Integrand size = 26, antiderivative size = 161 \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {a^2 (2 A b-a B) e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 b^{5/2}} \]
1/9*B*(e*x)^(9/2)*(b*x^3+a)^(3/2)/b/e-1/24*a^2*(2*A*b-B*a)*e^(7/2)*arctanh ((e*x)^(3/2)*b^(1/2)/e^(3/2)/(b*x^3+a)^(1/2))/b^(5/2)+1/24*a*(2*A*b-B*a)*e ^2*(e*x)^(3/2)*(b*x^3+a)^(1/2)/b^2+1/12*(2*A*b-B*a)*(e*x)^(9/2)*(b*x^3+a)^ (1/2)/b/e
Time = 1.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.76 \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {(e x)^{7/2} \sqrt {a+b x^3} \left (6 a A b-3 a^2 B+12 A b^2 x^3+2 a b B x^3+8 b^2 B x^6\right )}{72 b^2 x^2}+\frac {a^2 (-2 A b+a B) (e x)^{7/2} \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )}{24 b^{5/2} x^{7/2}} \]
((e*x)^(7/2)*Sqrt[a + b*x^3]*(6*a*A*b - 3*a^2*B + 12*A*b^2*x^3 + 2*a*b*B*x ^3 + 8*b^2*B*x^6))/(72*b^2*x^2) + (a^2*(-2*A*b + a*B)*(e*x)^(7/2)*Log[Sqrt [b]*x^(3/2) + Sqrt[a + b*x^3]])/(24*b^(5/2)*x^(7/2))
Time = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {959, 811, 843, 851, 807, 224, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(2 A b-a B) \int (e x)^{7/2} \sqrt {b x^3+a}dx}{2 b}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {(2 A b-a B) \left (\frac {1}{4} a \int \frac {(e x)^{7/2}}{\sqrt {b x^3+a}}dx+\frac {(e x)^{9/2} \sqrt {a+b x^3}}{6 e}\right )}{2 b}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {(2 A b-a B) \left (\frac {1}{4} a \left (\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a e^3 \int \frac {\sqrt {e x}}{\sqrt {b x^3+a}}dx}{2 b}\right )+\frac {(e x)^{9/2} \sqrt {a+b x^3}}{6 e}\right )}{2 b}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {(2 A b-a B) \left (\frac {1}{4} a \left (\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a e^2 \int \frac {e x}{\sqrt {b x^3+a}}d\sqrt {e x}}{b}\right )+\frac {(e x)^{9/2} \sqrt {a+b x^3}}{6 e}\right )}{2 b}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {(2 A b-a B) \left (\frac {1}{4} a \left (\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a e^2 \int \frac {1}{\sqrt {a+\frac {b x}{e^2}}}d(e x)^{3/2}}{3 b}\right )+\frac {(e x)^{9/2} \sqrt {a+b x^3}}{6 e}\right )}{2 b}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {(2 A b-a B) \left (\frac {1}{4} a \left (\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a e^2 \int \frac {1}{1-\frac {b x}{e^2}}d\frac {(e x)^{3/2}}{\sqrt {a+\frac {b x}{e^2}}}}{3 b}\right )+\frac {(e x)^{9/2} \sqrt {a+b x^3}}{6 e}\right )}{2 b}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(2 A b-a B) \left (\frac {1}{4} a \left (\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+\frac {b x}{e^2}}}\right )}{3 b^{3/2}}\right )+\frac {(e x)^{9/2} \sqrt {a+b x^3}}{6 e}\right )}{2 b}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}\) |
(B*(e*x)^(9/2)*(a + b*x^3)^(3/2))/(9*b*e) + ((2*A*b - a*B)*(((e*x)^(9/2)*S qrt[a + b*x^3])/(6*e) + (a*((e^2*(e*x)^(3/2)*Sqrt[a + b*x^3])/(3*b) - (a*e ^(7/2)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + (b*x)/e^2])])/(3*b^ (3/2))))/4))/(2*b)
3.6.16.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
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] + Simp[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] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
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] - Simp[ 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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/c Subst[Int[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]
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] - Simp[(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]
Time = 5.65 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {x^{2} \left (8 b^{2} B \,x^{6}+12 A \,b^{2} x^{3}+2 B a b \,x^{3}+6 a b A -3 a^{2} B \right ) \sqrt {b \,x^{3}+a}\, e^{4}}{72 b^{2} \sqrt {e x}}-\frac {a^{2} \left (2 A b -B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) e^{4} \sqrt {\left (b \,x^{3}+a \right ) e x}}{24 b^{2} \sqrt {b e}\, \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\) | \(138\) |
default | \(-\frac {e^{3} \sqrt {e x}\, \sqrt {b \,x^{3}+a}\, \left (-8 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b^{2} x^{7}-12 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b^{2} x^{4}-2 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a b \,x^{4}+6 A \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{2} b e -6 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a b x -3 B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{3} e +3 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a^{2} x \right )}{72 \sqrt {\left (b \,x^{3}+a \right ) e x}\, b^{2} \sqrt {b e}}\) | \(224\) |
elliptic | \(\text {Expression too large to display}\) | \(1165\) |
1/72*x^2*(8*B*b^2*x^6+12*A*b^2*x^3+2*B*a*b*x^3+6*A*a*b-3*B*a^2)*(b*x^3+a)^ (1/2)/b^2*e^4/(e*x)^(1/2)-1/24*a^2*(2*A*b-B*a)/b^2/(b*e)^(1/2)*arctanh(((b *x^3+a)*e*x)^(1/2)/x^2/(b*e)^(1/2))*e^4*((b*x^3+a)*e*x)^(1/2)/(e*x)^(1/2)/ (b*x^3+a)^(1/2)
Time = 0.57 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.83 \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\left [-\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} b\right )} e^{3} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e + 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right ) - 4 \, {\left (8 \, B b^{2} e^{3} x^{7} + 2 \, {\left (B a b + 6 \, A b^{2}\right )} e^{3} x^{4} - 3 \, {\left (B a^{2} - 2 \, A a b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{288 \, b^{2}}, -\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} b\right )} e^{3} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right ) - 2 \, {\left (8 \, B b^{2} e^{3} x^{7} + 2 \, {\left (B a b + 6 \, A b^{2}\right )} e^{3} x^{4} - 3 \, {\left (B a^{2} - 2 \, A a b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{144 \, b^{2}}\right ] \]
[-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)*sqrt(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)*sqrt(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]
Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (141) = 282\).
Time = 18.80 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.85 \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\begin {cases} \frac {2 \left (\begin {cases} \text {NaN} & \text {for}\: e^{3} = 0 \\\frac {A e^{3} \left (\begin {cases} - \frac {a^{2} e^{3} \left (\begin {cases} \frac {\log {\left (\frac {2 b \left (e x\right )^{\frac {3}{2}}}{e^{3}} + 2 \sqrt {\frac {b}{e^{3}}} \sqrt {a + b x^{3}} \right )}}{\sqrt {\frac {b}{e^{3}}}} & \text {for}\: a \neq 0 \\\frac {\left (e x\right )^{\frac {3}{2}} \log {\left (\left (e x\right )^{\frac {3}{2}} \right )}}{\sqrt {b x^{3}}} & \text {otherwise} \end {cases}\right )}{8 b} + \sqrt {a + b x^{3}} \left (\frac {a e^{3} \left (e x\right )^{\frac {3}{2}}}{8 b} + \frac {\left (e x\right )^{\frac {9}{2}}}{4}\right ) & \text {for}\: \frac {b}{e^{3}} \neq 0 \\\frac {\sqrt {a} \left (e x\right )^{\frac {9}{2}}}{3} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} \frac {a^{3} e^{6} \left (\begin {cases} \frac {\log {\left (\frac {2 b \left (e x\right )^{\frac {3}{2}}}{e^{3}} + 2 \sqrt {\frac {b}{e^{3}}} \sqrt {a + b x^{3}} \right )}}{\sqrt {\frac {b}{e^{3}}}} & \text {for}\: a \neq 0 \\\frac {\left (e x\right )^{\frac {3}{2}} \log {\left (\left (e x\right )^{\frac {3}{2}} \right )}}{\sqrt {b x^{3}}} & \text {otherwise} \end {cases}\right )}{16 b^{2}} + \sqrt {a + b x^{3}} \left (- \frac {a^{2} e^{6} \left (e x\right )^{\frac {3}{2}}}{16 b^{2}} + \frac {a e^{3} \left (e x\right )^{\frac {9}{2}}}{24 b} + \frac {\left (e x\right )^{\frac {15}{2}}}{6}\right ) & \text {for}\: \frac {b}{e^{3}} \neq 0 \\\frac {\sqrt {a} \left (e x\right )^{\frac {15}{2}}}{5} & \text {otherwise} \end {cases}\right )}{3 e^{3}} & \text {otherwise} \end {cases}\right )}{e} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((2*Piecewise((nan, Eq(e**3, 0)), ((A*e**3*Piecewise((-a**2*e**3* Piecewise((log(2*b*(e*x)**(3/2)/e**3 + 2*sqrt(b/e**3)*sqrt(a + b*x**3))/sq rt(b/e**3), Ne(a, 0)), ((e*x)**(3/2)*log((e*x)**(3/2))/sqrt(b*x**3), True) )/(8*b) + sqrt(a + b*x**3)*(a*e**3*(e*x)**(3/2)/(8*b) + (e*x)**(9/2)/4), N e(b/e**3, 0)), (sqrt(a)*(e*x)**(9/2)/3, True)) + B*Piecewise((a**3*e**6*Pi ecewise((log(2*b*(e*x)**(3/2)/e**3 + 2*sqrt(b/e**3)*sqrt(a + b*x**3))/sqrt (b/e**3), Ne(a, 0)), ((e*x)**(3/2)*log((e*x)**(3/2))/sqrt(b*x**3), True))/ (16*b**2) + sqrt(a + b*x**3)*(-a**2*e**6*(e*x)**(3/2)/(16*b**2) + a*e**3*( e*x)**(9/2)/(24*b) + (e*x)**(15/2)/6), Ne(b/e**3, 0)), (sqrt(a)*(e*x)**(15 /2)/5, True)))/(3*e**3), True))/e, Ne(e, 0)), (0, True))
\[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} \sqrt {b x^{3} + a} \left (e x\right )^{\frac {7}{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (126) = 252\).
Time = 0.43 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.03 \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {1}{12} \, \sqrt {b e^{4} x^{3} + a e^{4}} \sqrt {e x} {\left (\frac {2 \, x^{3}}{e} + \frac {a}{b e}\right )} A x {\left | e \right |}^{2} + \frac {1}{72} \, \sqrt {b e^{4} x^{3} + a e^{4}} {\left (2 \, e^{3} x^{3} {\left (\frac {4 \, x^{3}}{e^{4}} + \frac {a}{b e^{4}}\right )} - \frac {3 \, a^{2}}{b^{2} e}\right )} \sqrt {e x} B x {\left | e \right |}^{2} - \frac {{\left (B^{2} a^{6} e - 4 \, A B a^{5} b e + 4 \, A^{2} a^{4} b^{2} e\right )}^{2} e^{5} \log \left ({\left | -{\left (\sqrt {e x} B a^{3} e^{2} x - 2 \, \sqrt {e x} A a^{2} b e^{2} x\right )} \sqrt {b e} + \sqrt {B^{2} a^{7} e^{6} - 4 \, A B a^{6} b e^{6} + 4 \, A^{2} a^{5} b^{2} e^{6} + {\left (\sqrt {e x} B a^{3} e^{2} x - 2 \, \sqrt {e x} A a^{2} b e^{2} x\right )}^{2} b e} \right |}\right )}{24 \, \sqrt {b e} b^{2} {\left | B^{2} a^{6} e - 4 \, A B a^{5} b e + 4 \, A^{2} a^{4} b^{2} e \right |} {\left | -B a^{3} + 2 \, A a^{2} b \right |} {\left | e \right |}^{2}} \]
1/12*sqrt(b*e^4*x^3 + a*e^4)*sqrt(e*x)*(2*x^3/e + a/(b*e))*A*x*abs(e)^2 + 1/72*sqrt(b*e^4*x^3 + a*e^4)*(2*e^3*x^3*(4*x^3/e^4 + a/(b*e^4)) - 3*a^2/(b ^2*e))*sqrt(e*x)*B*x*abs(e)^2 - 1/24*(B^2*a^6*e - 4*A*B*a^5*b*e + 4*A^2*a^ 4*b^2*e)^2*e^5*log(abs(-(sqrt(e*x)*B*a^3*e^2*x - 2*sqrt(e*x)*A*a^2*b*e^2*x )*sqrt(b*e) + sqrt(B^2*a^7*e^6 - 4*A*B*a^6*b*e^6 + 4*A^2*a^5*b^2*e^6 + (sq rt(e*x)*B*a^3*e^2*x - 2*sqrt(e*x)*A*a^2*b*e^2*x)^2*b*e)))/(sqrt(b*e)*b^2*a bs(B^2*a^6*e - 4*A*B*a^5*b*e + 4*A^2*a^4*b^2*e)*abs(-B*a^3 + 2*A*a^2*b)*ab s(e)^2)
Timed out. \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\int \left (B\,x^3+A\right )\,{\left (e\,x\right )}^{7/2}\,\sqrt {b\,x^3+a} \,d x \]