + 1)*a**6*x**6 - 7224* 
sqrt( - a**2*x**2 + 1)*a**5*x**5 - 8448*sqrt( - a**2*x**2 + 1)*a**4*x**4 + 
 3066*sqrt( - a**2*x**2 + 1)*a**3*x**3 + 10240*sqrt( - a**2*x**2 + 1)*a**2 
*x**2 + 4599*sqrt( - a**2*x**2 + 1)*a*x - 3712*sqrt( - a**2*x**2 + 1) + 37 
12))/(8064*a)