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

Optimal. Leaf size=188 \[ \frac {5 a^2 (a B+6 A b) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 \sqrt {b} e^{5/2}}+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2} (a B+6 A b)}{9 a e^4}+\frac {5 (e x)^{3/2} \left (a+b x^3\right )^{3/2} (a B+6 A b)}{36 e^4}+\frac {5 a (e x)^{3/2} \sqrt {a+b x^3} (a B+6 A b)}{24 e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{3 a e (e x)^{3/2}} \]

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Rubi [A]  time = 0.13, antiderivative size = 188, 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 = {453, 279, 329, 275, 217, 206} \begin {gather*} \frac {5 a^2 (a B+6 A b) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 \sqrt {b} e^{5/2}}+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2} (a B+6 A b)}{9 a e^4}+\frac {5 (e x)^{3/2} \left (a+b x^3\right )^{3/2} (a B+6 A b)}{36 e^4}+\frac {5 a (e x)^{3/2} \sqrt {a+b x^3} (a B+6 A b)}{24 e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{3 a e (e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*(6*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^3])/(24*e^4) + (5*(6*A*b + a*B)*(e*x)^(3/2)*(a + b*x^3)^(3/2))/(36
*e^4) + ((6*A*b + a*B)*(e*x)^(3/2)*(a + b*x^3)^(5/2))/(9*a*e^4) - (2*A*(a + b*x^3)^(7/2))/(3*a*e*(e*x)^(3/2))
+ (5*a^2*(6*A*b + a*B)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + b*x^3])])/(24*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 {\left (a+b x^3\right )^{5/2} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx &=-\frac {2 A \left (a+b x^3\right )^{7/2}}{3 a e (e x)^{3/2}}+\frac {(6 A b+a B) \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \, dx}{a e^3}\\ &=\frac {(6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{3 a e (e x)^{3/2}}+\frac {(5 (6 A b+a B)) \int \sqrt {e x} \left (a+b x^3\right )^{3/2} \, dx}{6 e^3}\\ &=\frac {5 (6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 e^4}+\frac {(6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{3 a e (e x)^{3/2}}+\frac {(5 a (6 A b+a B)) \int \sqrt {e x} \sqrt {a+b x^3} \, dx}{8 e^3}\\ &=\frac {5 a (6 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{24 e^4}+\frac {5 (6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 e^4}+\frac {(6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{3 a e (e x)^{3/2}}+\frac {\left (5 a^2 (6 A b+a B)\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{16 e^3}\\ &=\frac {5 a (6 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{24 e^4}+\frac {5 (6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 e^4}+\frac {(6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{3 a e (e x)^{3/2}}+\frac {\left (5 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 e^4}\\ &=\frac {5 a (6 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{24 e^4}+\frac {5 (6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 e^4}+\frac {(6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{3 a e (e x)^{3/2}}+\frac {\left (5 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 e^4}\\ &=\frac {5 a (6 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{24 e^4}+\frac {5 (6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 e^4}+\frac {(6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{3 a e (e x)^{3/2}}+\frac {\left (5 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 e^4}\\ &=\frac {5 a (6 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{24 e^4}+\frac {5 (6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{36 e^4}+\frac {(6 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{3 a e (e x)^{3/2}}+\frac {5 a^2 (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 \sqrt {b} e^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 150, normalized size = 0.80 \begin {gather*} \frac {x \sqrt {a+b x^3} \left (15 a^{3/2} x^{3/2} (a B+6 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )+\sqrt {b} \sqrt {\frac {b x^3}{a}+1} \left (a^2 \left (33 B x^3-48 A\right )+a \left (54 A b x^3+26 b B x^6\right )+4 b^2 x^6 \left (3 A+2 B x^3\right )\right )\right )}{72 \sqrt {b} (e x)^{5/2} \sqrt {\frac {b x^3}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[a + b*x^3]*(Sqrt[b]*Sqrt[1 + (b*x^3)/a]*(4*b^2*x^6*(3*A + 2*B*x^3) + a^2*(-48*A + 33*B*x^3) + a*(54*A*
b*x^3 + 26*b*B*x^6)) + 15*a^(3/2)*(6*A*b + a*B)*x^(3/2)*ArcSinh[(Sqrt[b]*x^(3/2))/Sqrt[a]]))/(72*Sqrt[b]*(e*x)
^(5/2)*Sqrt[1 + (b*x^3)/a])

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IntegrateAlgebraic [A]  time = 0.85, size = 157, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a+b x^3} \left (-48 a^2 A e^9+33 a^2 B e^9 x^3+54 a A b e^9 x^3+26 a b B e^9 x^6+12 A b^2 e^9 x^6+8 b^2 B e^9 x^9\right )}{72 e^{10} (e x)^{3/2}}-\frac {5 \sqrt {\frac {b}{e^3}} \left (a^3 B+6 a^2 A b\right ) \log \left (\sqrt {a+b x^3}-\sqrt {\frac {b}{e^3}} (e x)^{3/2}\right )}{24 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^3]*(-48*a^2*A*e^9 + 54*a*A*b*e^9*x^3 + 33*a^2*B*e^9*x^3 + 12*A*b^2*e^9*x^6 + 26*a*b*B*e^9*x^6 +
8*b^2*B*e^9*x^9))/(72*e^10*(e*x)^(3/2)) - (5*(6*a^2*A*b + a^3*B)*Sqrt[b/e^3]*Log[-(Sqrt[b/e^3]*(e*x)^(3/2)) +
Sqrt[a + b*x^3]])/(24*b*e)

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fricas [A]  time = 1.28, size = 309, normalized size = 1.64 \begin {gather*} \left [\frac {15 \, {\left (B a^{3} + 6 \, A a^{2} 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 (8 \, B b^{3} x^{9} + 2 \, {\left (13 \, B a b^{2} + 6 \, A b^{3}\right )} x^{6} - 48 \, A a^{2} b + 3 \, {\left (11 \, B a^{2} b + 18 \, A a b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{288 \, b e^{3} x^{2}}, -\frac {15 \, {\left (B a^{3} + 6 \, A a^{2} 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 (8 \, B b^{3} x^{9} + 2 \, {\left (13 \, B a b^{2} + 6 \, A b^{3}\right )} x^{6} - 48 \, A a^{2} b + 3 \, {\left (11 \, B a^{2} b + 18 \, A a b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{144 \, b e^{3} x^{2}}\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)^(5/2),x, algorithm="fricas")

[Out]

[1/288*(15*(B*a^3 + 6*A*a^2*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*(8*B*b^3*x^9 + 2*(13*B*a*b^2 + 6*A*b^3)*x^6 - 48*A*a^2*b + 3*(11*B*a^2*b +
18*A*a*b^2)*x^3)*sqrt(b*x^3 + a)*sqrt(e*x))/(b*e^3*x^2), -1/144*(15*(B*a^3 + 6*A*a^2*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*(8*B*b^3*x^9 + 2*(13*B*a*b^2 + 6*A*b^3)*x^6 -
48*A*a^2*b + 3*(11*B*a^2*b + 18*A*a*b^2)*x^3)*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 )} {\left (b x^{3} + a\right )}^{\frac {5}{2}}}{\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)^(5/2)*(B*x^3+A)/(e*x)^(5/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 1.16, size = 7544, normalized size = 40.13 \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)^(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 (b x^{3} + a\right )}^{\frac {5}{2}}}{\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)^(5/2)*(B*x^3+A)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

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

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sympy [B]  time = 63.94, size = 403, normalized size = 2.14 \begin {gather*} - \frac {2 A a^{\frac {5}{2}}}{3 e^{\frac {5}{2}} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {2 A a^{\frac {3}{2}} b x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e^{\frac {5}{2}}} - \frac {7 A a^{\frac {3}{2}} b x^{\frac {3}{2}}}{12 e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {A \sqrt {a} b^{2} x^{\frac {9}{2}}}{4 e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {5 A a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{4 e^{\frac {5}{2}}} + \frac {A b^{3} x^{\frac {15}{2}}}{6 \sqrt {a} e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B a^{\frac {5}{2}} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e^{\frac {5}{2}}} + \frac {B a^{\frac {5}{2}} x^{\frac {3}{2}}}{8 e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {35 B a^{\frac {3}{2}} b x^{\frac {9}{2}}}{72 e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {17 B \sqrt {a} b^{2} x^{\frac {15}{2}}}{36 e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {5 B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{24 \sqrt {b} e^{\frac {5}{2}}} + \frac {B b^{3} x^{\frac {21}{2}}}{9 \sqrt {a} e^{\frac {5}{2}} \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)**(5/2),x)

[Out]

-2*A*a**(5/2)/(3*e**(5/2)*x**(3/2)*sqrt(1 + b*x**3/a)) + 2*A*a**(3/2)*b*x**(3/2)*sqrt(1 + b*x**3/a)/(3*e**(5/2
)) - 7*A*a**(3/2)*b*x**(3/2)/(12*e**(5/2)*sqrt(1 + b*x**3/a)) + A*sqrt(a)*b**2*x**(9/2)/(4*e**(5/2)*sqrt(1 + b
*x**3/a)) + 5*A*a**2*sqrt(b)*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(4*e**(5/2)) + A*b**3*x**(15/2)/(6*sqrt(a)*e**(5/
2)*sqrt(1 + b*x**3/a)) + B*a**(5/2)*x**(3/2)*sqrt(1 + b*x**3/a)/(3*e**(5/2)) + B*a**(5/2)*x**(3/2)/(8*e**(5/2)
*sqrt(1 + b*x**3/a)) + 35*B*a**(3/2)*b*x**(9/2)/(72*e**(5/2)*sqrt(1 + b*x**3/a)) + 17*B*sqrt(a)*b**2*x**(15/2)
/(36*e**(5/2)*sqrt(1 + b*x**3/a)) + 5*B*a**3*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(24*sqrt(b)*e**(5/2)) + B*b**3*x*
*(21/2)/(9*sqrt(a)*e**(5/2)*sqrt(1 + b*x**3/a))

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