3.276 \(\int x^5 (\frac {e (a+b x^2)}{c+d x^2})^{3/2} \, dx\)

Optimal. Leaf size=282 \[ \frac {e \left (c+d x^2\right ) \left (-5 a^2 d^2-50 a b c d+79 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{48 b d^4}-\frac {e^{3/2} (b c-a d) \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{16 b^{3/2} d^{9/2}}+\frac {c^2 e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^4}-\frac {e \left (c+d x^2\right )^2 (a d+11 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{24 d^4}+\frac {\left (c+d x^2\right )^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 b d^2 e} \]

[Out]

1/6*(e*(b*x^2+a)/(d*x^2+c))^(5/2)*(d*x^2+c)^3/b/d^2/e-1/16*(-a*d+b*c)*(-a^2*d^2-10*a*b*c*d+35*b^2*c^2)*e^(3/2)
*arctanh(d^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^(1/2)/e^(1/2))/b^(3/2)/d^(9/2)+c^2*(-a*d+b*c)*e*(e*(b*x^2+a)/
(d*x^2+c))^(1/2)/d^4+1/48*(-5*a^2*d^2-50*a*b*c*d+79*b^2*c^2)*e*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b/d^4-1
/24*(a*d+11*b*c)*e*(d*x^2+c)^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/d^4

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Rubi [A]  time = 0.38, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1960, 463, 455, 1157, 388, 208} \[ -\frac {e^{3/2} (b c-a d) \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{16 b^{3/2} d^{9/2}}+\frac {e \left (c+d x^2\right ) \left (-5 a^2 d^2-50 a b c d+79 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{48 b d^4}+\frac {c^2 e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^4}+\frac {\left (c+d x^2\right )^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 b d^2 e}-\frac {e \left (c+d x^2\right )^2 (a d+11 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{24 d^4} \]

Antiderivative was successfully verified.

[In]

Int[x^5*((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]

[Out]

(c^2*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/d^4 + ((79*b^2*c^2 - 50*a*b*c*d - 5*a^2*d^2)*e*Sqrt[(e*(
a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(48*b*d^4) - ((11*b*c + a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*
x^2)^2)/(24*d^4) + (((e*(a + b*x^2))/(c + d*x^2))^(5/2)*(c + d*x^2)^3)/(6*b*d^2*e) - ((b*c - a*d)*(35*b^2*c^2
- 10*a*b*c*d - a^2*d^2)*e^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^
(3/2)*d^(9/2))

Rule 208

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

Rule 388

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

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 463

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

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1960

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(Simplify[(m + 1)/n] - 1
))/(b*e - d*x^q)^(Simplify[(m + 1)/n] + 1), x], x, ((e*(a + b*x^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, m, n}, x] && FractionQ[p] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int x^5 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx &=((b c-a d) e) \operatorname {Subst}\left (\int \frac {x^4 \left (-a e+c x^2\right )^2}{\left (b e-d x^2\right )^4} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {x^4 \left (-6 a^2 d^2 e^2+5 (b c e-a d e)^2+6 b c^2 d e x^2\right )}{\left (b e-d x^2\right )^3} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{6 b d^2}\\ &=-\frac {(11 b c+a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {-b d (b c-a d) (11 b c+a d) e^3-4 d^2 (b c-a d) (11 b c+a d) e^2 x^2-24 b c^2 d^3 e x^4}{\left (b e-d x^2\right )^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{24 b d^5}\\ &=\frac {\left (79 b^2 c^2-50 a b c d-5 a^2 d^2\right ) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{48 b d^4}-\frac {(11 b c+a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}+\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {-3 b d \left (19 b^2 c^2-10 a b c d-a^2 d^2\right ) e^3-48 b^2 c^2 d^2 e^2 x^2}{b e-d x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{48 b^2 d^5 e}\\ &=\frac {c^2 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^4}+\frac {\left (79 b^2 c^2-50 a b c d-5 a^2 d^2\right ) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{48 b d^4}-\frac {(11 b c+a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {\left ((b c-a d) \left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) e^2\right ) \operatorname {Subst}\left (\int \frac {1}{b e-d x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{16 b d^4}\\ &=\frac {c^2 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^4}+\frac {\left (79 b^2 c^2-50 a b c d-5 a^2 d^2\right ) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{48 b d^4}-\frac {(11 b c+a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac {(b c-a d) \left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{16 b^{3/2} d^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 294, normalized size = 1.04 \[ \frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (b \sqrt {d} \sqrt {b c-a d} \left (3 a^3 d^2 \left (c+d x^2\right )+a^2 b d \left (-100 c^2-35 c d x^2+17 d^2 x^4\right )+a b^2 \left (105 c^3-65 c^2 d x^2-52 c d^2 x^4+22 d^3 x^6\right )+b^3 x^2 \left (105 c^3+35 c^2 d x^2-14 c d^2 x^4+8 d^3 x^6\right )\right )-3 \sqrt {a+b x^2} (b c-a d)^2 \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )\right )}{48 b^2 d^{9/2} \left (a+b x^2\right ) \sqrt {b c-a d}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]

[Out]

(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(b*Sqrt[d]*Sqrt[b*c - a*d]*(3*a^3*d^2*(c + d*x^2) + a^2*b*d*(-100*c^2 - 3
5*c*d*x^2 + 17*d^2*x^4) + b^3*x^2*(105*c^3 + 35*c^2*d*x^2 - 14*c*d^2*x^4 + 8*d^3*x^6) + a*b^2*(105*c^3 - 65*c^
2*d*x^2 - 52*c*d^2*x^4 + 22*d^3*x^6)) - 3*(b*c - a*d)^2*(35*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*Sqrt[a + b*x^2]*Sq
rt[(b*(c + d*x^2))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]]))/(48*b^2*d^(9/2)*Sqrt[b*c
- a*d]*(a + b*x^2))

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fricas [A]  time = 1.85, size = 553, normalized size = 1.96 \[ \left [\frac {3 \, {\left (35 \, b^{3} c^{3} - 45 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} e \sqrt {\frac {e}{b d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e - 4 \, {\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} + {\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{b d}}\right ) + 4 \, {\left (8 \, b^{2} d^{3} e x^{6} - 14 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} e x^{4} + {\left (35 \, b^{2} c^{2} d - 38 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} e x^{2} + {\left (105 \, b^{2} c^{3} - 100 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{192 \, b d^{4}}, \frac {3 \, {\left (35 \, b^{3} c^{3} - 45 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} e \sqrt {-\frac {e}{b d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{b d}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) + 2 \, {\left (8 \, b^{2} d^{3} e x^{6} - 14 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} e x^{4} + {\left (35 \, b^{2} c^{2} d - 38 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} e x^{2} + {\left (105 \, b^{2} c^{3} - 100 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{96 \, b d^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="fricas")

[Out]

[1/192*(3*(35*b^3*c^3 - 45*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*e*sqrt(e/(b*d))*log(8*b^2*d^2*e*x^4 + 8*(b^2
*c*d + a*b*d^2)*e*x^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*e - 4*(2*b^2*d^3*x^4 + b^2*c^2*d + a*b*c*d^2 + (3*b^2*
c*d^2 + a*b*d^3)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(e/(b*d))) + 4*(8*b^2*d^3*e*x^6 - 14*(b^2*c*d^2 -
a*b*d^3)*e*x^4 + (35*b^2*c^2*d - 38*a*b*c*d^2 + 3*a^2*d^3)*e*x^2 + (105*b^2*c^3 - 100*a*b*c^2*d + 3*a^2*c*d^2)
*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b*d^4), 1/96*(3*(35*b^3*c^3 - 45*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3
)*e*sqrt(-e/(b*d))*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(-e/(b*d))/(b*e*x^
2 + a*e)) + 2*(8*b^2*d^3*e*x^6 - 14*(b^2*c*d^2 - a*b*d^3)*e*x^4 + (35*b^2*c^2*d - 38*a*b*c*d^2 + 3*a^2*d^3)*e*
x^2 + (105*b^2*c^3 - 100*a*b*c^2*d + 3*a^2*c*d^2)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b*d^4)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x^2*d+c)]Evaluation time: 0.84Unable to divide, perhaps due to rounding error%%%{%%%{2,[0,5,0]%%%},[2,0,0,0]
%%%}+%%%{%%{[%%%{-4,[0,4,0]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,0,0,1]%%%}+%%%{%%%{2,[1,4,1]%%%},[0,0,0,2]%%
%} / %%%{%%%{1,[0,2,2]%%%},[2,0,0,0]%%%}+%%%{%%{[%%%{-2,[0,1,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,0,0,1]%%
%}+%%%{%%%{1,[1,1,3]%%%},[0,0,0,2]%%%} Error: Bad Argument Value

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maple [B]  time = 0.08, size = 1027, normalized size = 3.64 \[ \frac {\left (-3 a^{3} d^{4} x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-27 a^{2} b c \,d^{3} x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+135 a \,b^{2} c^{2} d^{2} x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-105 b^{3} c^{3} d \,x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+12 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b \,d^{3} x^{4}-60 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c \,d^{2} x^{4}-3 a^{3} c \,d^{3} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-27 a^{2} b \,c^{2} d^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+135 a \,b^{2} c^{3} d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-105 b^{3} c^{4} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} d^{3} x^{2}-108 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b c \,d^{2} x^{2}+54 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c^{2} d \,x^{2}+6 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} c \,d^{2}-120 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b \,c^{2} d -96 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a b \,c^{2} d +114 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c^{3}+96 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b^{2} c^{3}+16 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {b d}\, b \,d^{2} x^{2}+16 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {b d}\, b c d \right ) \left (d \,x^{2}+c \right ) \left (\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}\right )^{\frac {3}{2}}}{96 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (b \,x^{2}+a \right ) b \,d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*((b*x^2+a)/(d*x^2+c)*e)^(3/2),x)

[Out]

1/96*(12*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^4*a*b*d^3-60*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b
*d)^(1/2)*x^4*b^2*c*d^2-3*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(
1/2))*x^2*a^3*d^4-27*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))
*x^2*a^2*b*c*d^3+135*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))
*x^2*a*b^2*c^2*d^2-105*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2
))*x^2*b^3*c^3*d+16*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(b*d)^(1/2)*x^2*b*d^2+6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^
(1/2)*(b*d)^(1/2)*x^2*a^2*d^3-108*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*a*b*c*d^2+54*(b*d*x^4+a*
d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*b^2*c^2*d-3*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)
^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a^3*c*d^3-27*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*
(b*d)^(1/2))/(b*d)^(1/2))*a^2*b*c^2*d^2+135*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b
*d)^(1/2))/(b*d)^(1/2))*a*b^2*c^3*d-105*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^
(1/2))/(b*d)^(1/2))*b^3*c^4+16*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(b*d)^(1/2)*b*c*d+6*(b*d*x^4+a*d*x^2+b*c*x^
2+a*c)^(1/2)*(b*d)^(1/2)*a^2*c*d^2-120*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*b*c^2*d+114*(b*d*x^4+
a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b^2*c^3-96*(b*d)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*a*b*c^2*d+96*(b*d)^(
1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*b^2*c^3)/d^4/b*(d*x^2+c)*((b*x^2+a)/(d*x^2+c)*e)^(3/2)/(b*d)^(1/2)/((d*x^2+c)
*(b*x^2+a))^(1/2)/(b*x^2+a)

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maxima [A]  time = 2.12, size = 454, normalized size = 1.61 \[ \frac {1}{96} \, e {\left (\frac {2 \, {\left (3 \, {\left (29 \, b^{3} c^{3} d^{2} - 51 \, a b^{2} c^{2} d^{3} + 23 \, a^{2} b c d^{4} - a^{3} d^{5}\right )} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {5}{2}} e - 8 \, {\left (17 \, b^{4} c^{3} d - 27 \, a b^{3} c^{2} d^{2} + 9 \, a^{2} b^{2} c d^{3} + a^{3} b d^{4}\right )} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} e^{2} + 3 \, {\left (19 \, b^{5} c^{3} - 29 \, a b^{4} c^{2} d + 9 \, a^{2} b^{3} c d^{2} + a^{3} b^{2} d^{3}\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e^{3}\right )}}{b^{4} d^{4} e^{3} - \frac {3 \, {\left (b x^{2} + a\right )} b^{3} d^{5} e^{3}}{d x^{2} + c} + \frac {3 \, {\left (b x^{2} + a\right )}^{2} b^{2} d^{6} e^{3}}{{\left (d x^{2} + c\right )}^{2}} - \frac {{\left (b x^{2} + a\right )}^{3} b d^{7} e^{3}}{{\left (d x^{2} + c\right )}^{3}}} + \frac {96 \, {\left (b c^{3} - a c^{2} d\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{d^{4}} + \frac {3 \, {\left (35 \, b^{3} c^{3} - 45 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} e \log \left (\frac {d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {b d e}}{d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {b d e}}\right )}{\sqrt {b d e} b d^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

1/96*e*(2*(3*(29*b^3*c^3*d^2 - 51*a*b^2*c^2*d^3 + 23*a^2*b*c*d^4 - a^3*d^5)*((b*x^2 + a)*e/(d*x^2 + c))^(5/2)*
e - 8*(17*b^4*c^3*d - 27*a*b^3*c^2*d^2 + 9*a^2*b^2*c*d^3 + a^3*b*d^4)*((b*x^2 + a)*e/(d*x^2 + c))^(3/2)*e^2 +
3*(19*b^5*c^3 - 29*a*b^4*c^2*d + 9*a^2*b^3*c*d^2 + a^3*b^2*d^3)*sqrt((b*x^2 + a)*e/(d*x^2 + c))*e^3)/(b^4*d^4*
e^3 - 3*(b*x^2 + a)*b^3*d^5*e^3/(d*x^2 + c) + 3*(b*x^2 + a)^2*b^2*d^6*e^3/(d*x^2 + c)^2 - (b*x^2 + a)^3*b*d^7*
e^3/(d*x^2 + c)^3) + 96*(b*c^3 - a*c^2*d)*sqrt((b*x^2 + a)*e/(d*x^2 + c))/d^4 + 3*(35*b^3*c^3 - 45*a*b^2*c^2*d
+ 9*a^2*b*c*d^2 + a^3*d^3)*e*log((d*sqrt((b*x^2 + a)*e/(d*x^2 + c)) - sqrt(b*d*e))/(d*sqrt((b*x^2 + a)*e/(d*x
^2 + c)) + sqrt(b*d*e)))/(sqrt(b*d*e)*b*d^4))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^5\,{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*((e*(a + b*x^2))/(c + d*x^2))^(3/2),x)

[Out]

int(x^5*((e*(a + b*x^2))/(c + d*x^2))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(e*(b*x**2+a)/(d*x**2+c))**(3/2),x)

[Out]

Timed out

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