3.6.51 \(\int \frac {(e x)^{7/2} (A+B x^3)}{(a+b x^3)^{3/2}} \, dx\) [551]

Optimal. Leaf size=120 \[ -\frac {(2 A b-3 a B) e^2 (e x)^{3/2}}{3 b^2 \sqrt {a+b x^3}}+\frac {B (e x)^{9/2}}{3 b e \sqrt {a+b x^3}}+\frac {(2 A b-3 a B) e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 120, 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, 294, 335, 281, 223, 212} \begin {gather*} \frac {e^{7/2} (2 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{5/2}}-\frac {e^2 (e x)^{3/2} (2 A b-3 a B)}{3 b^2 \sqrt {a+b x^3}}+\frac {B (e x)^{9/2}}{3 b e \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)^(3/2),x]

[Out]

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

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Mathematica [A]
time = 0.39, size = 92, normalized size = 0.77 \begin {gather*} \frac {(e x)^{7/2} \left (\frac {\sqrt {b} x^{3/2} \left (-2 A b+3 a B+b B x^3\right )}{\sqrt {a+b x^3}}+(2 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {b} x^{3/2}}\right )\right )}{3 b^{5/2} x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(7/2)*((Sqrt[b]*x^(3/2)*(-2*A*b + 3*a*B + b*B*x^3))/Sqrt[a + b*x^3] + (2*A*b - 3*a*B)*ArcTanh[Sqrt[a +
b*x^3]/(Sqrt[b]*x^(3/2))]))/(3*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 = 0.38, size = 7016, normalized size = 58.47

method	result	size
risch	\(\text {Expression too large to display}\)	\(1079\)
elliptic	\(\text {Expression too large to display}\)	\(1097\)
default	\(\text {Expression too large to display}\)	\(7016\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(3/2),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. 175 vs. \(2 (82) = 164\).
time = 0.54, size = 175, normalized size = 1.46 \begin {gather*} \frac {1}{6} \, {\left (B {\left (\frac {2 \, {\left (2 \, a b - \frac {3 \, {\left (b x^{3} + a\right )} a}{x^{3}}\right )}}{\frac {\sqrt {b x^{3} + a} b^{3}}{x^{\frac {3}{2}}} - \frac {{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {9}{2}}}} + \frac {3 \, 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 {5}{2}}}\right )} - 2 \, A {\left (\frac {2 \, x^{\frac {3}{2}}}{\sqrt {b x^{3} + a} b} + \frac {\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}}}\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)^(3/2),x, algorithm="maxima")

[Out]

1/6*(B*(2*(2*a*b - 3*(b*x^3 + a)*a/x^3)/(sqrt(b*x^3 + a)*b^3/x^(3/2) - (b*x^3 + a)^(3/2)*b^2/x^(9/2)) + 3*a*lo
g(-(sqrt(b) - sqrt(b*x^3 + a)/x^(3/2))/(sqrt(b) + sqrt(b*x^3 + a)/x^(3/2)))/b^(5/2)) - 2*A*(2*x^(3/2)/(sqrt(b*
x^3 + a)*b) + log(-(sqrt(b) - sqrt(b*x^3 + a)/x^(3/2))/(sqrt(b) + sqrt(b*x^3 + a)/x^(3/2)))/b^(3/2)))*e^(7/2)

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Fricas [A]
time = 2.42, size = 264, normalized size = 2.20 \begin {gather*} \left [-\frac {{\left ({\left (3 \, B a b - 2 \, A b^{2}\right )} x^{3} + 3 \, B a^{2} - 2 \, 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 (B b^{2} x^{4} + {\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {7}{2}}}{12 \, {\left (b^{4} x^{3} + a b^{3}\right )}}, \frac {{\left ({\left (3 \, B a b - 2 \, A b^{2}\right )} x^{3} + 3 \, B a^{2} - 2 \, 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 (B b^{2} x^{4} + {\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\frac {7}{2}}}{6 \, {\left (b^{4} x^{3} + a 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)^(3/2),x, algorithm="fricas")

[Out]

[-1/12*(((3*B*a*b - 2*A*b^2)*x^3 + 3*B*a^2 - 2*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)*sqrt(b)*sqrt(x) - a^2) - 4*(B*b^2*x^4 + (3*B*a*b - 2*A*b^2)*x)*sqrt(b*x^3 + a)*sqrt(x)*
e^(7/2))/(b^4*x^3 + a*b^3), 1/6*(((3*B*a*b - 2*A*b^2)*x^3 + 3*B*a^2 - 2*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*(B*b^2*x^4 + (3*B*a*b - 2*A*b^2)*x)*sqrt(b*x^3 + a)*sqrt(x)*e^(
7/2))/(b^4*x^3 + a*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)**(3/2),x)

[Out]

Timed out

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Giac [A]
time = 1.59, size = 85, normalized size = 0.71 \begin {gather*} \frac {{\left (\frac {B x^{3}}{b} + \frac {3 \, B a b^{3} - 2 \, A b^{4}}{b^{5}}\right )} x^{\frac {3}{2}} e^{\frac {7}{2}}}{3 \, \sqrt {b x^{3} + a}} + \frac {{\left (3 \, B a b^{3} - 2 \, A b^{4}\right )} e^{\frac {7}{2}} \log \left ({\left | -\sqrt {b} x^{\frac {3}{2}} + \sqrt {b x^{3} + a} \right |}\right )}{3 \, 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)^(3/2),x, algorithm="giac")

[Out]

1/3*(B*x^3/b + (3*B*a*b^3 - 2*A*b^4)/b^5)*x^(3/2)*e^(7/2)/sqrt(b*x^3 + a) + 1/3*(3*B*a*b^3 - 2*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}}{{\left (b\,x^3+a\right )}^{3/2}} \,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)^(3/2),x)

[Out]

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

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