\(\int x^4 \sqrt {-c+d x} \sqrt {c+d x} (a+b x^2) \, dx\) [7]

Optimal result

Integrand size = 31, antiderivative size = 207 \[ \int x^4 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=-\frac {c^4 \left (5 b c^2+8 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{128 d^6}-\frac {c^2 \left (5 b c^2+8 a d^2\right ) x^3 \sqrt {-c+d x} \sqrt {c+d x}}{192 d^4}+\frac {1}{48} \left (8 a+\frac {5 b c^2}{d^2}\right ) x^5 \sqrt {-c+d x} \sqrt {c+d x}+\frac {b x^5 (-c+d x)^{3/2} (c+d x)^{3/2}}{8 d^2}-\frac {c^6 \left (5 b c^2+8 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{64 d^7} \] Output:

-1/128*c^4*(8*a*d^2+5*b*c^2)*x*(d*x-c)^(1/2)*(d*x+c)^(1/2)/d^6-1/192*c^2*( 
8*a*d^2+5*b*c^2)*x^3*(d*x-c)^(1/2)*(d*x+c)^(1/2)/d^4+1/48*(8*a+5*b*c^2/d^2 
)*x^5*(d*x-c)^(1/2)*(d*x+c)^(1/2)+1/8*b*x^5*(d*x-c)^(3/2)*(d*x+c)^(3/2)/d^ 
2-1/64*c^6*(8*a*d^2+5*b*c^2)*arctanh((d*x-c)^(1/2)/(d*x+c)^(1/2))/d^7
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.69 \[ \int x^4 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {d x \sqrt {-c+d x} \sqrt {c+d x} \left (8 a d^2 \left (-3 c^4-2 c^2 d^2 x^2+8 d^4 x^4\right )-b \left (15 c^6+10 c^4 d^2 x^2+8 c^2 d^4 x^4-48 d^6 x^6\right )\right )-6 c^6 \left (5 b c^2+8 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{384 d^7} \] Input:

Integrate[x^4*Sqrt[-c + d*x]*Sqrt[c + d*x]*(a + b*x^2),x]
 

Output:

(d*x*Sqrt[-c + d*x]*Sqrt[c + d*x]*(8*a*d^2*(-3*c^4 - 2*c^2*d^2*x^2 + 8*d^4 
*x^4) - b*(15*c^6 + 10*c^4*d^2*x^2 + 8*c^2*d^4*x^4 - 48*d^6*x^6)) - 6*c^6* 
(5*b*c^2 + 8*a*d^2)*ArcTanh[Sqrt[-c + d*x]/Sqrt[c + d*x]])/(384*d^7)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.89, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {960, 111, 27, 101, 27, 40, 45, 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.

\(\displaystyle \int x^4 \left (a+b x^2\right ) \sqrt {d x-c} \sqrt {c+d x} \, dx\)

\(\Big \downarrow \) 960

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

\(\Big \downarrow \) 111

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 40

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

\(\Big \downarrow \) 45

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{8} \left (8 a+\frac {5 b c^2}{d^2}\right ) \left (\frac {c^2 \left (\frac {c^2 \left (\frac {1}{2} x \sqrt {d x-c} \sqrt {c+d x}-\frac {c^2 \text {arctanh}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )}{d}\right )}{4 d^2}+\frac {x (d x-c)^{3/2} (c+d x)^{3/2}}{4 d^2}\right )}{2 d^2}+\frac {x^3 (d x-c)^{3/2} (c+d x)^{3/2}}{6 d^2}\right )+\frac {b x^5 (d x-c)^{3/2} (c+d x)^{3/2}}{8 d^2}\)

Input:

Int[x^4*Sqrt[-c + d*x]*Sqrt[c + d*x]*(a + b*x^2),x]
 

Output:

(b*x^5*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(8*d^2) + ((8*a + (5*b*c^2)/d^2)* 
((x^3*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(6*d^2) + (c^2*((x*(-c + d*x)^(3/2 
)*(c + d*x)^(3/2))/(4*d^2) + (c^2*((x*Sqrt[-c + d*x]*Sqrt[c + d*x])/2 - (c 
^2*ArcTanh[Sqrt[-c + d*x]/Sqrt[c + d*x]])/d))/(4*d^2)))/(2*d^2)))/8
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x* 
(a + b*x)^m*((c + d*x)^m/(2*m + 1)), x] + Simp[2*a*c*(m/(2*m + 1))   Int[(a 
 + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ 
b*c + a*d, 0] && IGtQ[m + 1/2, 0]
 

rule 45

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] &&  !GtQ[c, 0]
 

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

rule 111

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1))   Int[(a + b*x) 
^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m 
 - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m 
 + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & 
& GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
 

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 960

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.) 
*(x_)^(non2_.))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^( 
m + 1)*(a1 + b1*x^(n/2))^(p + 1)*((a2 + b2*x^(n/2))^(p + 1)/(b1*b2*e*(m + n 
*(p + 1) + 1))), x] - Simp[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/ 
(b1*b2*(m + n*(p + 1) + 1))   Int[(e*x)^m*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n 
/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n, p}, x] && EqQ[non2, 
 n/2] && EqQ[a2*b1 + a1*b2, 0] && NeQ[m + n*(p + 1) + 1, 0]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {x \left (-48 b \,x^{6} d^{6}-64 a \,d^{6} x^{4}+8 b \,c^{2} d^{4} x^{4}+16 a \,c^{2} d^{4} x^{2}+10 b \,c^{4} d^{2} x^{2}+24 a \,c^{4} d^{2}+15 b \,c^{6}\right ) \left (-d x +c \right ) \sqrt {d x +c}}{384 d^{6} \sqrt {d x -c}}-\frac {c^{6} \left (8 a \,d^{2}+5 b \,c^{2}\right ) \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-c^{2}}\right ) \sqrt {\left (d x -c \right ) \left (d x +c \right )}}{128 d^{6} \sqrt {d^{2}}\, \sqrt {d x -c}\, \sqrt {d x +c}}\)	\(184\)
default	\(-\frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (-48 \,\operatorname {csgn}\left (d \right ) b \,d^{7} x^{7} \sqrt {d^{2} x^{2}-c^{2}}-64 \,\operatorname {csgn}\left (d \right ) a \,d^{7} x^{5} \sqrt {d^{2} x^{2}-c^{2}}+8 \,\operatorname {csgn}\left (d \right ) b \,c^{2} d^{5} x^{5} \sqrt {d^{2} x^{2}-c^{2}}+16 \,\operatorname {csgn}\left (d \right ) a \,c^{2} d^{5} x^{3} \sqrt {d^{2} x^{2}-c^{2}}+10 \,\operatorname {csgn}\left (d \right ) b \,c^{4} d^{3} x^{3} \sqrt {d^{2} x^{2}-c^{2}}+24 \sqrt {d^{2} x^{2}-c^{2}}\, \operatorname {csgn}\left (d \right ) d^{3} a \,c^{4} x +15 \sqrt {d^{2} x^{2}-c^{2}}\, \operatorname {csgn}\left (d \right ) d b \,c^{6} x +24 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) a \,c^{6} d^{2}+15 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) b \,c^{8}\right ) \operatorname {csgn}\left (d \right )}{384 \sqrt {d^{2} x^{2}-c^{2}}\, d^{7}}\)	\(298\)

Input:

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

Output:

1/384*x*(-48*b*d^6*x^6-64*a*d^6*x^4+8*b*c^2*d^4*x^4+16*a*c^2*d^4*x^2+10*b* 
c^4*d^2*x^2+24*a*c^4*d^2+15*b*c^6)*(-d*x+c)*(d*x+c)^(1/2)/d^6/(d*x-c)^(1/2 
)-1/128*c^6*(8*a*d^2+5*b*c^2)/d^6*ln(d^2*x/(d^2)^(1/2)+(d^2*x^2-c^2)^(1/2) 
)/(d^2)^(1/2)*((d*x-c)*(d*x+c))^(1/2)/(d*x-c)^(1/2)/(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.67 \[ \int x^4 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {{\left (48 \, b d^{7} x^{7} - 8 \, {\left (b c^{2} d^{5} - 8 \, a d^{7}\right )} x^{5} - 2 \, {\left (5 \, b c^{4} d^{3} + 8 \, a c^{2} d^{5}\right )} x^{3} - 3 \, {\left (5 \, b c^{6} d + 8 \, a c^{4} d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {d x - c} + 3 \, {\left (5 \, b c^{8} + 8 \, a c^{6} d^{2}\right )} \log \left (-d x + \sqrt {d x + c} \sqrt {d x - c}\right )}{384 \, d^{7}} \] Input:

integrate(x^4*(d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a),x, algorithm="fricas")
 

Output:

1/384*((48*b*d^7*x^7 - 8*(b*c^2*d^5 - 8*a*d^7)*x^5 - 2*(5*b*c^4*d^3 + 8*a* 
c^2*d^5)*x^3 - 3*(5*b*c^6*d + 8*a*c^4*d^3)*x)*sqrt(d*x + c)*sqrt(d*x - c) 
+ 3*(5*b*c^8 + 8*a*c^6*d^2)*log(-d*x + sqrt(d*x + c)*sqrt(d*x - c)))/d^7
 

Sympy [F]

\[ \int x^4 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\int x^{4} \left (a + b x^{2}\right ) \sqrt {- c + d x} \sqrt {c + d x}\, dx \] Input:

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

Output:

Integral(x**4*(a + b*x**2)*sqrt(-c + d*x)*sqrt(c + d*x), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.19 \[ \int x^4 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {{\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} b x^{5}}{8 \, d^{2}} + \frac {5 \, {\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} b c^{2} x^{3}}{48 \, d^{4}} + \frac {{\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} a x^{3}}{6 \, d^{2}} - \frac {5 \, b c^{8} \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{128 \, d^{7}} - \frac {a c^{6} \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{16 \, d^{5}} + \frac {5 \, \sqrt {d^{2} x^{2} - c^{2}} b c^{6} x}{128 \, d^{6}} + \frac {\sqrt {d^{2} x^{2} - c^{2}} a c^{4} x}{16 \, d^{4}} + \frac {5 \, {\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} b c^{4} x}{64 \, d^{6}} + \frac {{\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} a c^{2} x}{8 \, d^{4}} \] Input:

integrate(x^4*(d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a),x, algorithm="maxima")
 

Output:

1/8*(d^2*x^2 - c^2)^(3/2)*b*x^5/d^2 + 5/48*(d^2*x^2 - c^2)^(3/2)*b*c^2*x^3 
/d^4 + 1/6*(d^2*x^2 - c^2)^(3/2)*a*x^3/d^2 - 5/128*b*c^8*log(2*d^2*x + 2*s 
qrt(d^2*x^2 - c^2)*d)/d^7 - 1/16*a*c^6*log(2*d^2*x + 2*sqrt(d^2*x^2 - c^2) 
*d)/d^5 + 5/128*sqrt(d^2*x^2 - c^2)*b*c^6*x/d^6 + 1/16*sqrt(d^2*x^2 - c^2) 
*a*c^4*x/d^4 + 5/64*(d^2*x^2 - c^2)^(3/2)*b*c^4*x/d^6 + 1/8*(d^2*x^2 - c^2 
)^(3/2)*a*c^2*x/d^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 519 vs. \(2 (177) = 354\).

Time = 0.22 (sec) , antiderivative size = 519, normalized size of antiderivative = 2.51 \[ \int x^4 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/13440*(56*(((2*((d*x + c)*(4*(d*x + c)*(5*(d*x + c)/d^5 - 31*c/d^5) + 32 
1*c^2/d^5) - 451*c^3/d^5)*(d*x + c) + 745*c^4/d^5)*(d*x + c) - 405*c^5/d^5 
)*sqrt(d*x + c)*sqrt(d*x - c) - 150*c^6*log(abs(-sqrt(d*x + c) + sqrt(d*x 
- c)))/d^5)*a*d + (((2*((4*(5*(d*x + c)*(6*(d*x + c)*(7*(d*x + c)/d^7 - 57 
*c/d^7) + 1219*c^2/d^7) - 12463*c^3/d^7)*(d*x + c) + 64233*c^4/d^7)*(d*x + 
 c) - 53963*c^5/d^7)*(d*x + c) + 59465*c^6/d^7)*(d*x + c) - 23205*c^7/d^7) 
*sqrt(d*x + c)*sqrt(d*x - c) - 7350*c^8*log(abs(-sqrt(d*x + c) + sqrt(d*x 
- c)))/d^7)*b*d + 112*(90*c^5*log(abs(-sqrt(d*x + c) + sqrt(d*x - c))) + ( 
195*c^4 - (295*c^3 - 2*(3*(4*d*x - 17*c)*(d*x + c) + 133*c^2)*(d*x + c))*( 
d*x + c))*sqrt(d*x + c)*sqrt(d*x - c))*a*c/d^4 + 8*(1050*c^7*log(abs(-sqrt 
(d*x + c) + sqrt(d*x - c))) + (2835*c^6 - (6335*c^5 - 2*(4781*c^4 - (4551* 
c^3 - 4*(5*(6*d*x - 37*c)*(d*x + c) + 661*c^2)*(d*x + c))*(d*x + c))*(d*x 
+ c))*(d*x + c))*sqrt(d*x + c)*sqrt(d*x - c))*b*c/d^6)/d
 

Mupad [B] (verification not implemented)

Time = 42.42 (sec) , antiderivative size = 2314, normalized size of antiderivative = 11.18 \[ \int x^4 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

((35*a*c^6*((c + d*x)^(1/2) - c^(1/2))^3)/(12*((-c)^(1/2) - (d*x - c)^(1/2 
))^3) - (a*c^6*((c + d*x)^(1/2) - c^(1/2)))/(4*((-c)^(1/2) - (d*x - c)^(1/ 
2))) + (757*a*c^6*((c + d*x)^(1/2) - c^(1/2))^5)/(4*((-c)^(1/2) - (d*x - c 
)^(1/2))^5) + (7339*a*c^6*((c + d*x)^(1/2) - c^(1/2))^7)/(4*((-c)^(1/2) - 
(d*x - c)^(1/2))^7) + (41929*a*c^6*((c + d*x)^(1/2) - c^(1/2))^9)/(6*((-c) 
^(1/2) - (d*x - c)^(1/2))^9) + (25661*a*c^6*((c + d*x)^(1/2) - c^(1/2))^11 
)/(2*((-c)^(1/2) - (d*x - c)^(1/2))^11) + (25661*a*c^6*((c + d*x)^(1/2) - 
c^(1/2))^13)/(2*((-c)^(1/2) - (d*x - c)^(1/2))^13) + (41929*a*c^6*((c + d* 
x)^(1/2) - c^(1/2))^15)/(6*((-c)^(1/2) - (d*x - c)^(1/2))^15) + (7339*a*c^ 
6*((c + d*x)^(1/2) - c^(1/2))^17)/(4*((-c)^(1/2) - (d*x - c)^(1/2))^17) + 
(757*a*c^6*((c + d*x)^(1/2) - c^(1/2))^19)/(4*((-c)^(1/2) - (d*x - c)^(1/2 
))^19) + (35*a*c^6*((c + d*x)^(1/2) - c^(1/2))^21)/(12*((-c)^(1/2) - (d*x 
- c)^(1/2))^21) - (a*c^6*((c + d*x)^(1/2) - c^(1/2))^23)/(4*((-c)^(1/2) - 
(d*x - c)^(1/2))^23))/(d^5 - (12*d^5*((c + d*x)^(1/2) - c^(1/2))^2)/((-c)^ 
(1/2) - (d*x - c)^(1/2))^2 + (66*d^5*((c + d*x)^(1/2) - c^(1/2))^4)/((-c)^ 
(1/2) - (d*x - c)^(1/2))^4 - (220*d^5*((c + d*x)^(1/2) - c^(1/2))^6)/((-c) 
^(1/2) - (d*x - c)^(1/2))^6 + (495*d^5*((c + d*x)^(1/2) - c^(1/2))^8)/((-c 
)^(1/2) - (d*x - c)^(1/2))^8 - (792*d^5*((c + d*x)^(1/2) - c^(1/2))^10)/(( 
-c)^(1/2) - (d*x - c)^(1/2))^10 + (924*d^5*((c + d*x)^(1/2) - c^(1/2))^12) 
/((-c)^(1/2) - (d*x - c)^(1/2))^12 - (792*d^5*((c + d*x)^(1/2) - c^(1/2...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.16 \[ \int x^4 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {-24 \sqrt {d x +c}\, \sqrt {d x -c}\, a \,c^{4} d^{3} x -16 \sqrt {d x +c}\, \sqrt {d x -c}\, a \,c^{2} d^{5} x^{3}+64 \sqrt {d x +c}\, \sqrt {d x -c}\, a \,d^{7} x^{5}-15 \sqrt {d x +c}\, \sqrt {d x -c}\, b \,c^{6} d x -10 \sqrt {d x +c}\, \sqrt {d x -c}\, b \,c^{4} d^{3} x^{3}-8 \sqrt {d x +c}\, \sqrt {d x -c}\, b \,c^{2} d^{5} x^{5}+48 \sqrt {d x +c}\, \sqrt {d x -c}\, b \,d^{7} x^{7}-48 \,\mathrm {log}\left (\frac {\sqrt {d x -c}+\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right ) a \,c^{6} d^{2}-30 \,\mathrm {log}\left (\frac {\sqrt {d x -c}+\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right ) b \,c^{8}}{384 d^{7}} \] Input:

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

Output:

( - 24*sqrt(c + d*x)*sqrt( - c + d*x)*a*c**4*d**3*x - 16*sqrt(c + d*x)*sqr 
t( - c + d*x)*a*c**2*d**5*x**3 + 64*sqrt(c + d*x)*sqrt( - c + d*x)*a*d**7* 
x**5 - 15*sqrt(c + d*x)*sqrt( - c + d*x)*b*c**6*d*x - 10*sqrt(c + d*x)*sqr 
t( - c + d*x)*b*c**4*d**3*x**3 - 8*sqrt(c + d*x)*sqrt( - c + d*x)*b*c**2*d 
**5*x**5 + 48*sqrt(c + d*x)*sqrt( - c + d*x)*b*d**7*x**7 - 48*log((sqrt( - 
 c + d*x) + sqrt(c + d*x))/(sqrt(c)*sqrt(2)))*a*c**6*d**2 - 30*log((sqrt( 
- c + d*x) + sqrt(c + d*x))/(sqrt(c)*sqrt(2)))*b*c**8)/(384*d**7)