3.6.35 \(\int (e x)^{7/2} (a+b x^3)^{5/2} (A+B x^3) \, dx\) [535]

Optimal. Leaf size=241 \[ \frac {a^3 (10 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{384 b^2}+\frac {a^2 (10 A b-3 a B) (e x)^{9/2} \sqrt {a+b x^3}}{192 b e}+\frac {a (10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{144 b e}+\frac {(10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{5/2}}{120 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{7/2}}{15 b e}-\frac {a^4 (10 A b-3 a B) e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{384 b^{5/2}} \]

[Out]

1/144*a*(10*A*b-3*B*a)*(e*x)^(9/2)*(b*x^3+a)^(3/2)/b/e+1/120*(10*A*b-3*B*a)*(e*x)^(9/2)*(b*x^3+a)^(5/2)/b/e+1/
15*B*(e*x)^(9/2)*(b*x^3+a)^(7/2)/b/e-1/384*a^4*(10*A*b-3*B*a)*e^(7/2)*arctanh((e*x)^(3/2)*b^(1/2)/e^(3/2)/(b*x
^3+a)^(1/2))/b^(5/2)+1/384*a^3*(10*A*b-3*B*a)*e^2*(e*x)^(3/2)*(b*x^3+a)^(1/2)/b^2+1/192*a^2*(10*A*b-3*B*a)*(e*
x)^(9/2)*(b*x^3+a)^(1/2)/b/e

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Rubi [A]
time = 0.12, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {470, 285, 327, 335, 281, 223, 212} \begin {gather*} -\frac {a^4 e^{7/2} (10 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{384 b^{5/2}}+\frac {a^3 e^2 (e x)^{3/2} \sqrt {a+b x^3} (10 A b-3 a B)}{384 b^2}+\frac {a^2 (e x)^{9/2} \sqrt {a+b x^3} (10 A b-3 a B)}{192 b e}+\frac {(e x)^{9/2} \left (a+b x^3\right )^{5/2} (10 A b-3 a B)}{120 b e}+\frac {a (e x)^{9/2} \left (a+b x^3\right )^{3/2} (10 A b-3 a B)}{144 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{7/2}}{15 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^(7/2)*(a + b*x^3)^(5/2)*(A + B*x^3),x]

[Out]

(a^3*(10*A*b - 3*a*B)*e^2*(e*x)^(3/2)*Sqrt[a + b*x^3])/(384*b^2) + (a^2*(10*A*b - 3*a*B)*(e*x)^(9/2)*Sqrt[a +
b*x^3])/(192*b*e) + (a*(10*A*b - 3*a*B)*(e*x)^(9/2)*(a + b*x^3)^(3/2))/(144*b*e) + ((10*A*b - 3*a*B)*(e*x)^(9/
2)*(a + b*x^3)^(5/2))/(120*b*e) + (B*(e*x)^(9/2)*(a + b*x^3)^(7/2))/(15*b*e) - (a^4*(10*A*b - 3*a*B)*e^(7/2)*A
rcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + b*x^3])])/(384*b^(5/2))

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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 285

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

Rule 327

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 470

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

Rubi steps

\begin {align*} \int (e x)^{7/2} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx &=\frac {B (e x)^{9/2} \left (a+b x^3\right )^{7/2}}{15 b e}-\frac {\left (-15 A b+\frac {9 a B}{2}\right ) \int (e x)^{7/2} \left (a+b x^3\right )^{5/2} \, dx}{15 b}\\ &=\frac {(10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{5/2}}{120 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{7/2}}{15 b e}+\frac {(a (10 A b-3 a B)) \int (e x)^{7/2} \left (a+b x^3\right )^{3/2} \, dx}{16 b}\\ &=\frac {a (10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{144 b e}+\frac {(10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{5/2}}{120 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{7/2}}{15 b e}+\frac {\left (a^2 (10 A b-3 a B)\right ) \int (e x)^{7/2} \sqrt {a+b x^3} \, dx}{32 b}\\ &=\frac {a^2 (10 A b-3 a B) (e x)^{9/2} \sqrt {a+b x^3}}{192 b e}+\frac {a (10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{144 b e}+\frac {(10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{5/2}}{120 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{7/2}}{15 b e}+\frac {\left (a^3 (10 A b-3 a B)\right ) \int \frac {(e x)^{7/2}}{\sqrt {a+b x^3}} \, dx}{128 b}\\ &=\frac {a^3 (10 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{384 b^2}+\frac {a^2 (10 A b-3 a B) (e x)^{9/2} \sqrt {a+b x^3}}{192 b e}+\frac {a (10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{144 b e}+\frac {(10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{5/2}}{120 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{7/2}}{15 b e}-\frac {\left (a^4 (10 A b-3 a B) e^3\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{256 b^2}\\ &=\frac {a^3 (10 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{384 b^2}+\frac {a^2 (10 A b-3 a B) (e x)^{9/2} \sqrt {a+b x^3}}{192 b e}+\frac {a (10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{144 b e}+\frac {(10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{5/2}}{120 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{7/2}}{15 b e}-\frac {\left (a^4 (10 A b-3 a B) e^2\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{128 b^2}\\ &=\frac {a^3 (10 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{384 b^2}+\frac {a^2 (10 A b-3 a B) (e x)^{9/2} \sqrt {a+b x^3}}{192 b e}+\frac {a (10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{144 b e}+\frac {(10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{5/2}}{120 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{7/2}}{15 b e}-\frac {\left (a^4 (10 A b-3 a B) e^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^2}{e^3}}} \, dx,x,(e x)^{3/2}\right )}{384 b^2}\\ &=\frac {a^3 (10 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{384 b^2}+\frac {a^2 (10 A b-3 a B) (e x)^{9/2} \sqrt {a+b x^3}}{192 b e}+\frac {a (10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{144 b e}+\frac {(10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{5/2}}{120 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{7/2}}{15 b e}-\frac {\left (a^4 (10 A b-3 a B) e^2\right ) \text {Subst}\left (\int \frac {1}{1-\frac {b x^2}{e^3}} \, dx,x,\frac {(e x)^{3/2}}{\sqrt {a+b x^3}}\right )}{384 b^2}\\ &=\frac {a^3 (10 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{384 b^2}+\frac {a^2 (10 A b-3 a B) (e x)^{9/2} \sqrt {a+b x^3}}{192 b e}+\frac {a (10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{144 b e}+\frac {(10 A b-3 a B) (e x)^{9/2} \left (a+b x^3\right )^{5/2}}{120 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{7/2}}{15 b e}-\frac {a^4 (10 A b-3 a B) e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{384 b^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.42, size = 164, normalized size = 0.68 \begin {gather*} \frac {e^3 \sqrt {e x} \left (\sqrt {b} x^{3/2} \sqrt {a+b x^3} \left (-45 a^4 B+30 a^3 b \left (5 A+B x^3\right )+96 b^4 x^9 \left (5 A+4 B x^3\right )+16 a b^3 x^6 \left (85 A+63 B x^3\right )+4 a^2 b^2 x^3 \left (295 A+186 B x^3\right )\right )+15 a^4 (-10 A b+3 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {b} x^{3/2}}\right )\right )}{5760 b^{5/2} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^(7/2)*(a + b*x^3)^(5/2)*(A + B*x^3),x]

[Out]

(e^3*Sqrt[e*x]*(Sqrt[b]*x^(3/2)*Sqrt[a + b*x^3]*(-45*a^4*B + 30*a^3*b*(5*A + B*x^3) + 96*b^4*x^9*(5*A + 4*B*x^
3) + 16*a*b^3*x^6*(85*A + 63*B*x^3) + 4*a^2*b^2*x^3*(295*A + 186*B*x^3)) + 15*a^4*(-10*A*b + 3*a*B)*ArcTanh[Sq
rt[a + b*x^3]/(Sqrt[b]*x^(3/2))]))/(5760*b^(5/2)*Sqrt[x])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.35, size = 8117, normalized size = 33.68

method	result	size
risch	\(\text {Expression too large to display}\)	\(1135\)
elliptic	\(\text {Expression too large to display}\)	\(1443\)
default	\(\text {Expression too large to display}\)	\(8117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(7/2)*(b*x^3+a)^(5/2)*(B*x^3+A),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 443 vs. \(2 (176) = 352\).
time = 0.50, size = 443, normalized size = 1.84 \begin {gather*} \frac {1}{11520} \, {\left (10 \, {\left (\frac {15 \, a^{4} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x^{3} + a}}{x^{\frac {3}{2}}}}{\sqrt {b} + \frac {\sqrt {b x^{3} + a}}{x^{\frac {3}{2}}}}\right )}{b^{\frac {3}{2}}} + \frac {2 \, {\left (\frac {15 \, \sqrt {b x^{3} + a} a^{4} b^{3}}{x^{\frac {3}{2}}} - \frac {55 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{4} b^{2}}{x^{\frac {9}{2}}} + \frac {73 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{4} b}{x^{\frac {15}{2}}} + \frac {15 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} a^{4}}{x^{\frac {21}{2}}}\right )}}{b^{5} - \frac {4 \, {\left (b x^{3} + a\right )} b^{4}}{x^{3}} + \frac {6 \, {\left (b x^{3} + a\right )}^{2} b^{3}}{x^{6}} - \frac {4 \, {\left (b x^{3} + a\right )}^{3} b^{2}}{x^{9}} + \frac {{\left (b x^{3} + a\right )}^{4} b}{x^{12}}}\right )} A - 3 \, {\left (\frac {15 \, a^{5} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x^{3} + a}}{x^{\frac {3}{2}}}}{\sqrt {b} + \frac {\sqrt {b x^{3} + a}}{x^{\frac {3}{2}}}}\right )}{b^{\frac {5}{2}}} + \frac {2 \, {\left (\frac {15 \, \sqrt {b x^{3} + a} a^{5} b^{4}}{x^{\frac {3}{2}}} - \frac {70 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{5} b^{3}}{x^{\frac {9}{2}}} + \frac {128 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{5} b^{2}}{x^{\frac {15}{2}}} + \frac {70 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} a^{5} b}{x^{\frac {21}{2}}} - \frac {15 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}} a^{5}}{x^{\frac {27}{2}}}\right )}}{b^{7} - \frac {5 \, {\left (b x^{3} + a\right )} b^{6}}{x^{3}} + \frac {10 \, {\left (b x^{3} + a\right )}^{2} b^{5}}{x^{6}} - \frac {10 \, {\left (b x^{3} + a\right )}^{3} b^{4}}{x^{9}} + \frac {5 \, {\left (b x^{3} + a\right )}^{4} b^{3}}{x^{12}} - \frac {{\left (b x^{3} + a\right )}^{5} b^{2}}{x^{15}}}\right )} B\right )} e^{\frac {7}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(b*x^3+a)^(5/2)*(B*x^3+A),x, algorithm="maxima")

[Out]

1/11520*(10*(15*a^4*log(-(sqrt(b) - sqrt(b*x^3 + a)/x^(3/2))/(sqrt(b) + sqrt(b*x^3 + a)/x^(3/2)))/b^(3/2) + 2*
(15*sqrt(b*x^3 + a)*a^4*b^3/x^(3/2) - 55*(b*x^3 + a)^(3/2)*a^4*b^2/x^(9/2) + 73*(b*x^3 + a)^(5/2)*a^4*b/x^(15/
2) + 15*(b*x^3 + a)^(7/2)*a^4/x^(21/2))/(b^5 - 4*(b*x^3 + a)*b^4/x^3 + 6*(b*x^3 + a)^2*b^3/x^6 - 4*(b*x^3 + a)
^3*b^2/x^9 + (b*x^3 + a)^4*b/x^12))*A - 3*(15*a^5*log(-(sqrt(b) - sqrt(b*x^3 + a)/x^(3/2))/(sqrt(b) + sqrt(b*x
^3 + a)/x^(3/2)))/b^(5/2) + 2*(15*sqrt(b*x^3 + a)*a^5*b^4/x^(3/2) - 70*(b*x^3 + a)^(3/2)*a^5*b^3/x^(9/2) + 128
*(b*x^3 + a)^(5/2)*a^5*b^2/x^(15/2) + 70*(b*x^3 + a)^(7/2)*a^5*b/x^(21/2) - 15*(b*x^3 + a)^(9/2)*a^5/x^(27/2))
/(b^7 - 5*(b*x^3 + a)*b^6/x^3 + 10*(b*x^3 + a)^2*b^5/x^6 - 10*(b*x^3 + a)^3*b^4/x^9 + 5*(b*x^3 + a)^4*b^3/x^12
- (b*x^3 + a)^5*b^2/x^15))*B)*e^(7/2)

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Fricas [A]
time = 2.17, size = 358, normalized size = 1.49 \begin {gather*} \left [-\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {b} e^{\frac {7}{2}} \log \left (-8 \, b^{2} x^{6} - 8 \, a b x^{3} + 4 \, {\left (2 \, b x^{4} + a x\right )} \sqrt {b x^{3} + a} \sqrt {b} \sqrt {x} - a^{2}\right ) - 4 \, {\left (384 \, B b^{5} x^{13} + 48 \, {\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{10} + 8 \, {\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{7} + 10 \, {\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x^{4} - 15 \, {\left (3 \, B a^{4} b - 10 \, A a^{3} b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {7}{2}}}{23040 \, b^{3}}, -\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {-b} x^{\frac {3}{2}}}{2 \, b x^{3} + a}\right ) e^{\frac {7}{2}} - 2 \, {\left (384 \, B b^{5} x^{13} + 48 \, {\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{10} + 8 \, {\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{7} + 10 \, {\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x^{4} - 15 \, {\left (3 \, B a^{4} b - 10 \, A a^{3} b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {7}{2}}}{11520 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(b*x^3+a)^(5/2)*(B*x^3+A),x, algorithm="fricas")

[Out]

[-1/23040*(15*(3*B*a^5 - 10*A*a^4*b)*sqrt(b)*e^(7/2)*log(-8*b^2*x^6 - 8*a*b*x^3 + 4*(2*b*x^4 + a*x)*sqrt(b*x^3
+ a)*sqrt(b)*sqrt(x) - a^2) - 4*(384*B*b^5*x^13 + 48*(21*B*a*b^4 + 10*A*b^5)*x^10 + 8*(93*B*a^2*b^3 + 170*A*a
*b^4)*x^7 + 10*(3*B*a^3*b^2 + 118*A*a^2*b^3)*x^4 - 15*(3*B*a^4*b - 10*A*a^3*b^2)*x)*sqrt(b*x^3 + a)*sqrt(x)*e^
(7/2))/b^3, -1/11520*(15*(3*B*a^5 - 10*A*a^4*b)*sqrt(-b)*arctan(2*sqrt(b*x^3 + a)*sqrt(-b)*x^(3/2)/(2*b*x^3 +
a))*e^(7/2) - 2*(384*B*b^5*x^13 + 48*(21*B*a*b^4 + 10*A*b^5)*x^10 + 8*(93*B*a^2*b^3 + 170*A*a*b^4)*x^7 + 10*(3
*B*a^3*b^2 + 118*A*a^2*b^3)*x^4 - 15*(3*B*a^4*b - 10*A*a^3*b^2)*x)*sqrt(b*x^3 + a)*sqrt(x)*e^(7/2))/b^3]

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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((e*x)**(7/2)*(b*x**3+a)**(5/2)*(B*x**3+A),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (176) = 352\).
time = 1.50, size = 437, normalized size = 1.81 \begin {gather*} \frac {1}{5760} \, {\left (480 \, \sqrt {b x^{3} + a} {\left (2 \, x^{3} + \frac {a}{b}\right )} A a^{2} x^{\frac {3}{2}} + 80 \, {\left (2 \, {\left (4 \, x^{3} + \frac {a}{b}\right )} x^{3} - \frac {3 \, a^{2}}{b^{2}}\right )} \sqrt {b x^{3} + a} B a^{2} x^{\frac {3}{2}} + 160 \, {\left (2 \, {\left (4 \, x^{3} + \frac {a}{b}\right )} x^{3} - \frac {3 \, a^{2}}{b^{2}}\right )} \sqrt {b x^{3} + a} A a b x^{\frac {3}{2}} + 20 \, {\left (2 \, {\left (4 \, {\left (6 \, x^{3} + \frac {a}{b}\right )} x^{3} - \frac {5 \, a^{2}}{b^{2}}\right )} x^{3} + \frac {15 \, a^{3}}{b^{3}}\right )} \sqrt {b x^{3} + a} B a b x^{\frac {3}{2}} + 10 \, {\left (2 \, {\left (4 \, {\left (6 \, x^{3} + \frac {a}{b}\right )} x^{3} - \frac {5 \, a^{2}}{b^{2}}\right )} x^{3} + \frac {15 \, a^{3}}{b^{3}}\right )} \sqrt {b x^{3} + a} A b^{2} x^{\frac {3}{2}} + {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x^{3} + \frac {a}{b}\right )} x^{3} - \frac {7 \, a^{2}}{b^{2}}\right )} x^{3} + \frac {35 \, a^{3}}{b^{3}}\right )} x^{3} - \frac {105 \, a^{4}}{b^{4}}\right )} \sqrt {b x^{3} + a} B b^{2} x^{\frac {3}{2}}\right )} e^{\frac {7}{2}} - \frac {{\left (9 \, B^{2} a^{10} - 60 \, A B a^{9} b + 100 \, A^{2} a^{8} b^{2}\right )} e^{\frac {7}{2}} \log \left ({\left | -{\left (3 \, B a^{5} x^{\frac {3}{2}} - 10 \, A a^{4} b x^{\frac {3}{2}}\right )} \sqrt {b} + \sqrt {9 \, B^{2} a^{11} - 60 \, A B a^{10} b + 100 \, A^{2} a^{9} b^{2} + {\left (3 \, B a^{5} x^{\frac {3}{2}} - 10 \, A a^{4} b x^{\frac {3}{2}}\right )}^{2} b} \right |}\right )}{384 \, b^{\frac {5}{2}} {\left | -3 \, B a^{5} + 10 \, A a^{4} b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(b*x^3+a)^(5/2)*(B*x^3+A),x, algorithm="giac")

[Out]

1/5760*(480*sqrt(b*x^3 + a)*(2*x^3 + a/b)*A*a^2*x^(3/2) + 80*(2*(4*x^3 + a/b)*x^3 - 3*a^2/b^2)*sqrt(b*x^3 + a)
*B*a^2*x^(3/2) + 160*(2*(4*x^3 + a/b)*x^3 - 3*a^2/b^2)*sqrt(b*x^3 + a)*A*a*b*x^(3/2) + 20*(2*(4*(6*x^3 + a/b)*
x^3 - 5*a^2/b^2)*x^3 + 15*a^3/b^3)*sqrt(b*x^3 + a)*B*a*b*x^(3/2) + 10*(2*(4*(6*x^3 + a/b)*x^3 - 5*a^2/b^2)*x^3
+ 15*a^3/b^3)*sqrt(b*x^3 + a)*A*b^2*x^(3/2) + (2*(4*(6*(8*x^3 + a/b)*x^3 - 7*a^2/b^2)*x^3 + 35*a^3/b^3)*x^3 -
105*a^4/b^4)*sqrt(b*x^3 + a)*B*b^2*x^(3/2))*e^(7/2) - 1/384*(9*B^2*a^10 - 60*A*B*a^9*b + 100*A^2*a^8*b^2)*e^(
7/2)*log(abs(-(3*B*a^5*x^(3/2) - 10*A*a^4*b*x^(3/2))*sqrt(b) + sqrt(9*B^2*a^11 - 60*A*B*a^10*b + 100*A^2*a^9*b
^2 + (3*B*a^5*x^(3/2) - 10*A*a^4*b*x^(3/2))^2*b)))/(b^(5/2)*abs(-3*B*a^5 + 10*A*a^4*b))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (B\,x^3+A\right )\,{\left (e\,x\right )}^{7/2}\,{\left (b\,x^3+a\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^3)*(e*x)^(7/2)*(a + b*x^3)^(5/2),x)

[Out]

int((A + B*x^3)*(e*x)^(7/2)*(a + b*x^3)^(5/2), x)

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