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

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

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Rubi [A]  time = 0.09, 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 = {459, 288, 329, 275, 217, 206} \begin {gather*} -\frac {e^2 (e x)^{3/2} (2 A b-3 a B)}{3 b^2 \sqrt {a+b x^3}}+\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 {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]

-((2*A*b - 3*a*B)*e^2*(e*x)^(3/2))/(3*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 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 288

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 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 )}{\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 ) \operatorname {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 ) \operatorname {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 ) \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 )}{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 = 0.18, size = 109, normalized size = 0.91 \begin {gather*} \frac {e^3 \sqrt {e x} \left (\sqrt {b} x^{3/2} \left (3 a B-2 A b+b B x^3\right )-\sqrt {a} \sqrt {\frac {b x^3}{a}+1} (3 a B-2 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )\right )}{3 b^{5/2} \sqrt {x} \sqrt {a+b x^3}} \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^3*Sqrt[e*x]*(Sqrt[b]*x^(3/2)*(-2*A*b + 3*a*B + b*B*x^3) - Sqrt[a]*(-2*A*b + 3*a*B)*Sqrt[1 + (b*x^3)/a]*ArcS
inh[(Sqrt[b]*x^(3/2))/Sqrt[a]]))/(3*b^(5/2)*Sqrt[x]*Sqrt[a + b*x^3])

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IntegrateAlgebraic [A]  time = 1.76, size = 136, normalized size = 1.13 \begin {gather*} \frac {\sqrt {a+b x^3} \left (3 a B e^5 (e x)^{3/2}-2 A b e^5 (e x)^{3/2}+b B e^2 (e x)^{9/2}\right )}{3 b^2 \left (a e^3+b e^3 x^3\right )}-\frac {e^5 \sqrt {\frac {b}{e^3}} (2 A b-3 a B) \log \left (\sqrt {a+b x^3}-\sqrt {\frac {b}{e^3}} (e x)^{3/2}\right )}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.92, size = 307, normalized size = 2.56 \begin {gather*} \left [-\frac {{\left ({\left (3 \, B a b - 2 \, A b^{2}\right )} e^{3} x^{3} + {\left (3 \, B a^{2} - 2 \, A a b\right )} e^{3}\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 (B b e^{3} x^{4} + {\left (3 \, B a - 2 \, A b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{12 \, {\left (b^{3} x^{3} + a b^{2}\right )}}, \frac {{\left ({\left (3 \, B a b - 2 \, A b^{2}\right )} e^{3} x^{3} + {\left (3 \, B a^{2} - 2 \, A a b\right )} e^{3}\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 (B b e^{3} x^{4} + {\left (3 \, B a - 2 \, A b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{6 \, {\left (b^{3} x^{3} + a b^{2}\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)*e^3*x^3 + (3*B*a^2 - 2*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*(B*b*e^3*x^4 + (3*B*a - 2*A*b)*e^3*x)*sqrt
(b*x^3 + a)*sqrt(e*x))/(b^3*x^3 + a*b^2), 1/6*(((3*B*a*b - 2*A*b^2)*e^3*x^3 + (3*B*a^2 - 2*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*(B*b*e^3*x^4 + (3*B*a - 2*A*b)*e^
3*x)*sqrt(b*x^3 + a)*sqrt(e*x))/(b^3*x^3 + a*b^2)]

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giac [A]  time = 0.37, size = 107, normalized size = 0.89 \begin {gather*} \frac {{\left (\frac {B x^{3} e^{4}}{b} + \frac {3 \, B a b^{3} e^{4} - 2 \, A b^{4} e^{4}}{b^{5}}\right )} x^{\frac {3}{2}} e^{\frac {3}{2}}}{3 \, \sqrt {b x^{3} e^{4} + a e^{4}}} + \frac {{\left (3 \, B a b^{3} e^{4} - 2 \, A b^{4} e^{4}\right )} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -\sqrt {b} x^{\frac {3}{2}} e^{2} + \sqrt {b x^{3} e^{4} + a e^{4}} \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*e^4/b + (3*B*a*b^3*e^4 - 2*A*b^4*e^4)/b^5)*x^(3/2)*e^(3/2)/sqrt(b*x^3*e^4 + a*e^4) + 1/3*(3*B*a*b^3
*e^4 - 2*A*b^4*e^4)*e^(-1/2)*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 = 1.13, size = 7016, normalized size = 58.47 \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)^(3/2),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}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}}\,{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)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)*(e*x)^(7/2)/(b*x^3 + a)^(3/2), 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}}{{\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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sympy [F(-1)]  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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