3.6.19 \(\int \sqrt {e x} \sqrt {a+b x^3} (A+B x^3) \, dx\) [519]

Optimal. Leaf size=121 \[ \frac {(4 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 b e}+\frac {a (4 A b-a B) \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{12 b^{3/2}} \]

[Out]

1/6*B*(e*x)^(3/2)*(b*x^3+a)^(3/2)/b/e+1/12*a*(4*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)+1/12*(4*A*b-B*a)*(e*x)^(3/2)*(b*x^3+a)^(1/2)/b/e

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Rubi [A]
time = 0.06, antiderivative size = 121, 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 = {470, 285, 335, 281, 223, 212} \begin {gather*} \frac {a \sqrt {e} (4 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{12 b^{3/2}}+\frac {(e x)^{3/2} \sqrt {a+b x^3} (4 A b-a B)}{12 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A*b - a*B)*(e*x)^(3/2)*Sqrt[a + b*x^3])/(12*b*e) + (B*(e*x)^(3/2)*(a + b*x^3)^(3/2))/(6*b*e) + (a*(4*A*b -
a*B)*Sqrt[e]*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + b*x^3])])/(12*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} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx &=\frac {B (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 b e}-\frac {\left (-6 A b+\frac {3 a B}{2}\right ) \int \sqrt {e x} \sqrt {a+b x^3} \, dx}{6 b}\\ &=\frac {(4 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 b e}+\frac {(a (4 A b-a B)) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{8 b}\\ &=\frac {(4 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 b e}+\frac {(a (4 A b-a B)) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{4 b e}\\ &=\frac {(4 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 b e}+\frac {(a (4 A b-a B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^2}{e^3}}} \, dx,x,(e x)^{3/2}\right )}{12 b e}\\ &=\frac {(4 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 b e}+\frac {(a (4 A b-a B)) \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 )}{12 b e}\\ &=\frac {(4 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 b e}+\frac {a (4 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 )}{12 b^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 95, normalized size = 0.79 \begin {gather*} \frac {x \sqrt {e x} \sqrt {a+b x^3} \left (4 A b+a B+2 b B x^3\right )}{12 b}-\frac {a (-4 A b+a B) \sqrt {e x} \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {b} x^{3/2}}\right )}{12 b^{3/2} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[e*x]*Sqrt[a + b*x^3]*(4*A*b + a*B + 2*b*B*x^3))/(12*b) - (a*(-4*A*b + a*B)*Sqrt[e*x]*ArcTanh[Sqrt[a +
b*x^3]/(Sqrt[b]*x^(3/2))])/(12*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 = 0.35, size = 6858, normalized size = 56.68

method	result	size
risch	\(\text {Expression too large to display}\)	\(1055\)
elliptic	\(\text {Expression too large to display}\)	\(1096\)
default	\(\text {Expression too large to display}\)	\(6858\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)*(e*x)^(1/2)*(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. 207 vs. \(2 (81) = 162\).
time = 0.50, size = 207, normalized size = 1.71 \begin {gather*} -\frac {1}{24} \, {\left (4 \, {\left (\frac {a \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 \, \sqrt {b x^{3} + a} a}{{\left (b - \frac {b x^{3} + a}{x^{3}}\right )} x^{\frac {3}{2}}}\right )} A - {\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 )} B\right )} e^{\frac {1}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(4*(a*log(-(sqrt(b) - sqrt(b*x^3 + a)/x^(3/2))/(sqrt(b) + sqrt(b*x^3 + a)/x^(3/2)))/sqrt(b) + 2*sqrt(b*x
^3 + a)*a/((b - (b*x^3 + a)/x^3)*x^(3/2)))*A - (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))*B)*e^(1/2)

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Fricas [A]
time = 4.34, size = 206, normalized size = 1.70 \begin {gather*} \left [-\frac {{\left (B a^{2} - 4 \, A a 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 (2 \, B b^{2} x^{4} + {\left (B a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {1}{2}}}{48 \, b^{2}}, \frac {{\left (B a^{2} - 4 \, A a 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 (2 \, B b^{2} x^{4} + {\left (B a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {1}{2}}}{24 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*((B*a^2 - 4*A*a*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*(2*B*b^2*x^4 + (B*a*b + 4*A*b^2)*x)*sqrt(b*x^3 + a)*sqrt(x)*e^(1/2))/b^2, 1/24*((B*a^2 -
4*A*a*b)*sqrt(-b)*arctan(2*sqrt(b*x^3 + a)*sqrt(-b)*x^(3/2)/(2*b*x^3 + a))*e^(1/2) + 2*(2*B*b^2*x^4 + (B*a*b
+ 4*A*b^2)*x)*sqrt(b*x^3 + a)*sqrt(x)*e^(1/2))/b^2]

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Sympy [A]
time = 4.95, size = 201, normalized size = 1.66 \begin {gather*} \frac {A \sqrt {a} \left (e x\right )^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e} + \frac {A a \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {b} \left (e x\right )^{\frac {3}{2}}}{\sqrt {a} e^{\frac {3}{2}}} \right )}}{3 \sqrt {b}} + \frac {B a^{\frac {3}{2}} \left (e x\right )^{\frac {3}{2}}}{12 b e \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B \sqrt {a} \left (e x\right )^{\frac {9}{2}}}{4 e^{4} \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {B a^{2} \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {b} \left (e x\right )^{\frac {3}{2}}}{\sqrt {a} e^{\frac {3}{2}}} \right )}}{12 b^{\frac {3}{2}}} + \frac {B b \left (e x\right )^{\frac {15}{2}}}{6 \sqrt {a} e^{7} \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)*(e*x)**(1/2)*(b*x**3+a)**(1/2),x)

[Out]

A*sqrt(a)*(e*x)**(3/2)*sqrt(1 + b*x**3/a)/(3*e) + A*a*sqrt(e)*asinh(sqrt(b)*(e*x)**(3/2)/(sqrt(a)*e**(3/2)))/(
3*sqrt(b)) + B*a**(3/2)*(e*x)**(3/2)/(12*b*e*sqrt(1 + b*x**3/a)) + B*sqrt(a)*(e*x)**(9/2)/(4*e**4*sqrt(1 + b*x
**3/a)) - B*a**2*sqrt(e)*asinh(sqrt(b)*(e*x)**(3/2)/(sqrt(a)*e**(3/2)))/(12*b**(3/2)) + B*b*(e*x)**(15/2)/(6*s
qrt(a)*e**7*sqrt(1 + b*x**3/a))

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Giac [A]
time = 2.02, size = 105, normalized size = 0.87 \begin {gather*} \frac {B a^{2} e^{\frac {1}{2}} \log \left ({\left | -\sqrt {b} x^{\frac {3}{2}} + \sqrt {b x^{3} + a} \right |}\right )}{12 \, b^{\frac {3}{2}}} + \frac {1}{12} \, {\left (\sqrt {b x^{3} + a} {\left (2 \, x^{3} + \frac {a}{b}\right )} B x^{\frac {3}{2}} + 4 \, {\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\right )} e^{\frac {1}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*B*a^2*e^(1/2)*log(abs(-sqrt(b)*x^(3/2) + sqrt(b*x^3 + a)))/b^(3/2) + 1/12*(sqrt(b*x^3 + a)*(2*x^3 + a/b)*
B*x^(3/2) + 4*(sqrt(b*x^3 + a)*x^(3/2) - a*log(abs(-sqrt(b)*x^(3/2) + sqrt(b*x^3 + a)))/sqrt(b))*A)*e^(1/2)

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

[Out]

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

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