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

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

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Rubi [A]  time = 0.16, 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 = {459, 279, 321, 329, 275, 217, 206} \begin {gather*} \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^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^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 206

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

Rule 217

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 275

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 279

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 321

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 329

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 459

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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.31, size = 188, normalized size = 0.78 \begin {gather*} \frac {e^3 \sqrt {e x} \sqrt {a+b x^3} \left (15 a^{7/2} (3 a B-10 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )+\sqrt {b} x^{3/2} \sqrt {\frac {b x^3}{a}+1} \left (-45 a^4 B+30 a^3 b \left (5 A+B x^3\right )+4 a^2 b^2 x^3 \left (295 A+186 B x^3\right )+16 a b^3 x^6 \left (85 A+63 B x^3\right )+96 b^4 x^9 \left (5 A+4 B x^3\right )\right )\right )}{5760 b^{5/2} \sqrt {x} \sqrt {\frac {b x^3}{a}+1}} \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[a + b*x^3]*(Sqrt[b]*x^(3/2)*Sqrt[1 + (b*x^3)/a]*(-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^(7/2)*(-10
*A*b + 3*a*B)*ArcSinh[(Sqrt[b]*x^(3/2))/Sqrt[a]]))/(5760*b^(5/2)*Sqrt[x]*Sqrt[1 + (b*x^3)/a])

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IntegrateAlgebraic [A]  time = 0.95, size = 238, normalized size = 0.99 \begin {gather*} \frac {e^5 \sqrt {\frac {b}{e^3}} \left (10 a^4 A b-3 a^5 B\right ) \log \left (\sqrt {a+b x^3}-\sqrt {\frac {b}{e^3}} (e x)^{3/2}\right )}{384 b^3}+\frac {\sqrt {a+b x^3} \left (-45 a^4 B e^{12} (e x)^{3/2}+150 a^3 A b e^{12} (e x)^{3/2}+30 a^3 b B e^9 (e x)^{9/2}+1180 a^2 A b^2 e^9 (e x)^{9/2}+744 a^2 b^2 B e^6 (e x)^{15/2}+1360 a A b^3 e^6 (e x)^{15/2}+1008 a b^3 B e^3 (e x)^{21/2}+480 A b^4 e^3 (e x)^{21/2}+384 b^4 B (e x)^{27/2}\right )}{5760 b^2 e^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^3]*(150*a^3*A*b*e^12*(e*x)^(3/2) - 45*a^4*B*e^12*(e*x)^(3/2) + 1180*a^2*A*b^2*e^9*(e*x)^(9/2) +
30*a^3*b*B*e^9*(e*x)^(9/2) + 1360*a*A*b^3*e^6*(e*x)^(15/2) + 744*a^2*b^2*B*e^6*(e*x)^(15/2) + 480*A*b^4*e^3*(e
*x)^(21/2) + 1008*a*b^3*B*e^3*(e*x)^(21/2) + 384*b^4*B*(e*x)^(27/2)))/(5760*b^2*e^10) + ((10*a^4*A*b - 3*a^5*B
)*Sqrt[b/e^3]*e^5*Log[-(Sqrt[b/e^3]*(e*x)^(3/2)) + Sqrt[a + b*x^3]])/(384*b^3)

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fricas [A]  time = 1.33, size = 409, normalized size = 1.70 \begin {gather*} \left [-\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} e^{3} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e + 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right ) - 4 \, {\left (384 \, B b^{4} e^{3} x^{13} + 48 \, {\left (21 \, B a b^{3} + 10 \, A b^{4}\right )} e^{3} x^{10} + 8 \, {\left (93 \, B a^{2} b^{2} + 170 \, A a b^{3}\right )} e^{3} x^{7} + 10 \, {\left (3 \, B a^{3} b + 118 \, A a^{2} b^{2}\right )} e^{3} x^{4} - 15 \, {\left (3 \, B a^{4} - 10 \, A a^{3} b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{23040 \, b^{2}}, -\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} e^{3} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right ) - 2 \, {\left (384 \, B b^{4} e^{3} x^{13} + 48 \, {\left (21 \, B a b^{3} + 10 \, A b^{4}\right )} e^{3} x^{10} + 8 \, {\left (93 \, B a^{2} b^{2} + 170 \, A a b^{3}\right )} e^{3} x^{7} + 10 \, {\left (3 \, B a^{3} b + 118 \, A a^{2} b^{2}\right )} e^{3} x^{4} - 15 \, {\left (3 \, B a^{4} - 10 \, A a^{3} b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{11520 \, b^{2}}\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)*e^3*sqrt(e/b)*log(-8*b^2*e*x^6 - 8*a*b*e*x^3 - a^2*e + 4*(2*b^2*x^4 + a*b
*x)*sqrt(b*x^3 + a)*sqrt(e*x)*sqrt(e/b)) - 4*(384*B*b^4*e^3*x^13 + 48*(21*B*a*b^3 + 10*A*b^4)*e^3*x^10 + 8*(93
*B*a^2*b^2 + 170*A*a*b^3)*e^3*x^7 + 10*(3*B*a^3*b + 118*A*a^2*b^2)*e^3*x^4 - 15*(3*B*a^4 - 10*A*a^3*b)*e^3*x)*
sqrt(b*x^3 + a)*sqrt(e*x))/b^2, -1/11520*(15*(3*B*a^5 - 10*A*a^4*b)*e^3*sqrt(-e/b)*arctan(2*sqrt(b*x^3 + a)*sq
rt(e*x)*b*x*sqrt(-e/b)/(2*b*e*x^3 + a*e)) - 2*(384*B*b^4*e^3*x^13 + 48*(21*B*a*b^3 + 10*A*b^4)*e^3*x^10 + 8*(9
3*B*a^2*b^2 + 170*A*a*b^3)*e^3*x^7 + 10*(3*B*a^3*b + 118*A*a^2*b^2)*e^3*x^4 - 15*(3*B*a^4 - 10*A*a^3*b)*e^3*x)
*sqrt(b*x^3 + a)*sqrt(e*x))/b^2]

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giac [B]  time = 1.49, size = 563, normalized size = 2.34 \begin {gather*} \frac {1}{12} \, \sqrt {b x^{3} e^{4} + a e^{4}} {\left (2 \, x^{3} e^{\left (-1\right )} + \frac {a e^{\left (-1\right )}}{b}\right )} A a^{2} x^{\frac {3}{2}} e^{\frac {5}{2}} + \frac {1}{72} \, \sqrt {b x^{3} e^{4} + a e^{4}} {\left (2 \, {\left (4 \, x^{3} e^{\left (-4\right )} + \frac {a e^{\left (-4\right )}}{b}\right )} x^{3} e^{3} - \frac {3 \, a^{2} e^{\left (-1\right )}}{b^{2}}\right )} B a^{2} x^{\frac {3}{2}} e^{\frac {5}{2}} + \frac {1}{36} \, \sqrt {b x^{3} e^{4} + a e^{4}} {\left (2 \, {\left (4 \, x^{3} e^{\left (-4\right )} + \frac {a e^{\left (-4\right )}}{b}\right )} x^{3} e^{3} - \frac {3 \, a^{2} e^{\left (-1\right )}}{b^{2}}\right )} A a b x^{\frac {3}{2}} e^{\frac {5}{2}} + \frac {1}{288} \, \sqrt {b x^{3} e^{4} + a e^{4}} {\left (2 \, {\left (4 \, {\left (6 \, x^{3} e^{\left (-7\right )} + \frac {a e^{\left (-7\right )}}{b}\right )} x^{3} e^{3} - \frac {5 \, a^{2} e^{\left (-4\right )}}{b^{2}}\right )} x^{3} e^{3} + \frac {15 \, a^{3} e^{\left (-1\right )}}{b^{3}}\right )} B a b x^{\frac {3}{2}} e^{\frac {5}{2}} + \frac {1}{576} \, \sqrt {b x^{3} e^{4} + a e^{4}} {\left (2 \, {\left (4 \, {\left (6 \, x^{3} e^{\left (-7\right )} + \frac {a e^{\left (-7\right )}}{b}\right )} x^{3} e^{3} - \frac {5 \, a^{2} e^{\left (-4\right )}}{b^{2}}\right )} x^{3} e^{3} + \frac {15 \, a^{3} e^{\left (-1\right )}}{b^{3}}\right )} A b^{2} x^{\frac {3}{2}} e^{\frac {5}{2}} + \frac {1}{5760} \, \sqrt {b x^{3} e^{4} + a e^{4}} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x^{3} e^{\left (-10\right )} + \frac {a e^{\left (-10\right )}}{b}\right )} x^{3} e^{3} - \frac {7 \, a^{2} e^{\left (-7\right )}}{b^{2}}\right )} x^{3} e^{3} + \frac {35 \, a^{3} e^{\left (-4\right )}}{b^{3}}\right )} x^{3} e^{3} - \frac {105 \, a^{4} e^{\left (-1\right )}}{b^{4}}\right )} B b^{2} x^{\frac {3}{2}} e^{\frac {5}{2}} - \frac {{\left (9 \, B^{2} a^{10} e^{7} - 60 \, A B a^{9} b e^{7} + 100 \, A^{2} a^{8} b^{2} e^{7}\right )} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -{\left (3 \, B a^{5} x^{\frac {3}{2}} e^{\frac {11}{2}} - 10 \, A a^{4} b x^{\frac {3}{2}} e^{\frac {11}{2}}\right )} \sqrt {b} e^{\frac {1}{2}} + \sqrt {9 \, B^{2} a^{11} e^{12} - 60 \, A B a^{10} b e^{12} + 100 \, A^{2} a^{9} b^{2} e^{12} + {\left (3 \, B a^{5} x^{\frac {3}{2}} e^{\frac {11}{2}} - 10 \, A a^{4} b x^{\frac {3}{2}} e^{\frac {11}{2}}\right )}^{2} b e} \right |}\right )}{384 \, b^{\frac {5}{2}} {\left | -3 \, B a^{5} e^{3} + 10 \, A a^{4} b e^{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="giac")

[Out]

1/12*sqrt(b*x^3*e^4 + a*e^4)*(2*x^3*e^(-1) + a*e^(-1)/b)*A*a^2*x^(3/2)*e^(5/2) + 1/72*sqrt(b*x^3*e^4 + a*e^4)*
(2*(4*x^3*e^(-4) + a*e^(-4)/b)*x^3*e^3 - 3*a^2*e^(-1)/b^2)*B*a^2*x^(3/2)*e^(5/2) + 1/36*sqrt(b*x^3*e^4 + a*e^4
)*(2*(4*x^3*e^(-4) + a*e^(-4)/b)*x^3*e^3 - 3*a^2*e^(-1)/b^2)*A*a*b*x^(3/2)*e^(5/2) + 1/288*sqrt(b*x^3*e^4 + a*
e^4)*(2*(4*(6*x^3*e^(-7) + a*e^(-7)/b)*x^3*e^3 - 5*a^2*e^(-4)/b^2)*x^3*e^3 + 15*a^3*e^(-1)/b^3)*B*a*b*x^(3/2)*
e^(5/2) + 1/576*sqrt(b*x^3*e^4 + a*e^4)*(2*(4*(6*x^3*e^(-7) + a*e^(-7)/b)*x^3*e^3 - 5*a^2*e^(-4)/b^2)*x^3*e^3
+ 15*a^3*e^(-1)/b^3)*A*b^2*x^(3/2)*e^(5/2) + 1/5760*sqrt(b*x^3*e^4 + a*e^4)*(2*(4*(6*(8*x^3*e^(-10) + a*e^(-10
)/b)*x^3*e^3 - 7*a^2*e^(-7)/b^2)*x^3*e^3 + 35*a^3*e^(-4)/b^3)*x^3*e^3 - 105*a^4*e^(-1)/b^4)*B*b^2*x^(3/2)*e^(5
/2) - 1/384*(9*B^2*a^10*e^7 - 60*A*B*a^9*b*e^7 + 100*A^2*a^8*b^2*e^7)*e^(-1/2)*log(abs(-(3*B*a^5*x^(3/2)*e^(11
/2) - 10*A*a^4*b*x^(3/2)*e^(11/2))*sqrt(b)*e^(1/2) + sqrt(9*B^2*a^11*e^12 - 60*A*B*a^10*b*e^12 + 100*A^2*a^9*b
^2*e^12 + (3*B*a^5*x^(3/2)*e^(11/2) - 10*A*a^4*b*x^(3/2)*e^(11/2))^2*b*e)))/(b^(5/2)*abs(-3*B*a^5*e^3 + 10*A*a
^4*b*e^3))

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maple [C]  time = 1.06, size = 8117, normalized size = 33.68 \begin {gather*} \text {output too large to display} \end {gather*}

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)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {5}{2}} \left (e x\right )^{\frac {7}{2}}\,{d x} \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]

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

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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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sympy [F(-1)]  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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