3.1559 \(\int (a+b x)^{3/2} \sqrt [3]{c+d x} \, dx\)

Optimal. Leaf size=457 \[ -\frac{108\ 3^{3/4} \sqrt{2-\sqrt{3}} (b c-a d)^3 \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt{\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),4 \sqrt{3}-7\right )}{935 b^{4/3} d^3 \sqrt{a+b x} \sqrt{-\frac{\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}-\frac{108 \sqrt{a+b x} \sqrt [3]{c+d x} (b c-a d)^2}{935 b d^2}+\frac{12 (a+b x)^{3/2} \sqrt [3]{c+d x} (b c-a d)}{187 b d}+\frac{6 (a+b x)^{5/2} \sqrt [3]{c+d x}}{17 b} \]

[Out]

(-108*(b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(1/3))/(935*b*d^2) + (12*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(1/
3))/(187*b*d) + (6*(a + b*x)^(5/2)*(c + d*x)^(1/3))/(17*b) - (108*3^(3/4)*Sqrt[2 - Sqrt[3]]*(b*c - a*d)^3*((b*
c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))*Sqrt[((b*c - a*d)^(2/3) + b^(1/3)*(b*c - a*d)^(1/3)*(c + d*x)^(1/3)
+ b^(2/3)*(c + d*x)^(2/3))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2]*EllipticF[ArcSin[((1
+ Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^
(1/3))], -7 + 4*Sqrt[3]])/(935*b^(4/3)*d^3*Sqrt[a + b*x]*Sqrt[-(((b*c - a*d)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3
)*(c + d*x)^(1/3)))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2)])

________________________________________________________________________________________

Rubi [A]  time = 0.627175, antiderivative size = 457, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {50, 63, 219} \[ -\frac{108\ 3^{3/4} \sqrt{2-\sqrt{3}} (b c-a d)^3 \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt{\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right )|-7+4 \sqrt{3}\right )}{935 b^{4/3} d^3 \sqrt{a+b x} \sqrt{-\frac{\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}-\frac{108 \sqrt{a+b x} \sqrt [3]{c+d x} (b c-a d)^2}{935 b d^2}+\frac{12 (a+b x)^{3/2} \sqrt [3]{c+d x} (b c-a d)}{187 b d}+\frac{6 (a+b x)^{5/2} \sqrt [3]{c+d x}}{17 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)*(c + d*x)^(1/3),x]

[Out]

(-108*(b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(1/3))/(935*b*d^2) + (12*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(1/
3))/(187*b*d) + (6*(a + b*x)^(5/2)*(c + d*x)^(1/3))/(17*b) - (108*3^(3/4)*Sqrt[2 - Sqrt[3]]*(b*c - a*d)^3*((b*
c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))*Sqrt[((b*c - a*d)^(2/3) + b^(1/3)*(b*c - a*d)^(1/3)*(c + d*x)^(1/3)
+ b^(2/3)*(c + d*x)^(2/3))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2]*EllipticF[ArcSin[((1
+ Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^
(1/3))], -7 + 4*Sqrt[3]])/(935*b^(4/3)*d^3*Sqrt[a + b*x]*Sqrt[-(((b*c - a*d)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3
)*(c + d*x)^(1/3)))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2)])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 219

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 - Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 + Sqrt[3
])*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[-((s*(s + r*x))/((1 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rubi steps

\begin{align*} \int (a+b x)^{3/2} \sqrt [3]{c+d x} \, dx &=\frac{6 (a+b x)^{5/2} \sqrt [3]{c+d x}}{17 b}+\frac{(2 (b c-a d)) \int \frac{(a+b x)^{3/2}}{(c+d x)^{2/3}} \, dx}{17 b}\\ &=\frac{12 (b c-a d) (a+b x)^{3/2} \sqrt [3]{c+d x}}{187 b d}+\frac{6 (a+b x)^{5/2} \sqrt [3]{c+d x}}{17 b}-\frac{\left (18 (b c-a d)^2\right ) \int \frac{\sqrt{a+b x}}{(c+d x)^{2/3}} \, dx}{187 b d}\\ &=-\frac{108 (b c-a d)^2 \sqrt{a+b x} \sqrt [3]{c+d x}}{935 b d^2}+\frac{12 (b c-a d) (a+b x)^{3/2} \sqrt [3]{c+d x}}{187 b d}+\frac{6 (a+b x)^{5/2} \sqrt [3]{c+d x}}{17 b}+\frac{\left (54 (b c-a d)^3\right ) \int \frac{1}{\sqrt{a+b x} (c+d x)^{2/3}} \, dx}{935 b d^2}\\ &=-\frac{108 (b c-a d)^2 \sqrt{a+b x} \sqrt [3]{c+d x}}{935 b d^2}+\frac{12 (b c-a d) (a+b x)^{3/2} \sqrt [3]{c+d x}}{187 b d}+\frac{6 (a+b x)^{5/2} \sqrt [3]{c+d x}}{17 b}+\frac{\left (162 (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-\frac{b c}{d}+\frac{b x^3}{d}}} \, dx,x,\sqrt [3]{c+d x}\right )}{935 b d^3}\\ &=-\frac{108 (b c-a d)^2 \sqrt{a+b x} \sqrt [3]{c+d x}}{935 b d^2}+\frac{12 (b c-a d) (a+b x)^{3/2} \sqrt [3]{c+d x}}{187 b d}+\frac{6 (a+b x)^{5/2} \sqrt [3]{c+d x}}{17 b}-\frac{108\ 3^{3/4} \sqrt{2-\sqrt{3}} (b c-a d)^3 \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt{\frac{(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right )|-7+4 \sqrt{3}\right )}{935 b^{4/3} d^3 \sqrt{a+b x} \sqrt{-\frac{\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}\\ \end{align*}

Mathematica [C]  time = 0.0381882, size = 73, normalized size = 0.16 \[ \frac{2 (a+b x)^{5/2} \sqrt [3]{c+d x} \, _2F_1\left (-\frac{1}{3},\frac{5}{2};\frac{7}{2};\frac{d (a+b x)}{a d-b c}\right )}{5 b \sqrt [3]{\frac{b (c+d x)}{b c-a d}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)*(c + d*x)^(1/3),x]

[Out]

(2*(a + b*x)^(5/2)*(c + d*x)^(1/3)*Hypergeometric2F1[-1/3, 5/2, 7/2, (d*(a + b*x))/(-(b*c) + a*d)])/(5*b*((b*(
c + d*x))/(b*c - a*d))^(1/3))

________________________________________________________________________________________

Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{{\frac{3}{2}}}\sqrt [3]{dx+c}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(d*x+c)^(1/3),x)

[Out]

int((b*x+a)^(3/2)*(d*x+c)^(1/3),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{\frac{3}{2}}{\left (d x + c\right )}^{\frac{1}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(d*x+c)^(1/3),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(3/2)*(d*x + c)^(1/3), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x + a\right )}^{\frac{3}{2}}{\left (d x + c\right )}^{\frac{1}{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(d*x+c)^(1/3),x, algorithm="fricas")

[Out]

integral((b*x + a)^(3/2)*(d*x + c)^(1/3), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b x\right )^{\frac{3}{2}} \sqrt [3]{c + d x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(d*x+c)**(1/3),x)

[Out]

Integral((a + b*x)**(3/2)*(c + d*x)**(1/3), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{\frac{3}{2}}{\left (d x + c\right )}^{\frac{1}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(d*x+c)^(1/3),x, algorithm="giac")

[Out]

integrate((b*x + a)^(3/2)*(d*x + c)^(1/3), x)