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

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

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Rubi [A]  time = 0.08, 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 = {452, 288, 329, 275, 217, 206} \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^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 {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 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 452

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 ) \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-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 ) \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-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 ) \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-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 = 0.30, size = 119, normalized size = 1.04 \begin {gather*} \frac {2 e^3 \sqrt {e x} \left (3 a^{3/2} B \left (a+b x^3\right ) \sqrt {\frac {b x^3}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )+\sqrt {b} x^{3/2} \left (-3 a^2 B-4 a b B x^3+A b^2 x^3\right )\right )}{9 a b^{5/2} \sqrt {x} \left (a+b x^3\right )^{3/2}} \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) + 3*a^(3/2)*B*(a + b*x^3)*Sqrt[1 + (b*x
^3)/a]*ArcSinh[(Sqrt[b]*x^(3/2))/Sqrt[a]]))/(9*a*b^(5/2)*Sqrt[x]*(a + b*x^3)^(3/2))

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IntegrateAlgebraic [A]  time = 1.91, size = 137, normalized size = 1.20 \begin {gather*} -\frac {2 \sqrt {a+b x^3} \left (3 a^2 B e^8 (e x)^{3/2}+4 a b B e^5 (e x)^{9/2}-A b^2 e^5 (e x)^{9/2}\right )}{9 a b^2 \left (a e^3+b e^3 x^3\right )^2}-\frac {2 B e^5 \sqrt {\frac {b}{e^3}} \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)^(5/2),x]

[Out]

(-2*Sqrt[a + b*x^3]*(3*a^2*B*e^8*(e*x)^(3/2) - A*b^2*e^5*(e*x)^(9/2) + 4*a*b*B*e^5*(e*x)^(9/2)))/(9*a*b^2*(a*e
^3 + b*e^3*x^3)^2) - (2*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 = 0.84, size = 345, normalized size = 3.03 \begin {gather*} \left [\frac {3 \, {\left (B a b^{2} e^{3} x^{6} + 2 \, B a^{2} b e^{3} x^{3} + B a^{3} 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 ({\left (4 \, B a b - A b^{2}\right )} e^{3} x^{4} + 3 \, B a^{2} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{18 \, {\left (a b^{4} x^{6} + 2 \, a^{2} b^{3} x^{3} + a^{3} b^{2}\right )}}, -\frac {3 \, {\left (B a b^{2} e^{3} x^{6} + 2 \, B a^{2} b e^{3} x^{3} + B a^{3} 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 ({\left (4 \, B a b - A b^{2}\right )} e^{3} x^{4} + 3 \, B a^{2} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{9 \, {\left (a b^{4} x^{6} + 2 \, a^{2} b^{3} x^{3} + a^{3} 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)^(5/2),x, algorithm="fricas")

[Out]

[1/18*(3*(B*a*b^2*e^3*x^6 + 2*B*a^2*b*e^3*x^3 + B*a^3*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*((4*B*a*b - A*b^2)*e^3*x^4 + 3*B*a^2*e^3*x)*sqr
t(b*x^3 + a)*sqrt(e*x))/(a*b^4*x^6 + 2*a^2*b^3*x^3 + a^3*b^2), -1/9*(3*(B*a*b^2*e^3*x^6 + 2*B*a^2*b*e^3*x^3 +
B*a^3*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*((4*B*a*b - A*b
^2)*e^3*x^4 + 3*B*a^2*e^3*x)*sqrt(b*x^3 + a)*sqrt(e*x))/(a*b^4*x^6 + 2*a^2*b^3*x^3 + a^3*b^2)]

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giac [A]  time = 0.48, size = 102, normalized size = 0.89 \begin {gather*} -\frac {2 \, x^{\frac {3}{2}} {\left (\frac {3 \, B a e^{8}}{b^{2}} + \frac {{\left (4 \, B a^{5} b^{6} e^{24} - A a^{4} b^{7} e^{24}\right )} x^{3} e^{\left (-16\right )}}{a^{5} b^{7}}\right )} e^{\frac {3}{2}}}{9 \, {\left (b x^{3} e^{4} + a e^{4}\right )}^{\frac {3}{2}}} - \frac {2 \, B e^{\frac {7}{2}} \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 {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*e^8/b^2 + (4*B*a^5*b^6*e^24 - A*a^4*b^7*e^24)*x^3*e^(-16)/(a^5*b^7))*e^(3/2)/(b*x^3*e^4 +
a*e^4)^(3/2) - 2/3*B*e^(7/2)*log(abs(-sqrt(b)*x^(3/2)*e^2 + sqrt(b*x^3*e^4 + a*e^4)))/b^(5/2)

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maple [C]  time = 1.05, size = 7081, normalized size = 62.11 \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)^(5/2),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 {5}{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)^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

Timed out

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