3.4.68 \(\int \frac {x^4 (a+b x^2)}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\) [368]

Optimal. Leaf size=161 \[ -\frac {\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {b x^5}{4 d^2 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {3 \left (5 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^6}+\frac {3 c^2 \left (5 b c^2+4 a d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{4 d^7} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {471, 100, 21, 92, 12, 65, 223, 212} \begin {gather*} \frac {3 c^2 \left (4 a d^2+5 b c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )}{4 d^7}+\frac {3 x \sqrt {d x-c} \sqrt {c+d x} \left (4 a d^2+5 b c^2\right )}{8 d^6}-\frac {x^3 \left (4 a d^2+5 b c^2\right )}{4 d^4 \sqrt {d x-c} \sqrt {c+d x}}+\frac {b x^5}{4 d^2 \sqrt {d x-c} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*x^2))/((-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]

[Out]

-1/4*((5*b*c^2 + 4*a*d^2)*x^3)/(d^4*Sqrt[-c + d*x]*Sqrt[c + d*x]) + (b*x^5)/(4*d^2*Sqrt[-c + d*x]*Sqrt[c + d*x
]) + (3*(5*b*c^2 + 4*a*d^2)*x*Sqrt[-c + d*x]*Sqrt[c + d*x])/(8*d^6) + (3*c^2*(5*b*c^2 + 4*a*d^2)*ArcTanh[Sqrt[
-c + d*x]/Sqrt[c + d*x]])/(4*d^7)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[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 100

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

Rule 212

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

Rule 223

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 471

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] - Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(b1*b2*(m + n*(p + 1) + 1)), I
nt[(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]

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b x^2\right )}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx &=\frac {b x^5}{4 d^2 \sqrt {-c+d x} \sqrt {c+d x}}-\frac {1}{4} \left (-4 a-\frac {5 b c^2}{d^2}\right ) \int \frac {x^4}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\\ &=-\frac {\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {b x^5}{4 d^2 \sqrt {-c+d x} \sqrt {c+d x}}-\frac {\left (4 a+\frac {5 b c^2}{d^2}\right ) \int \frac {x^2 \left (-3 c^2-3 c d x\right )}{\sqrt {-c+d x} (c+d x)^{3/2}} \, dx}{4 c d^2}\\ &=-\frac {\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {b x^5}{4 d^2 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {\left (3 \left (5 b c^2+4 a d^2\right )\right ) \int \frac {x^2}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx}{4 d^4}\\ &=-\frac {\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {b x^5}{4 d^2 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {3 \left (5 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^6}+\frac {\left (3 \left (5 b c^2+4 a d^2\right )\right ) \int \frac {c^2}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx}{8 d^6}\\ &=-\frac {\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {b x^5}{4 d^2 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {3 \left (5 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^6}+\frac {\left (3 c^2 \left (5 b c^2+4 a d^2\right )\right ) \int \frac {1}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx}{8 d^6}\\ &=-\frac {\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {b x^5}{4 d^2 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {3 \left (5 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^6}+\frac {\left (3 c^2 \left (5 b c^2+4 a d^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2 c+x^2}} \, dx,x,\sqrt {-c+d x}\right )}{4 d^7}\\ &=-\frac {\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {b x^5}{4 d^2 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {3 \left (5 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^6}+\frac {\left (3 c^2 \left (5 b c^2+4 a d^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{4 d^7}\\ &=-\frac {\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {b x^5}{4 d^2 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {3 \left (5 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^6}+\frac {3 c^2 \left (5 b c^2+4 a d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{4 d^7}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 137, normalized size = 0.85 \begin {gather*} \frac {4 a d^3 x \left (-3 c^2+d^2 x^2\right )+b d x \left (-15 c^4+5 c^2 d^2 x^2+2 d^4 x^4\right )+6 c^2 \left (5 b c^2+4 a d^2\right ) \sqrt {-c+d x} \sqrt {c+d x} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {-c+d x}}\right )}{8 d^7 \sqrt {-c+d x} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*x^2))/((-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.32, size = 324, normalized size = 2.01

method	result	size
default	\(\frac {\sqrt {d x -c}\, \left (-2 \,\mathrm {csgn}\left (d \right ) b \,d^{5} x^{5} \sqrt {d^{2} x^{2}-c^{2}}-4 \,\mathrm {csgn}\left (d \right ) a \,d^{5} x^{3} \sqrt {d^{2} x^{2}-c^{2}}-5 \,\mathrm {csgn}\left (d \right ) b \,c^{2} d^{3} x^{3} \sqrt {d^{2} x^{2}-c^{2}}-12 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right )+d x \right ) \mathrm {csgn}\left (d \right )\right ) a \,c^{2} d^{4} x^{2}-15 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right )+d x \right ) \mathrm {csgn}\left (d \right )\right ) b \,c^{4} d^{2} x^{2}+12 \sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right ) d^{3} a \,c^{2} x +15 \sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right ) d b \,c^{4} x +12 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right )+d x \right ) \mathrm {csgn}\left (d \right )\right ) a \,c^{4} d^{2}+15 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right )+d x \right ) \mathrm {csgn}\left (d \right )\right ) b \,c^{6}\right ) \mathrm {csgn}\left (d \right )}{8 \left (-d x +c \right ) \sqrt {d^{2} x^{2}-c^{2}}\, d^{7} \sqrt {d x +c}}\)	\(324\)
risch	\(-\frac {x \left (2 b \,d^{2} x^{2}+4 a \,d^{2}+7 b \,c^{2}\right ) \left (-d x +c \right ) \sqrt {d x +c}}{8 d^{6} \sqrt {d x -c}}-\frac {\left (-\frac {3 c^{2} \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-c^{2}}\right ) a}{2 d^{4} \sqrt {d^{2}}}-\frac {15 c^{4} \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-c^{2}}\right ) b}{8 d^{6} \sqrt {d^{2}}}+\frac {c^{2} \sqrt {d^{2} \left (x +\frac {c}{d}\right )^{2}-2 c d \left (x +\frac {c}{d}\right )}\, a}{2 d^{6} \left (x +\frac {c}{d}\right )}+\frac {c^{4} \sqrt {d^{2} \left (x +\frac {c}{d}\right )^{2}-2 c d \left (x +\frac {c}{d}\right )}\, b}{2 d^{8} \left (x +\frac {c}{d}\right )}+\frac {c^{2} \sqrt {d^{2} \left (x -\frac {c}{d}\right )^{2}+2 c d \left (x -\frac {c}{d}\right )}\, a}{2 d^{6} \left (x -\frac {c}{d}\right )}+\frac {c^{4} \sqrt {d^{2} \left (x -\frac {c}{d}\right )^{2}+2 c d \left (x -\frac {c}{d}\right )}\, b}{2 d^{8} \left (x -\frac {c}{d}\right )}\right ) \sqrt {\left (d x -c \right ) \left (d x +c \right )}}{\sqrt {d x -c}\, \sqrt {d x +c}}\)	\(354\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(d*x-c)^(1/2)*(-2*csgn(d)*b*d^5*x^5*(d^2*x^2-c^2)^(1/2)-4*csgn(d)*a*d^5*x^3*(d^2*x^2-c^2)^(1/2)-5*csgn(d)*
b*c^2*d^3*x^3*(d^2*x^2-c^2)^(1/2)-12*ln(((d^2*x^2-c^2)^(1/2)*csgn(d)+d*x)*csgn(d))*a*c^2*d^4*x^2-15*ln(((d^2*x
^2-c^2)^(1/2)*csgn(d)+d*x)*csgn(d))*b*c^4*d^2*x^2+12*(d^2*x^2-c^2)^(1/2)*csgn(d)*d^3*a*c^2*x+15*(d^2*x^2-c^2)^
(1/2)*csgn(d)*d*b*c^4*x+12*ln(((d^2*x^2-c^2)^(1/2)*csgn(d)+d*x)*csgn(d))*a*c^4*d^2+15*ln(((d^2*x^2-c^2)^(1/2)*
csgn(d)+d*x)*csgn(d))*b*c^6)*csgn(d)/(-d*x+c)/(d^2*x^2-c^2)^(1/2)/d^7/(d*x+c)^(1/2)

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Maxima [A]
time = 0.31, size = 196, normalized size = 1.22 \begin {gather*} \frac {b x^{5}}{4 \, \sqrt {d^{2} x^{2} - c^{2}} d^{2}} + \frac {5 \, b c^{2} x^{3}}{8 \, \sqrt {d^{2} x^{2} - c^{2}} d^{4}} + \frac {a x^{3}}{2 \, \sqrt {d^{2} x^{2} - c^{2}} d^{2}} - \frac {15 \, b c^{4} x}{8 \, \sqrt {d^{2} x^{2} - c^{2}} d^{6}} - \frac {3 \, a c^{2} x}{2 \, \sqrt {d^{2} x^{2} - c^{2}} d^{4}} + \frac {15 \, b c^{4} \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{8 \, d^{7}} + \frac {3 \, a c^{2} \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{2 \, d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

1/4*b*x^5/(sqrt(d^2*x^2 - c^2)*d^2) + 5/8*b*c^2*x^3/(sqrt(d^2*x^2 - c^2)*d^4) + 1/2*a*x^3/(sqrt(d^2*x^2 - c^2)
*d^2) - 15/8*b*c^4*x/(sqrt(d^2*x^2 - c^2)*d^6) - 3/2*a*c^2*x/(sqrt(d^2*x^2 - c^2)*d^4) + 15/8*b*c^4*log(2*d^2*
x + 2*sqrt(d^2*x^2 - c^2)*d)/d^7 + 3/2*a*c^2*log(2*d^2*x + 2*sqrt(d^2*x^2 - c^2)*d)/d^5

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Fricas [A]
time = 3.14, size = 190, normalized size = 1.18 \begin {gather*} \frac {8 \, b c^{6} + 8 \, a c^{4} d^{2} - 8 \, {\left (b c^{4} d^{2} + a c^{2} d^{4}\right )} x^{2} + {\left (2 \, b d^{5} x^{5} + {\left (5 \, b c^{2} d^{3} + 4 \, a d^{5}\right )} x^{3} - 3 \, {\left (5 \, b c^{4} d + 4 \, a c^{2} d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {d x - c} + 3 \, {\left (5 \, b c^{6} + 4 \, a c^{4} d^{2} - {\left (5 \, b c^{4} d^{2} + 4 \, a c^{2} d^{4}\right )} x^{2}\right )} \log \left (-d x + \sqrt {d x + c} \sqrt {d x - c}\right )}{8 \, {\left (d^{9} x^{2} - c^{2} d^{7}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/8*(8*b*c^6 + 8*a*c^4*d^2 - 8*(b*c^4*d^2 + a*c^2*d^4)*x^2 + (2*b*d^5*x^5 + (5*b*c^2*d^3 + 4*a*d^5)*x^3 - 3*(5
*b*c^4*d + 4*a*c^2*d^3)*x)*sqrt(d*x + c)*sqrt(d*x - c) + 3*(5*b*c^6 + 4*a*c^4*d^2 - (5*b*c^4*d^2 + 4*a*c^2*d^4
)*x^2)*log(-d*x + sqrt(d*x + c)*sqrt(d*x - c)))/(d^9*x^2 - c^2*d^7)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.60, size = 214, normalized size = 1.33 \begin {gather*} \frac {{\left ({\left ({\left (d x + c\right )} {\left (2 \, {\left (d x + c\right )} {\left (\frac {{\left (d x + c\right )} b}{d^{7}} - \frac {5 \, b c}{d^{7}}\right )} + \frac {25 \, b c^{2} d^{35} + 4 \, a d^{37}}{d^{42}}\right )} - \frac {35 \, b c^{3} d^{35} + 12 \, a c d^{37}}{d^{42}}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (7 \, b c^{4} d^{35} + 2 \, a c^{2} d^{37}\right )}}{d^{42}}\right )} \sqrt {d x + c}}{8 \, \sqrt {d x - c}} - \frac {3 \, {\left (5 \, b c^{4} + 4 \, a c^{2} d^{2}\right )} \log \left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}\right )}{8 \, d^{7}} - \frac {2 \, {\left (b c^{5} + a c^{3} d^{2}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2} + 2 \, c\right )} d^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(((d*x + c)*(2*(d*x + c)*((d*x + c)*b/d^7 - 5*b*c/d^7) + (25*b*c^2*d^35 + 4*a*d^37)/d^42) - (35*b*c^3*d^35
+ 12*a*c*d^37)/d^42)*(d*x + c) + 2*(7*b*c^4*d^35 + 2*a*c^2*d^37)/d^42)*sqrt(d*x + c)/sqrt(d*x - c) - 3/8*(5*b
*c^4 + 4*a*c^2*d^2)*log((sqrt(d*x + c) - sqrt(d*x - c))^2)/d^7 - 2*(b*c^5 + a*c^3*d^2)/(((sqrt(d*x + c) - sqrt
(d*x - c))^2 + 2*c)*d^7)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,\left (b\,x^2+a\right )}{{\left (c+d\,x\right )}^{3/2}\,{\left (d\,x-c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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