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

Optimal. Leaf size=161 \[ -\frac {a^2 e^{7/2} (2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 b^{5/2}}+\frac {a e^2 (e x)^{3/2} \sqrt {a+b x^3} (2 A b-a B)}{24 b^2}+\frac {(e x)^{9/2} \sqrt {a+b x^3} (2 A b-a B)}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e} \]

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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 = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {459, 279, 321, 329, 275, 217, 206} \begin {gather*} -\frac {a^2 e^{7/2} (2 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^{5/2}}+\frac {a e^2 (e x)^{3/2} \sqrt {a+b x^3} (2 A b-a B)}{24 b^2}+\frac {(e x)^{9/2} \sqrt {a+b x^3} (2 A b-a B)}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*A*b - a*B)*e^2*(e*x)^(3/2)*Sqrt[a + b*x^3])/(24*b^2) + ((2*A*b - a*B)*(e*x)^(9/2)*Sqrt[a + b*x^3])/(12*b
*e) + (B*(e*x)^(9/2)*(a + b*x^3)^(3/2))/(9*b*e) - (a^2*(2*A*b - a*B)*e^(7/2)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^
(3/2)*Sqrt[a + b*x^3])])/(24*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 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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && 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 (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx &=\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {\left (-9 A b+\frac {9 a B}{2}\right ) \int (e x)^{7/2} \sqrt {a+b x^3} \, dx}{9 b}\\ &=\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}+\frac {(a (2 A b-a B)) \int \frac {(e x)^{7/2}}{\sqrt {a+b x^3}} \, dx}{8 b}\\ &=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {\left (a^2 (2 A b-a B) e^3\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{16 b^2}\\ &=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {\left (a^2 (2 A b-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 )}{8 b^2}\\ &=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {\left (a^2 (2 A b-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 )}{24 b^2}\\ &=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {\left (a^2 (2 A b-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 )}{24 b^2}\\ &=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {a^2 (2 A b-a B) e^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 145, normalized size = 0.90 \begin {gather*} \frac {e^3 \sqrt {e x} \sqrt {a+b x^3} \left (3 a^{3/2} (a B-2 A b) \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 (-3 a^2 B+2 a b \left (3 A+B x^3\right )+4 b^2 x^3 \left (3 A+2 B x^3\right )\right )\right )}{72 b^{5/2} \sqrt {x} \sqrt {\frac {b x^3}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*Sqrt[e*x]*Sqrt[a + b*x^3]*(Sqrt[b]*x^(3/2)*Sqrt[1 + (b*x^3)/a]*(-3*a^2*B + 2*a*b*(3*A + B*x^3) + 4*b^2*x^
3*(3*A + 2*B*x^3)) + 3*a^(3/2)*(-2*A*b + a*B)*ArcSinh[(Sqrt[b]*x^(3/2))/Sqrt[a]]))/(72*b^(5/2)*Sqrt[x]*Sqrt[1
+ (b*x^3)/a])

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IntegrateAlgebraic [A]  time = 0.60, 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}+6 a A b e^6 (e x)^{3/2}+2 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^2 e^4}+\frac {e^5 \sqrt {\frac {b}{e^3}} \left (2 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^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^3]*(6*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) + 2*a*b*B*e^3*(
e*x)^(9/2) + 8*b^2*B*(e*x)^(15/2)))/(72*b^2*e^4) + ((2*a^2*A*b - a^3*B)*Sqrt[b/e^3]*e^5*Log[-(Sqrt[b/e^3]*(e*x
)^(3/2)) + Sqrt[a + b*x^3]])/(24*b^3)

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fricas [A]  time = 1.26, size = 295, normalized size = 1.83 \begin {gather*} \left [-\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} b\right )} e^{3} \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} e^{3} x^{7} + 2 \, {\left (B a b + 6 \, A b^{2}\right )} e^{3} x^{4} - 3 \, {\left (B a^{2} - 2 \, A a b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{288 \, b^{2}}, -\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} b\right )} e^{3} \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} e^{3} x^{7} + 2 \, {\left (B a b + 6 \, A b^{2}\right )} e^{3} x^{4} - 3 \, {\left (B a^{2} - 2 \, A a b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{144 \, b^{2}}\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)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.75, size = 251, normalized size = 1.56 \begin {gather*} \frac {1}{12} \, \sqrt {b x^{3} e^{4} + a e^{4}} {\left (2 \, x^{3} e^{\left (-1\right )} + \frac {a e^{\left (-1\right )}}{b}\right )} A x^{\frac {3}{2}} e^{\frac {5}{2}} + \frac {1}{72} \, \sqrt {b x^{3} e^{4} + a e^{4}} {\left (2 \, {\left (4 \, x^{3} e^{\left (-4\right )} + \frac {a e^{\left (-4\right )}}{b}\right )} x^{3} e^{3} - \frac {3 \, a^{2} e^{\left (-1\right )}}{b^{2}}\right )} B x^{\frac {3}{2}} e^{\frac {5}{2}} - \frac {{\left (B^{2} a^{6} e^{7} - 4 \, A B a^{5} b e^{7} + 4 \, A^{2} a^{4} b^{2} e^{7}\right )} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -{\left (B a^{3} x^{\frac {3}{2}} e^{\frac {11}{2}} - 2 \, A a^{2} b x^{\frac {3}{2}} e^{\frac {11}{2}}\right )} \sqrt {b} e^{\frac {1}{2}} + \sqrt {B^{2} a^{7} e^{12} - 4 \, A B a^{6} b e^{12} + 4 \, A^{2} a^{5} b^{2} e^{12} + {\left (B a^{3} x^{\frac {3}{2}} e^{\frac {11}{2}} - 2 \, A a^{2} b x^{\frac {3}{2}} e^{\frac {11}{2}}\right )}^{2} b e} \right |}\right )}{24 \, b^{\frac {5}{2}} {\left | -B a^{3} e^{3} + 2 \, A a^{2} b e^{3} \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)^(1/2),x, algorithm="giac")

[Out]

1/12*sqrt(b*x^3*e^4 + a*e^4)*(2*x^3*e^(-1) + a*e^(-1)/b)*A*x^(3/2)*e^(5/2) + 1/72*sqrt(b*x^3*e^4 + a*e^4)*(2*(
4*x^3*e^(-4) + a*e^(-4)/b)*x^3*e^3 - 3*a^2*e^(-1)/b^2)*B*x^(3/2)*e^(5/2) - 1/24*(B^2*a^6*e^7 - 4*A*B*a^5*b*e^7
+ 4*A^2*a^4*b^2*e^7)*e^(-1/2)*log(abs(-(B*a^3*x^(3/2)*e^(11/2) - 2*A*a^2*b*x^(3/2)*e^(11/2))*sqrt(b)*e^(1/2)
+ sqrt(B^2*a^7*e^12 - 4*A*B*a^6*b*e^12 + 4*A^2*a^5*b^2*e^12 + (B*a^3*x^(3/2)*e^(11/2) - 2*A*a^2*b*x^(3/2)*e^(1
1/2))^2*b*e)))/(b^(5/2)*abs(-B*a^3*e^3 + 2*A*a^2*b*e^3))

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maple [C]  time = 8.02, size = 7293, normalized size = 45.30 \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)^(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 )} \sqrt {b x^{3} + a} \left (e x\right )^{\frac {7}{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)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)*sqrt(b*x^3 + a)*(e*x)^(7/2), 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 )\,{\left (e\,x\right )}^{7/2}\,\sqrt {b\,x^3+a} \,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)^(1/2),x)

[Out]

int((A + B*x^3)*(e*x)^(7/2)*(a + b*x^3)^(1/2), x)

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sympy [B]  time = 130.86, size = 292, normalized size = 1.81 \begin {gather*} \frac {A a^{\frac {3}{2}} e^{\frac {7}{2}} x^{\frac {3}{2}}}{12 b \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {A \sqrt {a} e^{\frac {7}{2}} x^{\frac {9}{2}}}{4 \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {A a^{2} e^{\frac {7}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{12 b^{\frac {3}{2}}} + \frac {A b e^{\frac {7}{2}} x^{\frac {15}{2}}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {B a^{\frac {5}{2}} e^{\frac {7}{2}} x^{\frac {3}{2}}}{24 b^{2} \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {B a^{\frac {3}{2}} e^{\frac {7}{2}} x^{\frac {9}{2}}}{72 b \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {5 B \sqrt {a} e^{\frac {7}{2}} x^{\frac {15}{2}}}{36 \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B a^{3} e^{\frac {7}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{24 b^{\frac {5}{2}}} + \frac {B b e^{\frac {7}{2}} x^{\frac {21}{2}}}{9 \sqrt {a} \sqrt {1 + \frac {b x^{3}}{a}}} \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)**(1/2),x)

[Out]

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

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