3.6.42 \(\int \frac {(a+b x^3)^{5/2} (A+B x^3)}{(e x)^{7/2}} \, dx\) [542]

Optimal. Leaf size=352 \[ \frac {27 a (16 A b+5 a B) \sqrt {e x} \sqrt {a+b x^3}}{320 e^4}+\frac {3 (16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{3/2}}{80 e^4}+\frac {(16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{5/2}}{40 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {27\ 3^{3/4} a^{5/3} (16 A b+5 a B) \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{640 e^4 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]

[Out]

-2/5*A*(b*x^3+a)^(7/2)/a/e/(e*x)^(5/2)+3/80*(16*A*b+5*B*a)*(b*x^3+a)^(3/2)*(e*x)^(1/2)/e^4+1/40*(16*A*b+5*B*a)
*(b*x^3+a)^(5/2)*(e*x)^(1/2)/a/e^4+27/320*a*(16*A*b+5*B*a)*(e*x)^(1/2)*(b*x^3+a)^(1/2)/e^4+27/640*3^(3/4)*a^(5
/3)*(16*A*b+5*B*a)*(a^(1/3)+b^(1/3)*x)*((a^(1/3)+b^(1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(
1/2)/(a^(1/3)+b^(1/3)*x*(1-3^(1/2)))*(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))*EllipticF((1-(a^(1/3)+b^(1/3)*x*(1-3^(1/2
)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(e*x)^(1/2)*((a^(2/3)-a^(1/3)*b^(1/3)*
x+b^(2/3)*x^2)/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)/e^4/(b*x^3+a)^(1/2)/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^
(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 352, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {464, 285, 335, 231} \begin {gather*} \frac {27\ 3^{3/4} a^{5/3} \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} (5 a B+16 A b) F\left (\text {ArcCos}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{640 e^4 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {\sqrt {e x} \left (a+b x^3\right )^{5/2} (5 a B+16 A b)}{40 a e^4}+\frac {3 \sqrt {e x} \left (a+b x^3\right )^{3/2} (5 a B+16 A b)}{80 e^4}+\frac {27 a \sqrt {e x} \sqrt {a+b x^3} (5 a B+16 A b)}{320 e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^(5/2)*(A + B*x^3))/(e*x)^(7/2),x]

[Out]

(27*a*(16*A*b + 5*a*B)*Sqrt[e*x]*Sqrt[a + b*x^3])/(320*e^4) + (3*(16*A*b + 5*a*B)*Sqrt[e*x]*(a + b*x^3)^(3/2))
/(80*e^4) + ((16*A*b + 5*a*B)*Sqrt[e*x]*(a + b*x^3)^(5/2))/(40*a*e^4) - (2*A*(a + b*x^3)^(7/2))/(5*a*e*(e*x)^(
5/2)) + (27*3^(3/4)*a^(5/3)*(16*A*b + 5*a*B)*Sqrt[e*x]*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x
+ b^(2/3)*x^2)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*EllipticF[ArcCos[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a
^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(640*e^4*Sqrt[(b^(1/3)*x*(a^(1/3) + b^(1/3)*x))/(a^(1/3)
+ (1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a + b*x^3])

Rule 231

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[x*(s +
r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*(
(s + r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^
2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x]

Rule 285

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 335

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 464

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)^{7/2}} \, dx &=-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {(16 A b+5 a B) \int \frac {\left (a+b x^3\right )^{5/2}}{\sqrt {e x}} \, dx}{5 a e^3}\\ &=\frac {(16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{5/2}}{40 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {(3 (16 A b+5 a B)) \int \frac {\left (a+b x^3\right )^{3/2}}{\sqrt {e x}} \, dx}{16 e^3}\\ &=\frac {3 (16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{3/2}}{80 e^4}+\frac {(16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{5/2}}{40 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {(27 a (16 A b+5 a B)) \int \frac {\sqrt {a+b x^3}}{\sqrt {e x}} \, dx}{160 e^3}\\ &=\frac {27 a (16 A b+5 a B) \sqrt {e x} \sqrt {a+b x^3}}{320 e^4}+\frac {3 (16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{3/2}}{80 e^4}+\frac {(16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{5/2}}{40 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {\left (81 a^2 (16 A b+5 a B)\right ) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^3}} \, dx}{640 e^3}\\ &=\frac {27 a (16 A b+5 a B) \sqrt {e x} \sqrt {a+b x^3}}{320 e^4}+\frac {3 (16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{3/2}}{80 e^4}+\frac {(16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{5/2}}{40 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {\left (81 a^2 (16 A b+5 a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{320 e^4}\\ &=\frac {27 a (16 A b+5 a B) \sqrt {e x} \sqrt {a+b x^3}}{320 e^4}+\frac {3 (16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{3/2}}{80 e^4}+\frac {(16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{5/2}}{40 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {27\ 3^{3/4} a^{5/3} (16 A b+5 a B) \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{640 e^4 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.05, size = 88, normalized size = 0.25 \begin {gather*} \frac {2 x \sqrt {a+b x^3} \left (-A \left (a+b x^3\right )^3+\frac {a^2 (16 A b+5 a B) x^3 \, _2F_1\left (-\frac {5}{2},\frac {1}{6};\frac {7}{6};-\frac {b x^3}{a}\right )}{\sqrt {1+\frac {b x^3}{a}}}\right )}{5 a (e x)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^(5/2)*(A + B*x^3))/(e*x)^(7/2),x]

[Out]

(2*x*Sqrt[a + b*x^3]*(-(A*(a + b*x^3)^3) + (a^2*(16*A*b + 5*a*B)*x^3*Hypergeometric2F1[-5/2, 1/6, 7/6, -((b*x^
3)/a)])/Sqrt[1 + (b*x^3)/a]))/(5*a*(e*x)^(7/2))

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 0.39, size = 4422, normalized size = 12.56

method	result	size
risch	\(-\frac {\sqrt {b \,x^{3}+a}\, \left (-40 b^{2} B \,x^{9}-64 A \,b^{2} x^{6}-140 B a b \,x^{6}-368 a A b \,x^{3}-235 a^{2} B \,x^{3}+128 a^{2} A \right )}{320 x^{2} e^{3} \sqrt {e x}}+\frac {81 a^{2} \left (16 A b +5 B a \right ) \left (\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {\frac {\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) x}{\left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}\, \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )^{2} \sqrt {\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}} \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}{b \left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}\, \sqrt {\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}} \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}{b \left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}\, b \EllipticF \left (\sqrt {\frac {\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) x}{\left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}, \sqrt {\frac {\left (\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}{\left (\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right ) \sqrt {\left (b \,x^{3}+a \right ) e x}}{320 \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {b e x \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right ) \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}\, e^{3} \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\)	\(786\)
elliptic	\(\text {Expression too large to display}\)	\(988\)
default	\(\text {Expression too large to display}\)	\(4422\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(5/2)*(B*x^3+A)/(e*x)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/320*(b*x^3+a)^(1/2)/x^2/b/(-a*b^2)^(1/3)*(810*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)
^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I
*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/
2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3
))^(1/2))*(-a*b^2)^(2/3)*a^3*e*x^3-235*I*B*3^(1/2)*((b*x^3+a)*e*x)^(1/2)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3
^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*(-a*b^2)^(1
/3)*a^2*b*x^3-40*I*B*3^(1/2)*((b*x^3+a)*e*x)^(1/2)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+
2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*(-a*b^2)^(1/3)*b^3*x^9+120*B*((b*
x^3+a)*e*x)^(1/2)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*
(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*(-a*b^2)^(1/3)*b^3*x^9+192*A*((b*x^3+a)*e*x)^(1/2)*(1/b^2*e*x*(-b*
x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/
3)))^(1/2)*(-a*b^2)^(1/3)*b^3*x^6-384*A*((b*x^3+a)*e*x)^(1/2)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*
b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*(-a*b^2)^(1/3)*a^2*b+2
592*I*A*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b
*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))
/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)
))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*a^2*b^3*e*x^5+1104*A*((b*x^3+a)*e*x
)^(1/2)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(
1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*(-a*b^2)^(1/3)*a*b^2*x^3+705*B*((b*x^3+a)*e*x)^(1/2)*(1/b^2*e*x*(-b*x+(-a*b^
2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/
2)*(-a*b^2)^(1/3)*a^2*b*x^3+420*B*((b*x^3+a)*e*x)^(1/2)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(
1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*(-a*b^2)^(1/3)*a*b^2*x^6-518
4*I*A*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x
+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(
-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))
^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(1/3)*a^2*b^2*e*x^4-64*I*A*3
^(1/2)*((b*x^3+a)*e*x)^(1/2)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*
(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*(-a*b^2)^(1/3)*b^3*x^6+5184*A*(-(I*3^(1/2)-3)*x*b/(-1+I
*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-
a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1
/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1
+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(1/3)*a^2*b^2*e*x^4+1620*B*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x
+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^
(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((
-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*
3^(1/2)-3))^(1/2))*(-a*b^2)^(1/3)*a^3*b*e*x^4-2592*A*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))
^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)
*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*
b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2)
)*(-a*b^2)^(2/3)*a^2*b*e*x^3+2592*I*A*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*
((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^
2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I
*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a*b
^2)^(2/3)*a^2*b*e*x^3-140*I*B*3^(1/2)*((b*x^3+a)*e*x)^(1/2)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^
2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*(-a*b^2)^(1/3)*a*b^2*x^6
-810*B*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x...

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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)^(7/2),x, algorithm="maxima")

[Out]

e^(-7/2)*integrate((B*x^3 + A)*(b*x^3 + a)^(5/2)/x^(7/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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)^(7/2),x, algorithm="fricas")

[Out]

integral((B*b^2*x^9 + (2*B*a*b + A*b^2)*x^6 + (B*a^2 + 2*A*a*b)*x^3 + A*a^2)*sqrt(b*x^3 + a)*e^(-7/2)/x^(7/2),
x)

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Sympy [C] Result contains complex when optimal does not.
time = 50.41, size = 311, normalized size = 0.88 \begin {gather*} \frac {A a^{\frac {5}{2}} \Gamma \left (- \frac {5}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{6}, - \frac {1}{2} \\ \frac {1}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {1}{6}\right )} + \frac {2 A a^{\frac {3}{2}} b \sqrt {x} \Gamma \left (\frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{6} \\ \frac {7}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} \Gamma \left (\frac {7}{6}\right )} + \frac {A \sqrt {a} b^{2} x^{\frac {7}{2}} \Gamma \left (\frac {7}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{6} \\ \frac {13}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} \Gamma \left (\frac {13}{6}\right )} + \frac {B a^{\frac {5}{2}} \sqrt {x} \Gamma \left (\frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{6} \\ \frac {7}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} \Gamma \left (\frac {7}{6}\right )} + \frac {2 B a^{\frac {3}{2}} b x^{\frac {7}{2}} \Gamma \left (\frac {7}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{6} \\ \frac {13}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} \Gamma \left (\frac {13}{6}\right )} + \frac {B \sqrt {a} b^{2} x^{\frac {13}{2}} \Gamma \left (\frac {13}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {13}{6} \\ \frac {19}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} \Gamma \left (\frac {19}{6}\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)**(7/2),x)

[Out]

A*a**(5/2)*gamma(-5/6)*hyper((-5/6, -1/2), (1/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(7/2)*x**(5/2)*gamma(1/6))
+ 2*A*a**(3/2)*b*sqrt(x)*gamma(1/6)*hyper((-1/2, 1/6), (7/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(7/2)*gamma(7/6
)) + A*sqrt(a)*b**2*x**(7/2)*gamma(7/6)*hyper((-1/2, 7/6), (13/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(7/2)*gamm
a(13/6)) + B*a**(5/2)*sqrt(x)*gamma(1/6)*hyper((-1/2, 1/6), (7/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(7/2)*gamm
a(7/6)) + 2*B*a**(3/2)*b*x**(7/2)*gamma(7/6)*hyper((-1/2, 7/6), (13/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(7/2)
*gamma(13/6)) + B*sqrt(a)*b**2*x**(13/2)*gamma(13/6)*hyper((-1/2, 13/6), (19/6,), b*x**3*exp_polar(I*pi)/a)/(3
*e**(7/2)*gamma(19/6))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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)^(7/2),x, algorithm="giac")

[Out]

integrate((B*x^3 + A)*(b*x^3 + a)^(5/2)*e^(-7/2)/x^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (B\,x^3+A\right )\,{\left (b\,x^3+a\right )}^{5/2}}{{\left (e\,x\right )}^{7/2}} \,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)^(7/2),x)

[Out]

int(((A + B*x^3)*(a + b*x^3)^(5/2))/(e*x)^(7/2), x)

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