3.6.38 \(\int \sqrt {e x} (a+b x^3)^{5/2} (A+B x^3) \, dx\) [538]

Optimal. Leaf size=201 \[ \frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {5 a^3 (8 A b-a B) \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{192 b^{3/2}} \]

[Out]

5/288*a*(8*A*b-B*a)*(e*x)^(3/2)*(b*x^3+a)^(3/2)/b/e+1/72*(8*A*b-B*a)*(e*x)^(3/2)*(b*x^3+a)^(5/2)/b/e+1/12*B*(e
*x)^(3/2)*(b*x^3+a)^(7/2)/b/e+5/192*a^3*(8*A*b-B*a)*arctanh((e*x)^(3/2)*b^(1/2)/e^(3/2)/(b*x^3+a)^(1/2))*e^(1/
2)/b^(3/2)+5/192*a^2*(8*A*b-B*a)*(e*x)^(3/2)*(b*x^3+a)^(1/2)/b/e

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Rubi [A]
time = 0.09, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {470, 285, 335, 281, 223, 212} \begin {gather*} \frac {5 a^3 \sqrt {e} (8 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{192 b^{3/2}}+\frac {5 a^2 (e x)^{3/2} \sqrt {a+b x^3} (8 A b-a B)}{192 b e}+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2} (8 A b-a B)}{72 b e}+\frac {5 a (e x)^{3/2} \left (a+b x^3\right )^{3/2} (8 A b-a B)}{288 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^2*(8*A*b - a*B)*(e*x)^(3/2)*Sqrt[a + b*x^3])/(192*b*e) + (5*a*(8*A*b - a*B)*(e*x)^(3/2)*(a + b*x^3)^(3/2)
)/(288*b*e) + ((8*A*b - a*B)*(e*x)^(3/2)*(a + b*x^3)^(5/2))/(72*b*e) + (B*(e*x)^(3/2)*(a + b*x^3)^(7/2))/(12*b
*e) + (5*a^3*(8*A*b - a*B)*Sqrt[e]*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + b*x^3])])/(192*b^(3/2))

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 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 \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx &=\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}-\frac {\left (-12 A b+\frac {3 a B}{2}\right ) \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \, dx}{12 b}\\ &=\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {(5 a (8 A b-a B)) \int \sqrt {e x} \left (a+b x^3\right )^{3/2} \, dx}{48 b}\\ &=\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {\left (5 a^2 (8 A b-a B)\right ) \int \sqrt {e x} \sqrt {a+b x^3} \, dx}{64 b}\\ &=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {\left (5 a^3 (8 A b-a B)\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{128 b}\\ &=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {\left (5 a^3 (8 A b-a B)\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{64 b e}\\ &=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {\left (5 a^3 (8 A b-a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^2}{e^3}}} \, dx,x,(e x)^{3/2}\right )}{192 b e}\\ &=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {\left (5 a^3 (8 A b-a B)\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 )}{192 b e}\\ &=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {5 a^3 (8 A b-a B) \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{192 b^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.35, size = 140, normalized size = 0.70 \begin {gather*} \frac {\sqrt {e x} \left (\sqrt {b} x^{3/2} \sqrt {a+b x^3} \left (15 a^3 B+16 b^3 x^6 \left (4 A+3 B x^3\right )+8 a b^2 x^3 \left (26 A+17 B x^3\right )+2 a^2 b \left (132 A+59 B x^3\right )\right )-15 a^3 (-8 A b+a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {b} x^{3/2}}\right )\right )}{576 b^{3/2} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e*x]*(Sqrt[b]*x^(3/2)*Sqrt[a + b*x^3]*(15*a^3*B + 16*b^3*x^6*(4*A + 3*B*x^3) + 8*a*b^2*x^3*(26*A + 17*B*
x^3) + 2*a^2*b*(132*A + 59*B*x^3)) - 15*a^3*(-8*A*b + a*B)*ArcTanh[Sqrt[a + b*x^3]/(Sqrt[b]*x^(3/2))]))/(576*b
^(3/2)*Sqrt[x])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.40, size = 7702, normalized size = 38.32

method	result	size
risch	\(\text {Expression too large to display}\)	\(1104\)
elliptic	\(\text {Expression too large to display}\)	\(1304\)
default	\(\text {Expression too large to display}\)	\(7702\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(5/2)*(B*x^3+A)*(e*x)^(1/2),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. 364 vs. \(2 (141) = 282\).
time = 0.51, size = 364, normalized size = 1.81 \begin {gather*} -\frac {1}{1152} \, {\left (8 \, {\left (\frac {15 \, a^{3} \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 )}{\sqrt {b}} + \frac {2 \, {\left (\frac {15 \, \sqrt {b x^{3} + a} a^{3} b^{2}}{x^{\frac {3}{2}}} - \frac {40 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{3} b}{x^{\frac {9}{2}}} + \frac {33 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{3}}{x^{\frac {15}{2}}}\right )}}{b^{3} - \frac {3 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {3 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {{\left (b x^{3} + a\right )}^{3}}{x^{9}}}\right )} A - {\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 )} B\right )} e^{\frac {1}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/2)*(B*x^3+A)*(e*x)^(1/2),x, algorithm="maxima")

[Out]

-1/1152*(8*(15*a^3*log(-(sqrt(b) - sqrt(b*x^3 + a)/x^(3/2))/(sqrt(b) + sqrt(b*x^3 + a)/x^(3/2)))/sqrt(b) + 2*(
15*sqrt(b*x^3 + a)*a^3*b^2/x^(3/2) - 40*(b*x^3 + a)^(3/2)*a^3*b/x^(9/2) + 33*(b*x^3 + a)^(5/2)*a^3/x^(15/2))/(
b^3 - 3*(b*x^3 + a)*b^2/x^3 + 3*(b*x^3 + a)^2*b/x^6 - (b*x^3 + a)^3/x^9))*A - (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))*B)*e^(1/
2)

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Fricas [A]
time = 2.83, size = 308, normalized size = 1.53 \begin {gather*} \left [-\frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \sqrt {b} e^{\frac {1}{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 (48 \, B b^{4} x^{10} + 8 \, {\left (17 \, B a b^{3} + 8 \, A b^{4}\right )} x^{7} + 2 \, {\left (59 \, B a^{2} b^{2} + 104 \, A a b^{3}\right )} x^{4} + 3 \, {\left (5 \, B a^{3} b + 88 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {1}{2}}}{2304 \, b^{2}}, \frac {15 \, {\left (B a^{4} - 8 \, A a^{3} 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 {1}{2}} + 2 \, {\left (48 \, B b^{4} x^{10} + 8 \, {\left (17 \, B a b^{3} + 8 \, A b^{4}\right )} x^{7} + 2 \, {\left (59 \, B a^{2} b^{2} + 104 \, A a b^{3}\right )} x^{4} + 3 \, {\left (5 \, B a^{3} b + 88 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {1}{2}}}{1152 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/2)*(B*x^3+A)*(e*x)^(1/2),x, algorithm="fricas")

[Out]

[-1/2304*(15*(B*a^4 - 8*A*a^3*b)*sqrt(b)*e^(1/2)*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*(48*B*b^4*x^10 + 8*(17*B*a*b^3 + 8*A*b^4)*x^7 + 2*(59*B*a^2*b^2 + 104*A*a*b^3)*x^
4 + 3*(5*B*a^3*b + 88*A*a^2*b^2)*x)*sqrt(b*x^3 + a)*sqrt(x)*e^(1/2))/b^2, 1/1152*(15*(B*a^4 - 8*A*a^3*b)*sqrt(
-b)*arctan(2*sqrt(b*x^3 + a)*sqrt(-b)*x^(3/2)/(2*b*x^3 + a))*e^(1/2) + 2*(48*B*b^4*x^10 + 8*(17*B*a*b^3 + 8*A*
b^4)*x^7 + 2*(59*B*a^2*b^2 + 104*A*a*b^3)*x^4 + 3*(5*B*a^3*b + 88*A*a^2*b^2)*x)*sqrt(b*x^3 + a)*sqrt(x)*e^(1/2
))/b^2]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 413 vs. \(2 (177) = 354\).
time = 49.81, size = 413, normalized size = 2.05 \begin {gather*} \frac {A a^{\frac {5}{2}} \left (e x\right )^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e} + \frac {A a^{\frac {5}{2}} \left (e x\right )^{\frac {3}{2}}}{8 e \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {35 A a^{\frac {3}{2}} b \left (e x\right )^{\frac {9}{2}}}{72 e^{4} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {17 A \sqrt {a} b^{2} \left (e x\right )^{\frac {15}{2}}}{36 e^{7} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {5 A a^{3} \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {b} \left (e x\right )^{\frac {3}{2}}}{\sqrt {a} e^{\frac {3}{2}}} \right )}}{24 \sqrt {b}} + \frac {A b^{3} \left (e x\right )^{\frac {21}{2}}}{9 \sqrt {a} e^{10} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {5 B a^{\frac {7}{2}} \left (e x\right )^{\frac {3}{2}}}{192 b e \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {133 B a^{\frac {5}{2}} \left (e x\right )^{\frac {9}{2}}}{576 e^{4} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {127 B a^{\frac {3}{2}} b \left (e x\right )^{\frac {15}{2}}}{288 e^{7} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {23 B \sqrt {a} b^{2} \left (e x\right )^{\frac {21}{2}}}{72 e^{10} \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {5 B a^{4} \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {b} \left (e x\right )^{\frac {3}{2}}}{\sqrt {a} e^{\frac {3}{2}}} \right )}}{192 b^{\frac {3}{2}}} + \frac {B b^{3} \left (e x\right )^{\frac {27}{2}}}{12 \sqrt {a} e^{13} \sqrt {1 + \frac {b x^{3}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(5/2)*(B*x**3+A)*(e*x)**(1/2),x)

[Out]

A*a**(5/2)*(e*x)**(3/2)*sqrt(1 + b*x**3/a)/(3*e) + A*a**(5/2)*(e*x)**(3/2)/(8*e*sqrt(1 + b*x**3/a)) + 35*A*a**
(3/2)*b*(e*x)**(9/2)/(72*e**4*sqrt(1 + b*x**3/a)) + 17*A*sqrt(a)*b**2*(e*x)**(15/2)/(36*e**7*sqrt(1 + b*x**3/a
)) + 5*A*a**3*sqrt(e)*asinh(sqrt(b)*(e*x)**(3/2)/(sqrt(a)*e**(3/2)))/(24*sqrt(b)) + A*b**3*(e*x)**(21/2)/(9*sq
rt(a)*e**10*sqrt(1 + b*x**3/a)) + 5*B*a**(7/2)*(e*x)**(3/2)/(192*b*e*sqrt(1 + b*x**3/a)) + 133*B*a**(5/2)*(e*x
)**(9/2)/(576*e**4*sqrt(1 + b*x**3/a)) + 127*B*a**(3/2)*b*(e*x)**(15/2)/(288*e**7*sqrt(1 + b*x**3/a)) + 23*B*s
qrt(a)*b**2*(e*x)**(21/2)/(72*e**10*sqrt(1 + b*x**3/a)) - 5*B*a**4*sqrt(e)*asinh(sqrt(b)*(e*x)**(3/2)/(sqrt(a)
*e**(3/2)))/(192*b**(3/2)) + B*b**3*(e*x)**(27/2)/(12*sqrt(a)*e**13*sqrt(1 + b*x**3/a))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 383 vs. \(2 (141) = 282\).
time = 2.12, size = 383, normalized size = 1.91 \begin {gather*} \frac {1}{576} \, {\left (48 \, \sqrt {b x^{3} + a} {\left (2 \, x^{3} + \frac {a}{b}\right )} B a^{2} x^{\frac {3}{2}} + 96 \, \sqrt {b x^{3} + a} {\left (2 \, x^{3} + \frac {a}{b}\right )} A a b x^{\frac {3}{2}} + 16 \, {\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 b x^{\frac {3}{2}} + 8 \, {\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 b^{2} x^{\frac {3}{2}} + {\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 b^{2} x^{\frac {3}{2}} + 192 \, {\left (\sqrt {b x^{3} + a} x^{\frac {3}{2}} - \frac {a \log \left ({\left | -\sqrt {b} x^{\frac {3}{2}} + \sqrt {b x^{3} + a} \right |}\right )}{\sqrt {b}}\right )} A a^{2}\right )} e^{\frac {1}{2}} - \frac {{\left (25 \, B^{2} a^{8} + 240 \, A B a^{7} b + 576 \, A^{2} a^{6} b^{2}\right )} e^{\frac {1}{2}} \log \left ({\left | {\left (5 \, B a^{4} x^{\frac {3}{2}} + 24 \, A a^{3} b x^{\frac {3}{2}}\right )} \sqrt {b} + \sqrt {25 \, B^{2} a^{9} + 240 \, A B a^{8} b + 576 \, A^{2} a^{7} b^{2} + {\left (5 \, B a^{4} x^{\frac {3}{2}} + 24 \, A a^{3} b x^{\frac {3}{2}}\right )}^{2} b} \right |}\right )}{192 \, b^{\frac {3}{2}} {\left | 5 \, B a^{4} + 24 \, A a^{3} b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/2)*(B*x^3+A)*(e*x)^(1/2),x, algorithm="giac")

[Out]

1/576*(48*sqrt(b*x^3 + a)*(2*x^3 + a/b)*B*a^2*x^(3/2) + 96*sqrt(b*x^3 + a)*(2*x^3 + a/b)*A*a*b*x^(3/2) + 16*(2
*(4*x^3 + a/b)*x^3 - 3*a^2/b^2)*sqrt(b*x^3 + a)*B*a*b*x^(3/2) + 8*(2*(4*x^3 + a/b)*x^3 - 3*a^2/b^2)*sqrt(b*x^3
+ a)*A*b^2*x^(3/2) + (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*b^2*x^(3/2) + 1
92*(sqrt(b*x^3 + a)*x^(3/2) - a*log(abs(-sqrt(b)*x^(3/2) + sqrt(b*x^3 + a)))/sqrt(b))*A*a^2)*e^(1/2) - 1/192*(
25*B^2*a^8 + 240*A*B*a^7*b + 576*A^2*a^6*b^2)*e^(1/2)*log(abs((5*B*a^4*x^(3/2) + 24*A*a^3*b*x^(3/2))*sqrt(b) +
sqrt(25*B^2*a^9 + 240*A*B*a^8*b + 576*A^2*a^7*b^2 + (5*B*a^4*x^(3/2) + 24*A*a^3*b*x^(3/2))^2*b)))/(b^(3/2)*ab
s(5*B*a^4 + 24*A*a^3*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 )\,\sqrt {e\,x}\,{\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)^(1/2)*(a + b*x^3)^(5/2),x)

[Out]

int((A + B*x^3)*(e*x)^(1/2)*(a + b*x^3)^(5/2), x)

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