3.6.59 \(\int \frac {(e x)^{7/2} (A+B x^3)}{(a+b x^3)^{5/2}} \, dx\) [559]

Optimal. Leaf size=114 \[ \frac {2 (A b-a B) (e x)^{9/2}}{9 a b e \left (a+b x^3\right )^{3/2}}-\frac {2 B e^2 (e x)^{3/2}}{3 b^2 \sqrt {a+b x^3}}+\frac {2 B e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{5/2}} \]

[Out]

2/9*(A*b-B*a)*(e*x)^(9/2)/a/b/e/(b*x^3+a)^(3/2)+2/3*B*e^(7/2)*arctanh((e*x)^(3/2)*b^(1/2)/e^(3/2)/(b*x^3+a)^(1
/2))/b^(5/2)-2/3*B*e^2*(e*x)^(3/2)/b^2/(b*x^3+a)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 114, 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 = {463, 294, 335, 281, 223, 212} \begin {gather*} \frac {2 (e x)^{9/2} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}}+\frac {2 B e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{5/2}}-\frac {2 B e^2 (e x)^{3/2}}{3 b^2 \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e*x)^(7/2)*(A + B*x^3))/(a + b*x^3)^(5/2),x]

[Out]

(2*(A*b - a*B)*(e*x)^(9/2))/(9*a*b*e*(a + b*x^3)^(3/2)) - (2*B*e^2*(e*x)^(3/2))/(3*b^2*Sqrt[a + b*x^3]) + (2*B
*e^(7/2)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + b*x^3])])/(3*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 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(b*c - a*d)*
(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*(m + 1))), x] + Dist[d/b, Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /;
FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx &=\frac {2 (A b-a B) (e x)^{9/2}}{9 a b e \left (a+b x^3\right )^{3/2}}+\frac {B \int \frac {(e x)^{7/2}}{\left (a+b x^3\right )^{3/2}} \, dx}{b}\\ &=\frac {2 (A b-a B) (e x)^{9/2}}{9 a b e \left (a+b x^3\right )^{3/2}}-\frac {2 B e^2 (e x)^{3/2}}{3 b^2 \sqrt {a+b x^3}}+\frac {\left (B e^3\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{b^2}\\ &=\frac {2 (A b-a B) (e x)^{9/2}}{9 a b e \left (a+b x^3\right )^{3/2}}-\frac {2 B e^2 (e x)^{3/2}}{3 b^2 \sqrt {a+b x^3}}+\frac {\left (2 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 )}{b^2}\\ &=\frac {2 (A b-a B) (e x)^{9/2}}{9 a b e \left (a+b x^3\right )^{3/2}}-\frac {2 B e^2 (e x)^{3/2}}{3 b^2 \sqrt {a+b x^3}}+\frac {\left (2 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 )}{3 b^2}\\ &=\frac {2 (A b-a B) (e x)^{9/2}}{9 a b e \left (a+b x^3\right )^{3/2}}-\frac {2 B e^2 (e x)^{3/2}}{3 b^2 \sqrt {a+b x^3}}+\frac {\left (2 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 )}{3 b^2}\\ &=\frac {2 (A b-a B) (e x)^{9/2}}{9 a b e \left (a+b x^3\right )^{3/2}}-\frac {2 B e^2 (e x)^{3/2}}{3 b^2 \sqrt {a+b x^3}}+\frac {2 B e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.50, size = 99, normalized size = 0.87 \begin {gather*} \frac {2 e^3 \sqrt {e x} \left (\frac {\sqrt {b} x^{3/2} \left (-3 a^2 B+A b^2 x^3-4 a b B x^3\right )}{a \left (a+b x^3\right )^{3/2}}+3 B \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {b} x^{3/2}}\right )\right )}{9 b^{5/2} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(7/2)*(A + B*x^3))/(a + b*x^3)^(5/2),x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.32, size = 7081, normalized size = 62.11

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(1098\)
default	\(\text {Expression too large to display}\)	\(7081\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [A]
time = 0.50, size = 103, normalized size = 0.90 \begin {gather*} \frac {1}{9} \, {\left (\frac {2 \, A x^{\frac {9}{2}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} a} - {\left (\frac {2 \, {\left (b + \frac {3 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} x^{\frac {9}{2}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2}} + \frac {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}}}\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)^(5/2),x, algorithm="maxima")

[Out]

1/9*(2*A*x^(9/2)/((b*x^3 + a)^(3/2)*a) - (2*(b + 3*(b*x^3 + a)/x^3)*x^(9/2)/((b*x^3 + a)^(3/2)*b^2) + 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))*B)*e^(7/2)

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Fricas [A]
time = 3.86, size = 296, normalized size = 2.60 \begin {gather*} \left [\frac {3 \, {\left (B a b^{2} x^{6} + 2 \, B a^{2} b x^{3} + B a^{3}\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 (3 \, B a^{2} b x + {\left (4 \, B a b^{2} - A b^{3}\right )} x^{4}\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {7}{2}}}{18 \, {\left (a b^{5} x^{6} + 2 \, a^{2} b^{4} x^{3} + a^{3} b^{3}\right )}}, -\frac {3 \, {\left (B a b^{2} x^{6} + 2 \, B a^{2} b x^{3} + B a^{3}\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 (3 \, B a^{2} b x + {\left (4 \, B a b^{2} - A b^{3}\right )} x^{4}\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {7}{2}}}{9 \, {\left (a b^{5} x^{6} + 2 \, a^{2} b^{4} x^{3} + a^{3} b^{3}\right )}}\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)^(5/2),x, algorithm="fricas")

[Out]

[1/18*(3*(B*a*b^2*x^6 + 2*B*a^2*b*x^3 + B*a^3)*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*(3*B*a^2*b*x + (4*B*a*b^2 - A*b^3)*x^4)*sqrt(b*x^3 + a)*sqrt(x)*e^(
7/2))/(a*b^5*x^6 + 2*a^2*b^4*x^3 + a^3*b^3), -1/9*(3*(B*a*b^2*x^6 + 2*B*a^2*b*x^3 + B*a^3)*sqrt(-b)*arctan(2*s
qrt(b*x^3 + a)*sqrt(-b)*x^(3/2)/(2*b*x^3 + a))*e^(7/2) + 2*(3*B*a^2*b*x + (4*B*a*b^2 - A*b^3)*x^4)*sqrt(b*x^3
+ a)*sqrt(x)*e^(7/2))/(a*b^5*x^6 + 2*a^2*b^4*x^3 + a^3*b^3)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 1.48, size = 82, normalized size = 0.72 \begin {gather*} -\frac {2 \, x^{\frac {3}{2}} {\left (\frac {3 \, B a}{b^{2}} + \frac {{\left (4 \, B a^{5} b^{6} - A a^{4} b^{7}\right )} x^{3}}{a^{5} b^{7}}\right )} e^{\frac {7}{2}}}{9 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}}} - \frac {2 \, B e^{\frac {7}{2}} \log \left ({\left | -\sqrt {b} x^{\frac {3}{2}} + \sqrt {b x^{3} + a} \right |}\right )}{3 \, b^{\frac {5}{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)^(5/2),x, algorithm="giac")

[Out]

-2/9*x^(3/2)*(3*B*a/b^2 + (4*B*a^5*b^6 - A*a^4*b^7)*x^3/(a^5*b^7))*e^(7/2)/(b*x^3 + a)^(3/2) - 2/3*B*e^(7/2)*l
og(abs(-sqrt(b)*x^(3/2) + sqrt(b*x^3 + a)))/b^(5/2)

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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}}{{\left (b\,x^3+a\right )}^{5/2}} \,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)^(5/2),x)

[Out]

int(((A + B*x^3)*(e*x)^(7/2))/(a + b*x^3)^(5/2), x)

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