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

Optimal. Leaf size=121 \[ -\frac {a e^{7/2} (4 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{12 b^{5/2}}+\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3} (4 A b-3 a B)}{12 b^2}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e} \]

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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, 321, 329, 275, 217, 206} \begin {gather*} \frac {e^2 (e x)^{3/2} \sqrt {a+b x^3} (4 A b-3 a B)}{12 b^2}-\frac {a e^{7/2} (4 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{12 b^{5/2}}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A*b - 3*a*B)*e^2*(e*x)^(3/2)*Sqrt[a + b*x^3])/(12*b^2) + (B*(e*x)^(9/2)*Sqrt[a + b*x^3])/(6*b*e) - (a*(4*A
*b - 3*a*B)*e^(7/2)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + b*x^3])])/(12*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 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 \frac {(e x)^{7/2} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx &=\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}-\frac {\left (-6 A b+\frac {9 a B}{2}\right ) \int \frac {(e x)^{7/2}}{\sqrt {a+b x^3}} \, dx}{6 b}\\ &=\frac {(4 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{12 b^2}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}-\frac {\left (a (4 A b-3 a B) e^3\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{8 b^2}\\ &=\frac {(4 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{12 b^2}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}-\frac {\left (a (4 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 )}{4 b^2}\\ &=\frac {(4 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{12 b^2}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}-\frac {\left (a (4 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 )}{12 b^2}\\ &=\frac {(4 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{12 b^2}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}-\frac {\left (a (4 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 )}{12 b^2}\\ &=\frac {(4 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{12 b^2}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}-\frac {a (4 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 )}{12 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 97, normalized size = 0.80 \begin {gather*} \frac {e^3 \sqrt {e x} \left (\sqrt {b} x^{3/2} \sqrt {a+b x^3} \left (-3 a B+4 A b+2 b B x^3\right )+a (3 a B-4 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a+b x^3}}\right )\right )}{12 b^{5/2} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*Sqrt[e*x]*(Sqrt[b]*x^(3/2)*Sqrt[a + b*x^3]*(4*A*b - 3*a*B + 2*b*B*x^3) + a*(-4*A*b + 3*a*B)*ArcTanh[(Sqrt
[b]*x^(3/2))/Sqrt[a + b*x^3]]))/(12*b^(5/2)*Sqrt[x])

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IntegrateAlgebraic [A]  time = 1.06, size = 124, normalized size = 1.02 \begin {gather*} \frac {e^5 \sqrt {\frac {b}{e^3}} \left (4 a A b-3 a^2 B\right ) \log \left (\sqrt {a+b x^3}-\sqrt {\frac {b}{e^3}} (e x)^{3/2}\right )}{12 b^3}+\frac {\sqrt {a+b x^3} \left (-3 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^2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^3]*(4*A*b*e^3*(e*x)^(3/2) - 3*a*B*e^3*(e*x)^(3/2) + 2*b*B*(e*x)^(9/2)))/(12*b^2*e) + ((4*a*A*b -
3*a^2*B)*Sqrt[b/e^3]*e^5*Log[-(Sqrt[b/e^3]*(e*x)^(3/2)) + Sqrt[a + b*x^3]])/(12*b^3)

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fricas [A]  time = 1.44, size = 245, normalized size = 2.02 \begin {gather*} \left [-\frac {{\left (3 \, B a^{2} - 4 \, A a 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 (2 \, B b e^{3} x^{4} - {\left (3 \, B a - 4 \, A b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{48 \, b^{2}}, -\frac {{\left (3 \, B a^{2} - 4 \, A a 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 (2 \, B b e^{3} x^{4} - {\left (3 \, B a - 4 \, A b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{24 \, 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)/(b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.36, size = 114, normalized size = 0.94 \begin {gather*} \frac {1}{12} \, \sqrt {b x^{3} e^{4} + a e^{4}} {\left (\frac {2 \, B x^{3} e^{\left (-2\right )}}{b} - \frac {{\left (3 \, B a b^{3} e^{5} - 4 \, A b^{4} e^{5}\right )} e^{\left (-7\right )}}{b^{5}}\right )} x^{\frac {3}{2}} e^{\frac {7}{2}} - \frac {{\left (3 \, B a^{2} b^{3} e^{9} - 4 \, A a b^{4} e^{9}\right )} e^{\left (-\frac {11}{2}\right )} \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 {11}{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="giac")

[Out]

1/12*sqrt(b*x^3*e^4 + a*e^4)*(2*B*x^3*e^(-2)/b - (3*B*a*b^3*e^5 - 4*A*b^4*e^5)*e^(-7)/b^5)*x^(3/2)*e^(7/2) - 1
/12*(3*B*a^2*b^3*e^9 - 4*A*a*b^4*e^9)*e^(-11/2)*log(abs(-sqrt(b)*x^(3/2)*e^2 + sqrt(b*x^3*e^4 + a*e^4)))/b^(11
/2)

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maple [C]  time = 1.01, size = 6861, normalized size = 56.70 \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)/(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 \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{\frac {7}{2}}}{\sqrt {b x^{3} + a}}\,{d x} \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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\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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sympy [A]  time = 95.30, size = 194, normalized size = 1.60 \begin {gather*} \frac {A \sqrt {a} e^{\frac {7}{2}} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 b} - \frac {A a e^{\frac {7}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{3 b^{\frac {3}{2}}} - \frac {B a^{\frac {3}{2}} e^{\frac {7}{2}} x^{\frac {3}{2}}}{4 b^{2} \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {B \sqrt {a} e^{\frac {7}{2}} x^{\frac {9}{2}}}{12 b \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B a^{2} e^{\frac {7}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{4 b^{\frac {5}{2}}} + \frac {B e^{\frac {7}{2}} x^{\frac {15}{2}}}{6 \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*sqrt(a)*e**(7/2)*x**(3/2)*sqrt(1 + b*x**3/a)/(3*b) - A*a*e**(7/2)*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(3*b**(3/2
)) - B*a**(3/2)*e**(7/2)*x**(3/2)/(4*b**2*sqrt(1 + b*x**3/a)) - B*sqrt(a)*e**(7/2)*x**(9/2)/(12*b*sqrt(1 + b*x
**3/a)) + B*a**2*e**(7/2)*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(4*b**(5/2)) + B*e**(7/2)*x**(15/2)/(6*sqrt(a)*sqrt(
1 + b*x**3/a))

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