3.4.26 \(\int \sqrt {e x} \sqrt {a+b x^3} (A+B x^3) \, dx\)

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

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Rubi [A]  time = 0.09, 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 = {459, 279, 329, 275, 217, 206} \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 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 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 \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)) \operatorname {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)) \operatorname {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)) \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 )}{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 = 0.15, size = 119, normalized size = 0.98 \begin {gather*} \frac {\sqrt {e x} \sqrt {a+b x^3} \left (\sqrt {b} x^{3/2} \sqrt {\frac {b x^3}{a}+1} \left (B \left (a+2 b x^3\right )+4 A b\right )-\sqrt {a} (a B-4 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )\right )}{12 b^{3/2} \sqrt {x} \sqrt {\frac {b x^3}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e*x]*Sqrt[a + b*x^3]*(Sqrt[b]*x^(3/2)*Sqrt[1 + (b*x^3)/a]*(4*A*b + B*(a + 2*b*x^3)) - Sqrt[a]*(-4*A*b +
a*B)*ArcSinh[(Sqrt[b]*x^(3/2))/Sqrt[a]]))/(12*b^(3/2)*Sqrt[x]*Sqrt[1 + (b*x^3)/a])

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IntegrateAlgebraic [A]  time = 0.52, size = 123, normalized size = 1.02 \begin {gather*} \frac {\sqrt {a+b x^3} \left (a B e^3 (e x)^{3/2}+4 A b e^3 (e x)^{3/2}+2 b B (e x)^{9/2}\right )}{12 b e^4}-\frac {e^2 \sqrt {\frac {b}{e^3}} \left (4 a A b-a^2 B\right ) \log \left (\sqrt {a+b x^3}-\sqrt {\frac {b}{e^3}} (e x)^{3/2}\right )}{12 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^3]*(4*A*b*e^3*(e*x)^(3/2) + a*B*e^3*(e*x)^(3/2) + 2*b*B*(e*x)^(9/2)))/(12*b*e^4) - ((4*a*A*b - a
^2*B)*Sqrt[b/e^3]*e^2*Log[-(Sqrt[b/e^3]*(e*x)^(3/2)) + Sqrt[a + b*x^3]])/(12*b^2)

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fricas [A]  time = 1.16, size = 221, normalized size = 1.83 \begin {gather*} \left [-\frac {{\left (B a^{2} - 4 \, A a b\right )} \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 (2 \, B b x^{4} + {\left (B a + 4 \, A b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{48 \, b}, \frac {{\left (B a^{2} - 4 \, A a b\right )} \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 (2 \, B b x^{4} + {\left (B a + 4 \, A b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{24 \, b}\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(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*(2*B*b*x^4 + (B*a + 4*A*b)*x)*sqrt(b*x^3 + a)*sqrt(e*x))/b, 1/24*((B*a^2 - 4*A*a
*b)*sqrt(-e/b)*arctan(2*sqrt(b*x^3 + a)*sqrt(e*x)*b*x*sqrt(-e/b)/(2*b*e*x^3 + a*e)) + 2*(2*B*b*x^4 + (B*a + 4*
A*b)*x)*sqrt(b*x^3 + a)*sqrt(e*x))/b]

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giac [A]  time = 0.42, size = 137, normalized size = 1.13 \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 )} B x^{\frac {3}{2}} e^{\left (-\frac {1}{2}\right )} + \frac {B a^{2} e^{\frac {1}{2}} \log \left ({\left | -\sqrt {b} x^{\frac {3}{2}} e^{2} + \sqrt {b x^{3} e^{4} + a e^{4}} \right |}\right )}{12 \, b^{\frac {3}{2}}} + \frac {1}{3} \, {\left (\sqrt {b x^{3} e^{4} + a e^{4}} x^{\frac {3}{2}} e^{\frac {3}{2}} - \frac {a e^{\frac {7}{2}} \log \left ({\left | -\sqrt {b} x^{\frac {3}{2}} e^{2} + \sqrt {b x^{3} e^{4} + a e^{4}} \right |}\right )}{\sqrt {b}}\right )} A e^{\left (-3\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="giac")

[Out]

1/12*sqrt(b*x^3*e^4 + a*e^4)*(2*x^3*e^(-1) + a*e^(-1)/b)*B*x^(3/2)*e^(-1/2) + 1/12*B*a^2*e^(1/2)*log(abs(-sqrt
(b)*x^(3/2)*e^2 + sqrt(b*x^3*e^4 + a*e^4)))/b^(3/2) + 1/3*(sqrt(b*x^3*e^4 + a*e^4)*x^(3/2)*e^(3/2) - a*e^(7/2)
*log(abs(-sqrt(b)*x^(3/2)*e^2 + sqrt(b*x^3*e^4 + a*e^4)))/sqrt(b))*A*e^(-3)

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maple [C]  time = 1.09, size = 6858, normalized size = 56.68 \begin {gather*} \text {output too large to display} \end {gather*}

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)

[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 )} \sqrt {b x^{3} + a} \sqrt {e x}\,{d x} \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]

integrate((B*x^3 + A)*sqrt(b*x^3 + a)*sqrt(e*x), x)

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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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sympy [A]  time = 9.67, 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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