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

Optimal. Leaf size=201 \[ \frac {5 a^3 \sqrt {e} (8 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{192 b^{3/2}}+\frac {5 a^2 (e x)^{3/2} \sqrt {a+b x^3} (8 A b-a B)}{192 b e}+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2} (8 A b-a B)}{72 b e}+\frac {5 a (e x)^{3/2} \left (a+b x^3\right )^{3/2} (8 A b-a B)}{288 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e} \]

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Rubi [A]  time = 0.13, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {5 a^3 \sqrt {e} (8 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{192 b^{3/2}}+\frac {5 a^2 (e x)^{3/2} \sqrt {a+b x^3} (8 A b-a B)}{192 b e}+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2} (8 A b-a B)}{72 b e}+\frac {5 a (e x)^{3/2} \left (a+b x^3\right )^{3/2} (8 A b-a B)}{288 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^2*(8*A*b - a*B)*(e*x)^(3/2)*Sqrt[a + b*x^3])/(192*b*e) + (5*a*(8*A*b - a*B)*(e*x)^(3/2)*(a + b*x^3)^(3/2)
)/(288*b*e) + ((8*A*b - a*B)*(e*x)^(3/2)*(a + b*x^3)^(5/2))/(72*b*e) + (B*(e*x)^(3/2)*(a + b*x^3)^(7/2))/(12*b
*e) + (5*a^3*(8*A*b - a*B)*Sqrt[e]*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + b*x^3])])/(192*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 )^{5/2} \left (A+B x^3\right ) \, dx &=\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}-\frac {\left (-12 A b+\frac {3 a B}{2}\right ) \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \, dx}{12 b}\\ &=\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {(5 a (8 A b-a B)) \int \sqrt {e x} \left (a+b x^3\right )^{3/2} \, dx}{48 b}\\ &=\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {\left (5 a^2 (8 A b-a B)\right ) \int \sqrt {e x} \sqrt {a+b x^3} \, dx}{64 b}\\ &=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {\left (5 a^3 (8 A b-a B)\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{128 b}\\ &=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {\left (5 a^3 (8 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 )}{64 b e}\\ &=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {\left (5 a^3 (8 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 )}{192 b e}\\ &=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {\left (5 a^3 (8 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 )}{192 b e}\\ &=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {5 a^3 (8 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 )}{192 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 146, normalized size = 0.73 \begin {gather*} \frac {x \sqrt {e x} \sqrt {a+b x^3} \left (\frac {(8 A b-a B) \left (15 a^{5/2} \sinh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )+\sqrt {b} x^{3/2} \sqrt {\frac {b x^3}{a}+1} \left (33 a^2+26 a b x^3+8 b^2 x^6\right )\right )}{48 \sqrt {b} x^{3/2} \sqrt {\frac {b x^3}{a}+1}}+B \left (a+b x^3\right )^3\right )}{12 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[e*x]*Sqrt[a + b*x^3]*(B*(a + b*x^3)^3 + ((8*A*b - a*B)*(Sqrt[b]*x^(3/2)*Sqrt[1 + (b*x^3)/a]*(33*a^2 +
26*a*b*x^3 + 8*b^2*x^6) + 15*a^(5/2)*ArcSinh[(Sqrt[b]*x^(3/2))/Sqrt[a]]))/(48*Sqrt[b]*x^(3/2)*Sqrt[1 + (b*x^3)
/a])))/(12*b)

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IntegrateAlgebraic [A]  time = 0.77, size = 200, normalized size = 1.00 \begin {gather*} \frac {\sqrt {a+b x^3} \left (15 a^3 B e^9 (e x)^{3/2}+264 a^2 A b e^9 (e x)^{3/2}+118 a^2 b B e^6 (e x)^{9/2}+208 a A b^2 e^6 (e x)^{9/2}+136 a b^2 B e^3 (e x)^{15/2}+64 A b^3 e^3 (e x)^{15/2}+48 b^3 B (e x)^{21/2}\right )}{576 b e^{10}}-\frac {5 e^2 \sqrt {\frac {b}{e^3}} \left (8 a^3 A b-a^4 B\right ) \log \left (\sqrt {a+b x^3}-\sqrt {\frac {b}{e^3}} (e x)^{3/2}\right )}{192 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^3]*(264*a^2*A*b*e^9*(e*x)^(3/2) + 15*a^3*B*e^9*(e*x)^(3/2) + 208*a*A*b^2*e^6*(e*x)^(9/2) + 118*a
^2*b*B*e^6*(e*x)^(9/2) + 64*A*b^3*e^3*(e*x)^(15/2) + 136*a*b^2*B*e^3*(e*x)^(15/2) + 48*b^3*B*(e*x)^(21/2)))/(5
76*b*e^10) - (5*(8*a^3*A*b - a^4*B)*Sqrt[b/e^3]*e^2*Log[-(Sqrt[b/e^3]*(e*x)^(3/2)) + Sqrt[a + b*x^3]])/(192*b^
2)

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fricas [A]  time = 1.39, size = 323, normalized size = 1.61 \begin {gather*} \left [-\frac {15 \, {\left (B a^{4} - 8 \, A a^{3} 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 (48 \, B b^{3} x^{10} + 8 \, {\left (17 \, B a b^{2} + 8 \, A b^{3}\right )} x^{7} + 2 \, {\left (59 \, B a^{2} b + 104 \, A a b^{2}\right )} x^{4} + 3 \, {\left (5 \, B a^{3} + 88 \, A a^{2} b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{2304 \, b}, \frac {15 \, {\left (B a^{4} - 8 \, A a^{3} 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 (48 \, B b^{3} x^{10} + 8 \, {\left (17 \, B a b^{2} + 8 \, A b^{3}\right )} x^{7} + 2 \, {\left (59 \, B a^{2} b + 104 \, A a b^{2}\right )} x^{4} + 3 \, {\left (5 \, B a^{3} + 88 \, A a^{2} b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{1152 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2304*(15*(B*a^4 - 8*A*a^3*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*(48*B*b^3*x^10 + 8*(17*B*a*b^2 + 8*A*b^3)*x^7 + 2*(59*B*a^2*b + 104*A*a*b
^2)*x^4 + 3*(5*B*a^3 + 88*A*a^2*b)*x)*sqrt(b*x^3 + a)*sqrt(e*x))/b, 1/1152*(15*(B*a^4 - 8*A*a^3*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*(48*B*b^3*x^10 + 8*(17*B*a*b^2 + 8*A*
b^3)*x^7 + 2*(59*B*a^2*b + 104*A*a*b^2)*x^4 + 3*(5*B*a^3 + 88*A*a^2*b)*x)*sqrt(b*x^3 + a)*sqrt(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)^(5/2)*(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^2*exp(1)
/exp(3)*2*(((26347216896000*b^16*exp(1)^7/1264666411008000/b^16/exp(1)^17*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x
*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))[phone]*b^15*exp(1)^10*a/1264666411008000/b^16/exp
(1)^17)*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))-548900352000
0*b^14*exp(1)^13*a^2/1264666411008000/b^16/exp(1)^17)*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(1))*sqrt(x*exp(
1))*sqrt(x*exp(1))*sqrt(x*exp(1))[phone]*b^13*exp(1)^16*a^3/1264666411008000/b^16/exp(1)^17)*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^2*exp(1)/exp(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*exp(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/exp(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*B*a^2*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*exp(1)^4+b*(x*exp(1
))^3*exp(1))+4*A*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*ex
p(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^2*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))))+4*B*a*b*exp(1)/exp(3)*2*((17740800*b^10*exp(1)^4/6
38668800/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*exp(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/exp(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))-1/2*exp(1)/32/exp(1)*exp(3)/b/exp(3)/abs(24*b*a^3*A
+5*B*a^4)/(576*exp(1)*b^2*a^6*A^2+240*exp(1)*b*B*a^7*A+25*exp(1)*B^2*a^8)^-1/3/sqrt(b*exp(1))*ln(abs(sqrt(576*
A^2*a^7*b^2*exp(1)^6+240*A*B*a^8*b*exp(1)^6+25*B^2*a^9*exp(1)^6+9*b*(-5*B*a^4*exp(1)*sqrt(x*exp(1))*x*exp(1)/3
-24*A*a^3*b*exp(1)*sqrt(x*exp(1))*x*exp(1)/3)^2*exp(1))-sqrt(9*b*exp(1))*(-5*B*a^4*exp(1)*sqrt(x*exp(1))*x*exp
(1)/3-24*A*a^3*b*exp(1)*sqrt(x*exp(1))*x*exp(1)/3)))

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maple [C]  time = 1.10, size = 7702, normalized size = 38.32 \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)^(5/2)*(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 {5}{2}} \sqrt {e x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (B\,x^3+A\right )\,\sqrt {e\,x}\,{\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)^(1/2)*(a + b*x^3)^(5/2),x)

[Out]

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

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sympy [B]  time = 56.28, size = 413, normalized size = 2.05 \begin {gather*} \frac {A a^{\frac {5}{2}} \left (e x\right )^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e} + \frac {A a^{\frac {5}{2}} \left (e x\right )^{\frac {3}{2}}}{8 e \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {35 A a^{\frac {3}{2}} b \left (e x\right )^{\frac {9}{2}}}{72 e^{4} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {17 A \sqrt {a} b^{2} \left (e x\right )^{\frac {15}{2}}}{36 e^{7} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {5 A 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 \sqrt {b}} + \frac {A b^{3} \left (e x\right )^{\frac {21}{2}}}{9 \sqrt {a} e^{10} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {5 B a^{\frac {7}{2}} \left (e x\right )^{\frac {3}{2}}}{192 b e \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {133 B a^{\frac {5}{2}} \left (e x\right )^{\frac {9}{2}}}{576 e^{4} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {127 B a^{\frac {3}{2}} b \left (e x\right )^{\frac {15}{2}}}{288 e^{7} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {23 B \sqrt {a} b^{2} \left (e x\right )^{\frac {21}{2}}}{72 e^{10} \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {5 B a^{4} \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {b} \left (e x\right )^{\frac {3}{2}}}{\sqrt {a} e^{\frac {3}{2}}} \right )}}{192 b^{\frac {3}{2}}} + \frac {B b^{3} \left (e x\right )^{\frac {27}{2}}}{12 \sqrt {a} e^{13} \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)**(5/2)*(B*x**3+A)*(e*x)**(1/2),x)

[Out]

A*a**(5/2)*(e*x)**(3/2)*sqrt(1 + b*x**3/a)/(3*e) + A*a**(5/2)*(e*x)**(3/2)/(8*e*sqrt(1 + b*x**3/a)) + 35*A*a**
(3/2)*b*(e*x)**(9/2)/(72*e**4*sqrt(1 + b*x**3/a)) + 17*A*sqrt(a)*b**2*(e*x)**(15/2)/(36*e**7*sqrt(1 + b*x**3/a
)) + 5*A*a**3*sqrt(e)*asinh(sqrt(b)*(e*x)**(3/2)/(sqrt(a)*e**(3/2)))/(24*sqrt(b)) + A*b**3*(e*x)**(21/2)/(9*sq
rt(a)*e**10*sqrt(1 + b*x**3/a)) + 5*B*a**(7/2)*(e*x)**(3/2)/(192*b*e*sqrt(1 + b*x**3/a)) + 133*B*a**(5/2)*(e*x
)**(9/2)/(576*e**4*sqrt(1 + b*x**3/a)) + 127*B*a**(3/2)*b*(e*x)**(15/2)/(288*e**7*sqrt(1 + b*x**3/a)) + 23*B*s
qrt(a)*b**2*(e*x)**(21/2)/(72*e**10*sqrt(1 + b*x**3/a)) - 5*B*a**4*sqrt(e)*asinh(sqrt(b)*(e*x)**(3/2)/(sqrt(a)
*e**(3/2)))/(192*b**(3/2)) + B*b**3*(e*x)**(27/2)/(12*sqrt(a)*e**13*sqrt(1 + b*x**3/a))

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