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

Optimal. Leaf size=161 \[ \frac {a^2 \sqrt {e} (6 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 b^{3/2}}+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2} (6 A b-a B)}{36 b e}+\frac {a (e x)^{3/2} \sqrt {a+b x^3} (6 A b-a B)}{24 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 b e} \]

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Rubi [A]  time = 0.11, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {459, 279, 329, 275, 217, 206} \begin {gather*} \frac {a^2 \sqrt {e} (6 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 b^{3/2}}+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2} (6 A b-a B)}{36 b e}+\frac {a (e x)^{3/2} \sqrt {a+b x^3} (6 A b-a B)}{24 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(6*A*b - a*B)*(e*x)^(3/2)*Sqrt[a + b*x^3])/(24*b*e) + ((6*A*b - a*B)*(e*x)^(3/2)*(a + b*x^3)^(3/2))/(36*b*e
) + (B*(e*x)^(3/2)*(a + b*x^3)^(5/2))/(9*b*e) + (a^2*(6*A*b - a*B)*Sqrt[e]*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3
/2)*Sqrt[a + b*x^3])])/(24*b^(3/2))

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 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 \sqrt {e x} \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx &=\frac {B (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 b e}-\frac {\left (-9 A b+\frac {3 a B}{2}\right ) \int \sqrt {e x} \left (a+b x^3\right )^{3/2} \, dx}{9 b}\\ &=\frac {(6 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 b e}+\frac {(a (6 A b-a B)) \int \sqrt {e x} \sqrt {a+b x^3} \, dx}{8 b}\\ &=\frac {a (6 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{24 b e}+\frac {(6 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 b e}+\frac {\left (a^2 (6 A b-a B)\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{16 b}\\ &=\frac {a (6 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{24 b e}+\frac {(6 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 b e}+\frac {\left (a^2 (6 A b-a B)\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{8 b e}\\ &=\frac {a (6 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{24 b e}+\frac {(6 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 b e}+\frac {\left (a^2 (6 A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^2}{e^3}}} \, dx,x,(e x)^{3/2}\right )}{24 b e}\\ &=\frac {a (6 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{24 b e}+\frac {(6 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 b e}+\frac {\left (a^2 (6 A b-a B)\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 )}{24 b e}\\ &=\frac {a (6 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{24 b e}+\frac {(6 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 b e}+\frac {a^2 (6 A b-a B) \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 143, normalized size = 0.89 \begin {gather*} \frac {\sqrt {e x} \sqrt {a+b x^3} \left (\sqrt {b} x^{3/2} \sqrt {\frac {b x^3}{a}+1} \left (3 a^2 B+2 a b \left (15 A+7 B x^3\right )+4 b^2 x^3 \left (3 A+2 B x^3\right )\right )-3 a^{3/2} (a B-6 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )\right )}{72 b^{3/2} \sqrt {x} \sqrt {\frac {b x^3}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e*x]*Sqrt[a + b*x^3]*(Sqrt[b]*x^(3/2)*Sqrt[1 + (b*x^3)/a]*(3*a^2*B + 4*b^2*x^3*(3*A + 2*B*x^3) + 2*a*b*(
15*A + 7*B*x^3)) - 3*a^(3/2)*(-6*A*b + a*B)*ArcSinh[(Sqrt[b]*x^(3/2))/Sqrt[a]]))/(72*b^(3/2)*Sqrt[x]*Sqrt[1 +
(b*x^3)/a])

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IntegrateAlgebraic [A]  time = 0.71, size = 162, normalized size = 1.01 \begin {gather*} \frac {\sqrt {a+b x^3} \left (3 a^2 B e^6 (e x)^{3/2}+30 a A b e^6 (e x)^{3/2}+14 a b B e^3 (e x)^{9/2}+12 A b^2 e^3 (e x)^{9/2}+8 b^2 B (e x)^{15/2}\right )}{72 b e^7}-\frac {e^2 \sqrt {\frac {b}{e^3}} \left (6 a^2 A b-a^3 B\right ) \log \left (\sqrt {a+b x^3}-\sqrt {\frac {b}{e^3}} (e x)^{3/2}\right )}{24 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^3]*(30*a*A*b*e^6*(e*x)^(3/2) + 3*a^2*B*e^6*(e*x)^(3/2) + 12*A*b^2*e^3*(e*x)^(9/2) + 14*a*b*B*e^3
*(e*x)^(9/2) + 8*b^2*B*(e*x)^(15/2)))/(72*b*e^7) - ((6*a^2*A*b - a^3*B)*Sqrt[b/e^3]*e^2*Log[-(Sqrt[b/e^3]*(e*x
)^(3/2)) + Sqrt[a + b*x^3]])/(24*b^2)

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fricas [A]  time = 1.56, size = 273, normalized size = 1.70 \begin {gather*} \left [-\frac {3 \, {\left (B a^{3} - 6 \, A a^{2} b\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 (8 \, B b^{2} x^{7} + 2 \, {\left (7 \, B a b + 6 \, A b^{2}\right )} x^{4} + 3 \, {\left (B a^{2} + 10 \, A a b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{288 \, b}, \frac {3 \, {\left (B a^{3} - 6 \, A a^{2} b\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 (8 \, B b^{2} x^{7} + 2 \, {\left (7 \, B a b + 6 \, A b^{2}\right )} x^{4} + 3 \, {\left (B a^{2} + 10 \, A a b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{144 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/288*(3*(B*a^3 - 6*A*a^2*b)*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*(8*B*b^2*x^7 + 2*(7*B*a*b + 6*A*b^2)*x^4 + 3*(B*a^2 + 10*A*a*b)*x)*sqrt(b*x
^3 + a)*sqrt(e*x))/b, 1/144*(3*(B*a^3 - 6*A*a^2*b)*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*(8*B*b^2*x^7 + 2*(7*B*a*b + 6*A*b^2)*x^4 + 3*(B*a^2 + 10*A*a*b)*x)*sqrt(b*x^3 + a)*sq
rt(e*x))/b]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-2,[1
,0,0,4]%%%}+%%%{-2,[0,1,6,1]%%%}+%%%{-2,[0,1,0,1]%%%},0,%%%{1,[2,0,0,8]%%%}+%%%{2,[1,1,6,5]%%%}+%%%{-2,[1,1,0,
5]%%%}+%%%{1,[0,2,12,2]%%%}+%%%{-2,[0,2,6,2]%%%}+%%%{1,[0,2,0,2]%%%}] at parameters values [91,88.2886286299,-
21,88]Warning, choosing root of [1,0,%%%{-2,[1,0,0,4]%%%}+%%%{-2,[0,1,6,1]%%%}+%%%{-2,[0,1,0,1]%%%},0,%%%{1,[2
,0,0,8]%%%}+%%%{2,[1,1,6,5]%%%}+%%%{-2,[1,1,0,5]%%%}+%%%{1,[0,2,12,2]%%%}+%%%{-2,[0,2,6,2]%%%}+%%%{1,[0,2,0,2]
%%%}] at parameters values [66,6.82230772497,-23,79]Warning, choosing root of [1,0,%%%{-2,[1,0,0,4]%%%}+%%%{-2
,[0,1,6,1]%%%}+%%%{-2,[0,1,0,1]%%%},0,%%%{1,[2,0,0,8]%%%}+%%%{2,[1,1,6,5]%%%}+%%%{-2,[1,1,0,5]%%%}+%%%{1,[0,2,
12,2]%%%}+%%%{-2,[0,2,6,2]%%%}+%%%{1,[0,2,0,2]%%%}] at parameters values [6,94.9264369817,-8,31]2*B*b*exp(1)/e
xp(3)*2*((17740800*b^10*exp(1)^4/638668800/b^10/exp(1)^11*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*
exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))+4435200*b^9*exp(1)^7*a/638668800/b^10/exp(1)^11)*sqrt(x*exp(1))*sqrt(x*e
xp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))-6652800*b^8*exp(1)^10*a^2/638668800/b^10/ex
p(1)^11)*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(a*exp(1)^4+b*(x*exp(1))^3*exp(1))+2*A*b*exp(1)/exp(
3)*2*(240*b^4*exp(1)/5760/b^4/exp(1)^5*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(
1))*sqrt(x*exp(1))+120*b^3*exp(1)^4*a/5760/b^4/exp(1)^5)*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(a*e
xp(1)^4+b*(x*exp(1))^3*exp(1))+2*B*a*exp(1)/exp(3)*2*(240*b^4*exp(1)/5760/b^4/exp(1)^5*sqrt(x*exp(1))*sqrt(x*e
xp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))+120*b^3*exp(1)^4*a/5760/b^4/exp(1)^5)*sqrt(
x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(a*exp(1)^4+b*(x*exp(1))^3*exp(1))+2*A*a*exp(1)/exp(3)/exp(1)/3*(1
/2*sqrt(x*exp(1))*x*exp(1)*sqrt(a*exp(1)^4+b*(x*exp(1))^3*exp(1))-2*a*exp(1)^4/4/sqrt(b*exp(1))*ln(abs(sqrt(a*
exp(1)^4+b*(x*exp(1))^3*exp(1))-sqrt(b*exp(1))*sqrt(x*exp(1))*x*exp(1))))-1/2*exp(1)/4/exp(1)*exp(3)/b/exp(3)/
abs(2*b*a^2*A+B*a^3)/(4*exp(1)*b^2*a^4*A^2+4*exp(1)*b*B*a^5*A+exp(1)*B^2*a^6)^-1/3/sqrt(b*exp(1))*ln(abs(sqrt(
4*A^2*a^5*b^2*exp(1)^6+4*A*B*a^6*b*exp(1)^6+B^2*a^7*exp(1)^6+9*b*(-B*a^3*exp(1)*sqrt(x*exp(1))*x*exp(1)/3-2*A*
a^2*b*exp(1)*sqrt(x*exp(1))*x*exp(1)/3)^2*exp(1))-sqrt(9*b*exp(1))*(-B*a^3*exp(1)*sqrt(x*exp(1))*x*exp(1)/3-2*
A*a^2*b*exp(1)*sqrt(x*exp(1))*x*exp(1)/3)))

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maple [C]  time = 1.06, size = 7290, normalized size = 45.28 \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)^(3/2)*(B*x^3+A)*(e*x)^(1/2),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 {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {3}{2}} \sqrt {e x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (B\,x^3+A\right )\,\sqrt {e\,x}\,{\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)^(1/2)*(a + b*x^3)^(3/2),x)

[Out]

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

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sympy [B]  time = 25.46, size = 335, normalized size = 2.08 \begin {gather*} \frac {A a^{\frac {3}{2}} \left (e x\right )^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e} + \frac {A a^{\frac {3}{2}} \left (e x\right )^{\frac {3}{2}}}{12 e \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {A \sqrt {a} b \left (e x\right )^{\frac {9}{2}}}{4 e^{4} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {A a^{2} \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {b} \left (e x\right )^{\frac {3}{2}}}{\sqrt {a} e^{\frac {3}{2}}} \right )}}{4 \sqrt {b}} + \frac {A b^{2} \left (e x\right )^{\frac {15}{2}}}{6 \sqrt {a} e^{7} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B a^{\frac {5}{2}} \left (e x\right )^{\frac {3}{2}}}{24 b e \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {17 B a^{\frac {3}{2}} \left (e x\right )^{\frac {9}{2}}}{72 e^{4} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {11 B \sqrt {a} b \left (e x\right )^{\frac {15}{2}}}{36 e^{7} \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {B a^{3} \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {b} \left (e x\right )^{\frac {3}{2}}}{\sqrt {a} e^{\frac {3}{2}}} \right )}}{24 b^{\frac {3}{2}}} + \frac {B b^{2} \left (e x\right )^{\frac {21}{2}}}{9 \sqrt {a} e^{10} \sqrt {1 + \frac {b x^{3}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**(3/2)*(e*x)**(3/2)*sqrt(1 + b*x**3/a)/(3*e) + A*a**(3/2)*(e*x)**(3/2)/(12*e*sqrt(1 + b*x**3/a)) + A*sqrt(
a)*b*(e*x)**(9/2)/(4*e**4*sqrt(1 + b*x**3/a)) + A*a**2*sqrt(e)*asinh(sqrt(b)*(e*x)**(3/2)/(sqrt(a)*e**(3/2)))/
(4*sqrt(b)) + A*b**2*(e*x)**(15/2)/(6*sqrt(a)*e**7*sqrt(1 + b*x**3/a)) + B*a**(5/2)*(e*x)**(3/2)/(24*b*e*sqrt(
1 + b*x**3/a)) + 17*B*a**(3/2)*(e*x)**(9/2)/(72*e**4*sqrt(1 + b*x**3/a)) + 11*B*sqrt(a)*b*(e*x)**(15/2)/(36*e*
*7*sqrt(1 + b*x**3/a)) - B*a**3*sqrt(e)*asinh(sqrt(b)*(e*x)**(3/2)/(sqrt(a)*e**(3/2)))/(24*b**(3/2)) + B*b**2*
(e*x)**(21/2)/(9*sqrt(a)*e**10*sqrt(1 + b*x**3/a))

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