3.16.59 \(\int (a+b x)^{3/2} \sqrt [3]{c+d x} \, dx\) [1559]

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

[Out]

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

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Rubi [A]
time = 0.60, antiderivative size = 457, normalized size of antiderivative = 1.00, 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 = {52, 65, 225} \begin {gather*} -\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 (\text {ArcSin}\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} \end {gather*}

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 52

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 65

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 225

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.06, size = 73, normalized size = 0.16 \begin {gather*} \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)}{-b c+a d}\right )}{5 b \sqrt [3]{\frac {b (c+d x)}{b c-a d}}} \end {gather*}

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))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {1}{3}}\, dx\]

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)

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b x\right )^{\frac {3}{2}} \sqrt [3]{c + d x}\, dx \end {gather*}

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)

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{1/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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