3.16.74 \(\int \frac {1}{(a+b x)^{5/2} (c+d x)^{2/3}} \, dx\) [1574]

Optimal. Leaf size=421 \[ -\frac {2 \sqrt [3]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}+\frac {14 d \sqrt [3]{c+d x}}{9 (b c-a d)^2 \sqrt {a+b x}}-\frac {14 \sqrt {2-\sqrt {3}} d \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right )|-7+4 \sqrt {3}\right )}{9 \sqrt [4]{3} \sqrt [3]{b} (b c-a d)^2 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \]

[Out]

-2/3*(d*x+c)^(1/3)/(-a*d+b*c)/(b*x+a)^(3/2)+14/9*d*(d*x+c)^(1/3)/(-a*d+b*c)^2/(b*x+a)^(1/2)-14/27*d*((-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))*3^(3/4)/b^(1/3)/(-a*d+b*c)^2/(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 = 0.27, antiderivative size = 421, 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 = {53, 65, 225} \begin {gather*} -\frac {14 \sqrt {2-\sqrt {3}} d \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 )}{9 \sqrt [4]{3} \sqrt [3]{b} \sqrt {a+b x} (b c-a d)^2 \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 {14 d \sqrt [3]{c+d x}}{9 \sqrt {a+b x} (b c-a d)^2}-\frac {2 \sqrt [3]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(c + d*x)^(1/3))/(3*(b*c - a*d)*(a + b*x)^(3/2)) + (14*d*(c + d*x)^(1/3))/(9*(b*c - a*d)^2*Sqrt[a + b*x])
- (14*Sqrt[2 - Sqrt[3]]*d*((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]])/(9*3^(1/4)*b^(1/3)*(b*c - a*d)^2*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 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 \frac {1}{(a+b x)^{5/2} (c+d x)^{2/3}} \, dx &=-\frac {2 \sqrt [3]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}-\frac {(7 d) \int \frac {1}{(a+b x)^{3/2} (c+d x)^{2/3}} \, dx}{9 (b c-a d)}\\ &=-\frac {2 \sqrt [3]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}+\frac {14 d \sqrt [3]{c+d x}}{9 (b c-a d)^2 \sqrt {a+b x}}+\frac {\left (7 d^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{2/3}} \, dx}{27 (b c-a d)^2}\\ &=-\frac {2 \sqrt [3]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}+\frac {14 d \sqrt [3]{c+d x}}{9 (b c-a d)^2 \sqrt {a+b x}}+\frac {(7 d) \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 )}{9 (b c-a d)^2}\\ &=-\frac {2 \sqrt [3]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}+\frac {14 d \sqrt [3]{c+d x}}{9 (b c-a d)^2 \sqrt {a+b x}}-\frac {14 \sqrt {2-\sqrt {3}} d \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 )}{9 \sqrt [4]{3} \sqrt [3]{b} (b c-a d)^2 \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 = 10.04, size = 73, normalized size = 0.17 \begin {gather*} -\frac {2 \left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} \, _2F_1\left (-\frac {3}{2},\frac {2}{3};-\frac {1}{2};\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} (c+d x)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(5/2)/(d*x+c)^(2/3),x)

[Out]

int(1/(b*x+a)^(5/2)/(d*x+c)^(2/3),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(1/(b*x+a)^(5/2)/(d*x+c)^(2/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(5/2)*(d*x + c)^(2/3)), 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(1/(b*x+a)^(5/2)/(d*x+c)^(2/3),x, algorithm="fricas")

[Out]

integral(sqrt(b*x + a)*(d*x + c)^(1/3)/(b^3*d*x^4 + a^3*c + (b^3*c + 3*a*b^2*d)*x^3 + 3*(a*b^2*c + a^2*b*d)*x^
2 + (3*a^2*b*c + a^3*d)*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {2}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(5/2)/(d*x+c)**(2/3),x)

[Out]

Integral(1/((a + b*x)**(5/2)*(c + d*x)**(2/3)), 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(1/(b*x+a)^(5/2)/(d*x+c)^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(5/2)*(d*x + c)^(2/3)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(5/2)*(c + d*x)^(2/3)),x)

[Out]

int(1/((a + b*x)^(5/2)*(c + d*x)^(2/3)), x)

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