3.1570 \(\int \frac {(a+b x)^{3/2}}{(c+d x)^{2/3}} \, dx\)

Optimal. Leaf size=416 \[ -\frac {54\ 3^{3/4} \sqrt {2-\sqrt {3}} (b c-a d)^2 \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),4 \sqrt {3}-7\right )}{55 \sqrt [3]{b} d^3 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}-\frac {54 \sqrt {a+b x} \sqrt [3]{c+d x} (b c-a d)}{55 d^2}+\frac {6 (a+b x)^{3/2} \sqrt [3]{c+d x}}{11 d} \]

[Out]

6/11*(b*x+a)^(3/2)*(d*x+c)^(1/3)/d-54/55*(-a*d+b*c)*(d*x+c)^(1/3)*(b*x+a)^(1/2)/d^2-54/55*3^(3/4)*(-a*d+b*c)^2
*((-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^(1/3)/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 = 0.39, antiderivative size = 416, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {50, 63, 219} \[ -\frac {54\ 3^{3/4} \sqrt {2-\sqrt {3}} (b c-a d)^2 \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right )|-7+4 \sqrt {3}\right )}{55 \sqrt [3]{b} d^3 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}-\frac {54 \sqrt {a+b x} \sqrt [3]{c+d x} (b c-a d)}{55 d^2}+\frac {6 (a+b x)^{3/2} \sqrt [3]{c+d x}}{11 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-54*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(1/3))/(55*d^2) + (6*(a + b*x)^(3/2)*(c + d*x)^(1/3))/(11*d) - (54*3^
(3/4)*Sqrt[2 - Sqrt[3]]*(b*c - a*d)^2*((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]])/(55*b^(1/3)*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 \frac {(a+b x)^{3/2}}{(c+d x)^{2/3}} \, dx &=\frac {6 (a+b x)^{3/2} \sqrt [3]{c+d x}}{11 d}-\frac {(9 (b c-a d)) \int \frac {\sqrt {a+b x}}{(c+d x)^{2/3}} \, dx}{11 d}\\ &=-\frac {54 (b c-a d) \sqrt {a+b x} \sqrt [3]{c+d x}}{55 d^2}+\frac {6 (a+b x)^{3/2} \sqrt [3]{c+d x}}{11 d}+\frac {\left (27 (b c-a d)^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{2/3}} \, dx}{55 d^2}\\ &=-\frac {54 (b c-a d) \sqrt {a+b x} \sqrt [3]{c+d x}}{55 d^2}+\frac {6 (a+b x)^{3/2} \sqrt [3]{c+d x}}{11 d}+\frac {\left (81 (b c-a d)^2\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 )}{55 d^3}\\ &=-\frac {54 (b c-a d) \sqrt {a+b x} \sqrt [3]{c+d x}}{55 d^2}+\frac {6 (a+b x)^{3/2} \sqrt [3]{c+d x}}{11 d}-\frac {54\ 3^{3/4} \sqrt {2-\sqrt {3}} (b c-a d)^2 \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 )}{55 \sqrt [3]{b} 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]  time = 0.03, size = 73, normalized size = 0.18 \[ \frac {2 (a+b x)^{5/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {5}{2};\frac {7}{2};\frac {d (a+b x)}{a d-b c}\right )}{5 b (c+d x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{\frac {2}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{\frac {3}{2}}}{\left (d x +c \right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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