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

Optimal. Leaf size=118 \[ \frac {(a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 \sqrt {b} e^{5/2}}+\frac {(e x)^{3/2} \sqrt {a+b x^3} (a B+2 A b)}{3 a e^4}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 118, 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 = {453, 279, 329, 275, 217, 206} \begin {gather*} \frac {(e x)^{3/2} \sqrt {a+b x^3} (a B+2 A b)}{3 a e^4}+\frac {(a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 \sqrt {b} e^{5/2}}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^3])/(3*a*e^4) - (2*A*(a + b*x^3)^(3/2))/(3*a*e*(e*x)^(3/2)) + ((2*A*b
+ a*B)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + b*x^3])])/(3*Sqrt[b]*e^(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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 453

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

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx &=-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {(2 A b+a B) \int \sqrt {e x} \sqrt {a+b x^3} \, dx}{a e^3}\\ &=\frac {(2 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{3 a e^4}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {(2 A b+a B) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{2 e^3}\\ &=\frac {(2 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{3 a e^4}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {(2 A b+a B) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{e^4}\\ &=\frac {(2 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{3 a e^4}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {(2 A b+a B) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^2}{e^3}}} \, dx,x,(e x)^{3/2}\right )}{3 e^4}\\ &=\frac {(2 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{3 a e^4}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {(2 A b+a B) \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 e^4}\\ &=\frac {(2 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{3 a e^4}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {(2 A b+a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 \sqrt {b} e^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 87, normalized size = 0.74 \begin {gather*} \frac {x \sqrt {a+b x^3} \left (\frac {x^{3/2} (a B+2 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \sqrt {\frac {b x^3}{a}+1}}-2 A+B x^3\right )}{3 (e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[a + b*x^3]*(-2*A + B*x^3 + ((2*A*b + a*B)*x^(3/2)*ArcSinh[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(Sqrt[a]*Sqrt[b]
*Sqrt[1 + (b*x^3)/a])))/(3*(e*x)^(5/2))

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IntegrateAlgebraic [A]  time = 0.56, size = 99, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a+b x^3} \left (B e^3 x^3-2 A e^3\right )}{3 e^4 (e x)^{3/2}}-\frac {\sqrt {\frac {b}{e^3}} (a B+2 A b) \log \left (\sqrt {a+b x^3}-\sqrt {\frac {b}{e^3}} (e x)^{3/2}\right )}{3 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^3]*(-2*A*e^3 + B*e^3*x^3))/(3*e^4*(e*x)^(3/2)) - ((2*A*b + a*B)*Sqrt[b/e^3]*Log[-(Sqrt[b/e^3]*(e
*x)^(3/2)) + Sqrt[a + b*x^3]])/(3*b*e)

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fricas [A]  time = 1.55, size = 207, normalized size = 1.75 \begin {gather*} \left [\frac {{\left (B a + 2 \, A b\right )} \sqrt {b e} x^{2} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b x^{4} + a x\right )} \sqrt {b x^{3} + a} \sqrt {b e} \sqrt {e x}\right ) + 4 \, {\left (B b x^{3} - 2 \, A b\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{12 \, b e^{3} x^{2}}, -\frac {{\left (B a + 2 \, A b\right )} \sqrt {-b e} x^{2} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {-b e} \sqrt {e x} x}{2 \, b e x^{3} + a e}\right ) - 2 \, {\left (B b x^{3} - 2 \, A b\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{6 \, b e^{3} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/(e*x)^(5/2),x, algorithm="fricas")

[Out]

[1/12*((B*a + 2*A*b)*sqrt(b*e)*x^2*log(-8*b^2*e*x^6 - 8*a*b*e*x^3 - a^2*e - 4*(2*b*x^4 + a*x)*sqrt(b*x^3 + a)*
sqrt(b*e)*sqrt(e*x)) + 4*(B*b*x^3 - 2*A*b)*sqrt(b*x^3 + a)*sqrt(e*x))/(b*e^3*x^2), -1/6*((B*a + 2*A*b)*sqrt(-b
*e)*x^2*arctan(2*sqrt(b*x^3 + a)*sqrt(-b*e)*sqrt(e*x)*x/(2*b*e*x^3 + a*e)) - 2*(B*b*x^3 - 2*A*b)*sqrt(b*x^3 +
a)*sqrt(e*x))/(b*e^3*x^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (B x^{3} + A\right )} \sqrt {b x^{3} + a}}{\left (e x\right )^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/(e*x)^(5/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 1.04, size = 6668, normalized size = 56.51 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)*(b*x^3+a)^(1/2)/(e*x)^(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 )} \sqrt {b x^{3} + a}}{\left (e x\right )^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)*sqrt(b*x^3 + a)/(e*x)^(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 )\,\sqrt {b\,x^3+a}}{{\left (e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^3)*(a + b*x^3)^(1/2))/(e*x)^(5/2),x)

[Out]

int(((A + B*x^3)*(a + b*x^3)^(1/2))/(e*x)^(5/2), x)

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sympy [A]  time = 9.72, size = 160, normalized size = 1.36 \begin {gather*} - \frac {2 A \sqrt {a}}{3 e^{\frac {5}{2}} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {2 A \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{3 e^{\frac {5}{2}}} - \frac {2 A b x^{\frac {3}{2}}}{3 \sqrt {a} e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B \sqrt {a} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e^{\frac {5}{2}}} + \frac {B a \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{3 \sqrt {b} e^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)*(b*x**3+a)**(1/2)/(e*x)**(5/2),x)

[Out]

-2*A*sqrt(a)/(3*e**(5/2)*x**(3/2)*sqrt(1 + b*x**3/a)) + 2*A*sqrt(b)*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(3*e**(5/2
)) - 2*A*b*x**(3/2)/(3*sqrt(a)*e**(5/2)*sqrt(1 + b*x**3/a)) + B*sqrt(a)*x**(3/2)*sqrt(1 + b*x**3/a)/(3*e**(5/2
)) + B*a*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(3*sqrt(b)*e**(5/2))

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