3.6.43 \(\int \frac {(e x)^{7/2} (A+B x^3)}{\sqrt {a+b x^3}} \, dx\) [543]

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

[Out]

-1/12*a*(4*A*b-3*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)+1/12*(4*A*b-3*B*a)*
e^2*(e*x)^(3/2)*(b*x^3+a)^(1/2)/b^2+1/6*B*(e*x)^(9/2)*(b*x^3+a)^(1/2)/b/e

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Rubi [A]
time = 0.18, 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, 327, 335, 281, 223, 212} \begin {gather*} -\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} \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 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 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 \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 ) \text {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 ) \text {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 ) \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^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 = 0.30, size = 99, normalized size = 0.82 \begin {gather*} \frac {(e x)^{7/2} \sqrt {a+b x^3} \left (4 A b-3 a B+2 b B x^3\right )}{12 b^2 x^2}+\frac {a (-4 A b+3 a B) (e x)^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {b} x^{3/2}}\right )}{12 b^{5/2} x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(7/2)*Sqrt[a + b*x^3]*(4*A*b - 3*a*B + 2*b*B*x^3))/(12*b^2*x^2) + (a*(-4*A*b + 3*a*B)*(e*x)^(7/2)*ArcTa
nh[Sqrt[a + b*x^3]/(Sqrt[b]*x^(3/2))])/(12*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 = 0.36, size = 6861, normalized size = 56.70

method	result	size
risch	\(\text {Expression too large to display}\)	\(1063\)
elliptic	\(\text {Expression too large to display}\)	\(1093\)
default	\(\text {Expression too large to display}\)	\(6861\)

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. 214 vs. \(2 (83) = 166\).
time = 0.51, size = 214, normalized size = 1.77 \begin {gather*} -\frac {1}{24} \, {\left (B {\left (\frac {3 \, 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 {5}{2}}} - \frac {2 \, {\left (\frac {5 \, \sqrt {b x^{3} + a} a^{2} b}{x^{\frac {3}{2}}} - \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2}}{x^{\frac {9}{2}}}\right )}}{b^{4} - \frac {2 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}}}\right )} - 4 \, A {\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 )}{b^{\frac {3}{2}}} - \frac {2 \, \sqrt {b x^{3} + a} a}{{\left (b^{2} - \frac {{\left (b x^{3} + a\right )} b}{x^{3}}\right )} x^{\frac {3}{2}}}\right )}\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]

-1/24*(B*(3*a^2*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*(5*s
qrt(b*x^3 + a)*a^2*b/x^(3/2) - 3*(b*x^3 + a)^(3/2)*a^2/x^(9/2))/(b^4 - 2*(b*x^3 + a)*b^3/x^3 + (b*x^3 + a)^2*b
^2/x^6)) - 4*A*(a*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*sq
rt(b*x^3 + a)*a/((b^2 - (b*x^3 + a)*b/x^3)*x^(3/2))))*e^(7/2)

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Fricas [A]
time = 1.30, size = 212, normalized size = 1.75 \begin {gather*} \left [-\frac {{\left (3 \, B a^{2} - 4 \, A a 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 (2 \, B b^{2} x^{4} - {\left (3 \, B a b - 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {7}{2}}}{48 \, b^{3}}, -\frac {{\left (3 \, 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 {7}{2}} - 2 \, {\left (2 \, B b^{2} x^{4} - {\left (3 \, B a b - 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {7}{2}}}{24 \, 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]

[-1/48*((3*B*a^2 - 4*A*a*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)*sqr
t(b)*sqrt(x) - a^2) - 4*(2*B*b^2*x^4 - (3*B*a*b - 4*A*b^2)*x)*sqrt(b*x^3 + a)*sqrt(x)*e^(7/2))/b^3, -1/24*((3*
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^(7/2) - 2*(2*B*b^2*x^4 -
(3*B*a*b - 4*A*b^2)*x)*sqrt(b*x^3 + a)*sqrt(x)*e^(7/2))/b^3]

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Sympy [A]
time = 59.46, 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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Giac [A]
time = 1.48, size = 90, normalized size = 0.74 \begin {gather*} \frac {1}{12} \, \sqrt {b x^{3} + a} {\left (\frac {2 \, B x^{3}}{b} - \frac {3 \, B a b^{3} - 4 \, A b^{4}}{b^{5}}\right )} x^{\frac {3}{2}} e^{\frac {7}{2}} - \frac {{\left (3 \, B a^{2} b^{3} - 4 \, A a b^{4}\right )} e^{\frac {7}{2}} \log \left ({\left | -\sqrt {b} x^{\frac {3}{2}} + \sqrt {b x^{3} + a} \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 + a)*(2*B*x^3/b - (3*B*a*b^3 - 4*A*b^4)/b^5)*x^(3/2)*e^(7/2) - 1/12*(3*B*a^2*b^3 - 4*A*a*b^4)*
e^(7/2)*log(abs(-sqrt(b)*x^(3/2) + sqrt(b*x^3 + a)))/b^(11/2)

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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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