3.6.16 \(\int (e x)^{7/2} \sqrt {a+b x^3} (A+B x^3) \, dx\) [516]

Optimal. Leaf size=161 \[ \frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {a^2 (2 A b-a B) e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 b^{5/2}} \]

[Out]

1/9*B*(e*x)^(9/2)*(b*x^3+a)^(3/2)/b/e-1/24*a^2*(2*A*b-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)+1/24*a*(2*A*b-B*a)*e^2*(e*x)^(3/2)*(b*x^3+a)^(1/2)/b^2+1/12*(2*A*b-B*a)*(e*x)^(9/2)*(b*x^3+a
)^(1/2)/b/e

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Rubi [A]
time = 0.08, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 7, 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^2 e^{7/2} (2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 b^{5/2}}+\frac {a e^2 (e x)^{3/2} \sqrt {a+b x^3} (2 A b-a B)}{24 b^2}+\frac {(e x)^{9/2} \sqrt {a+b x^3} (2 A b-a B)}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*A*b - a*B)*e^2*(e*x)^(3/2)*Sqrt[a + b*x^3])/(24*b^2) + ((2*A*b - a*B)*(e*x)^(9/2)*Sqrt[a + b*x^3])/(12*b
*e) + (B*(e*x)^(9/2)*(a + b*x^3)^(3/2))/(9*b*e) - (a^2*(2*A*b - a*B)*e^(7/2)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^
(3/2)*Sqrt[a + b*x^3])])/(24*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} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx &=\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {\left (-9 A b+\frac {9 a B}{2}\right ) \int (e x)^{7/2} \sqrt {a+b x^3} \, dx}{9 b}\\ &=\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}+\frac {(a (2 A b-a B)) \int \frac {(e x)^{7/2}}{\sqrt {a+b x^3}} \, dx}{8 b}\\ &=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {\left (a^2 (2 A b-a B) e^3\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{16 b^2}\\ &=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {\left (a^2 (2 A b-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 )}{8 b^2}\\ &=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {\left (a^2 (2 A b-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 )}{24 b^2}\\ &=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {\left (a^2 (2 A b-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 )}{24 b^2}\\ &=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {a^2 (2 A b-a B) e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 b^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 122, normalized size = 0.76 \begin {gather*} \frac {(e x)^{7/2} \sqrt {a+b x^3} \left (6 a A b-3 a^2 B+12 A b^2 x^3+2 a b B x^3+8 b^2 B x^6\right )}{72 b^2 x^2}+\frac {a^2 (-2 A b+a B) (e x)^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {b} x^{3/2}}\right )}{24 b^{5/2} x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(7/2)*Sqrt[a + b*x^3]*(6*a*A*b - 3*a^2*B + 12*A*b^2*x^3 + 2*a*b*B*x^3 + 8*b^2*B*x^6))/(72*b^2*x^2) + (a
^2*(-2*A*b + a*B)*(e*x)^(7/2)*ArcTanh[Sqrt[a + b*x^3]/(Sqrt[b]*x^(3/2))])/(24*b^(5/2)*x^(7/2))

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 1.61, size = 7293, normalized size = 45.30

method	result	size
risch	\(\text {Expression too large to display}\)	\(1087\)
elliptic	\(\text {Expression too large to display}\)	\(1165\)
default	\(\text {Expression too large to display}\)	\(7293\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(7/2)*(B*x^3+A)*(b*x^3+a)^(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. 292 vs. \(2 (111) = 222\).
time = 0.49, size = 292, normalized size = 1.81 \begin {gather*} \frac {1}{144} \, {\left (6 \, {\left (\frac {a^{2} \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 {\sqrt {b x^{3} + a} a^{2} b}{x^{\frac {3}{2}}} + \frac {{\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2}}{x^{\frac {9}{2}}}\right )}}{b^{3} - \frac {2 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b}{x^{6}}}\right )} A - {\left (\frac {3 \, 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 )}{b^{\frac {5}{2}}} + \frac {2 \, {\left (\frac {3 \, \sqrt {b x^{3} + a} a^{3} b^{2}}{x^{\frac {3}{2}}} + \frac {8 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{3} b}{x^{\frac {9}{2}}} - \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{3}}{x^{\frac {15}{2}}}\right )}}{b^{5} - \frac {3 \, {\left (b x^{3} + a\right )} b^{4}}{x^{3}} + \frac {3 \, {\left (b x^{3} + a\right )}^{2} b^{3}}{x^{6}} - \frac {{\left (b x^{3} + a\right )}^{3} b^{2}}{x^{9}}}\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)*(b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

1/144*(6*(a^2*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*(sqrt(
b*x^3 + a)*a^2*b/x^(3/2) + (b*x^3 + a)^(3/2)*a^2/x^(9/2))/(b^3 - 2*(b*x^3 + a)*b^2/x^3 + (b*x^3 + a)^2*b/x^6))
*A - (3*a^3*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*(3*sqrt(
b*x^3 + a)*a^3*b^2/x^(3/2) + 8*(b*x^3 + a)^(3/2)*a^3*b/x^(9/2) - 3*(b*x^3 + a)^(5/2)*a^3/x^(15/2))/(b^5 - 3*(b
*x^3 + a)*b^4/x^3 + 3*(b*x^3 + a)^2*b^3/x^6 - (b*x^3 + a)^3*b^2/x^9))*B)*e^(7/2)

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Fricas [A]
time = 2.28, size = 256, normalized size = 1.59 \begin {gather*} \left [-\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} 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 (8 \, B b^{3} x^{7} + 2 \, {\left (B a b^{2} + 6 \, A b^{3}\right )} x^{4} - 3 \, {\left (B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {7}{2}}}{288 \, b^{3}}, -\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} 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 (8 \, B b^{3} x^{7} + 2 \, {\left (B a b^{2} + 6 \, A b^{3}\right )} x^{4} - 3 \, {\left (B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {7}{2}}}{144 \, 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)*(b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/288*(3*(B*a^3 - 2*A*a^2*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*(8*B*b^3*x^7 + 2*(B*a*b^2 + 6*A*b^3)*x^4 - 3*(B*a^2*b - 2*A*a*b^2)*x)*sqrt(b*x^3 +
a)*sqrt(x)*e^(7/2))/b^3, -1/144*(3*(B*a^3 - 2*A*a^2*b)*sqrt(-b)*arctan(2*sqrt(b*x^3 + a)*sqrt(-b)*x^(3/2)/(2*b
*x^3 + a))*e^(7/2) - 2*(8*B*b^3*x^7 + 2*(B*a*b^2 + 6*A*b^3)*x^4 - 3*(B*a^2*b - 2*A*a*b^2)*x)*sqrt(b*x^3 + a)*s
qrt(x)*e^(7/2))/b^3]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (141) = 282\).
time = 82.27, size = 292, normalized size = 1.81 \begin {gather*} \frac {A a^{\frac {3}{2}} e^{\frac {7}{2}} x^{\frac {3}{2}}}{12 b \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {A \sqrt {a} e^{\frac {7}{2}} x^{\frac {9}{2}}}{4 \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {A a^{2} e^{\frac {7}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{12 b^{\frac {3}{2}}} + \frac {A b e^{\frac {7}{2}} x^{\frac {15}{2}}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {B a^{\frac {5}{2}} e^{\frac {7}{2}} x^{\frac {3}{2}}}{24 b^{2} \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {B a^{\frac {3}{2}} e^{\frac {7}{2}} x^{\frac {9}{2}}}{72 b \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {5 B \sqrt {a} e^{\frac {7}{2}} x^{\frac {15}{2}}}{36 \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B a^{3} e^{\frac {7}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{24 b^{\frac {5}{2}}} + \frac {B b e^{\frac {7}{2}} x^{\frac {21}{2}}}{9 \sqrt {a} \sqrt {1 + \frac {b x^{3}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(7/2)*(B*x**3+A)*(b*x**3+a)**(1/2),x)

[Out]

A*a**(3/2)*e**(7/2)*x**(3/2)/(12*b*sqrt(1 + b*x**3/a)) + A*sqrt(a)*e**(7/2)*x**(9/2)/(4*sqrt(1 + b*x**3/a)) -
A*a**2*e**(7/2)*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(12*b**(3/2)) + A*b*e**(7/2)*x**(15/2)/(6*sqrt(a)*sqrt(1 + b*x
**3/a)) - B*a**(5/2)*e**(7/2)*x**(3/2)/(24*b**2*sqrt(1 + b*x**3/a)) - B*a**(3/2)*e**(7/2)*x**(9/2)/(72*b*sqrt(
1 + b*x**3/a)) + 5*B*sqrt(a)*e**(7/2)*x**(15/2)/(36*sqrt(1 + b*x**3/a)) + B*a**3*e**(7/2)*asinh(sqrt(b)*x**(3/
2)/sqrt(a))/(24*b**(5/2)) + B*b*e**(7/2)*x**(21/2)/(9*sqrt(a)*sqrt(1 + b*x**3/a))

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Giac [A]
time = 1.08, size = 201, normalized size = 1.25 \begin {gather*} \frac {1}{72} \, {\left (6 \, \sqrt {b x^{3} + a} {\left (2 \, x^{3} + \frac {a}{b}\right )} A x^{\frac {3}{2}} + {\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 x^{\frac {3}{2}}\right )} e^{\frac {7}{2}} - \frac {{\left (B^{2} a^{6} - 4 \, A B a^{5} b + 4 \, A^{2} a^{4} b^{2}\right )} e^{\frac {7}{2}} \log \left ({\left | -{\left (B a^{3} x^{\frac {3}{2}} - 2 \, A a^{2} b x^{\frac {3}{2}}\right )} \sqrt {b} + \sqrt {B^{2} a^{7} - 4 \, A B a^{6} b + 4 \, A^{2} a^{5} b^{2} + {\left (B a^{3} x^{\frac {3}{2}} - 2 \, A a^{2} b x^{\frac {3}{2}}\right )}^{2} b} \right |}\right )}{24 \, b^{\frac {5}{2}} {\left | -B a^{3} + 2 \, A a^{2} b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(B*x^3+A)*(b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

1/72*(6*sqrt(b*x^3 + a)*(2*x^3 + a/b)*A*x^(3/2) + (2*(4*x^3 + a/b)*x^3 - 3*a^2/b^2)*sqrt(b*x^3 + a)*B*x^(3/2))
*e^(7/2) - 1/24*(B^2*a^6 - 4*A*B*a^5*b + 4*A^2*a^4*b^2)*e^(7/2)*log(abs(-(B*a^3*x^(3/2) - 2*A*a^2*b*x^(3/2))*s
qrt(b) + sqrt(B^2*a^7 - 4*A*B*a^6*b + 4*A^2*a^5*b^2 + (B*a^3*x^(3/2) - 2*A*a^2*b*x^(3/2))^2*b)))/(b^(5/2)*abs(
-B*a^3 + 2*A*a^2*b))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (B\,x^3+A\right )\,{\left (e\,x\right )}^{7/2}\,\sqrt {b\,x^3+a} \,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)^(1/2),x)

[Out]

int((A + B*x^3)*(e*x)^(7/2)*(a + b*x^3)^(1/2), x)

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