3.1754 \(\int \frac {(a+b x)^{5/2}}{(c+d x)^{5/6}} \, dx\)

Optimal. Leaf size=440 \[ -\frac {81\ 3^{3/4} \sqrt [6]{c+d x} (b c-a d)^{8/3} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\cos ^{-1}\left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{128 d^4 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}+\frac {81 \sqrt {a+b x} \sqrt [6]{c+d x} (b c-a d)^2}{64 d^3}-\frac {9 (a+b x)^{3/2} \sqrt [6]{c+d x} (b c-a d)}{16 d^2}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 d} \]

[Out]

-9/16*(-a*d+b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/6)/d^2+3/8*(b*x+a)^(5/2)*(d*x+c)^(1/6)/d+81/64*(-a*d+b*c)^2*(d*x+c)^
(1/6)*(b*x+a)^(1/2)/d^3-81/128*3^(3/4)*(-a*d+b*c)^(8/3)*(d*x+c)^(1/6)*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))
*(((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1-3^(1/2)))^2/((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1+3^(1/2)))^
2)^(1/2)/((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1-3^(1/2)))*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1+3^(1/
2)))*EllipticF((1-((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1-3^(1/2)))^2/((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/
3)*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(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))/((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1+3^(1/2)))^2)^(1/2)/d^4/(b*x+a)^(1/2)/(-b^(1/3)
*(d*x+c)^(1/3)*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))/((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1+3^(1/2)))^2
)^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 440, 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 = {50, 63, 225} \[ -\frac {81\ 3^{3/4} \sqrt [6]{c+d x} (b c-a d)^{8/3} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{128 d^4 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}+\frac {81 \sqrt {a+b x} \sqrt [6]{c+d x} (b c-a d)^2}{64 d^3}-\frac {9 (a+b x)^{3/2} \sqrt [6]{c+d x} (b c-a d)}{16 d^2}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/(c + d*x)^(5/6),x]

[Out]

(81*(b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(1/6))/(64*d^3) - (9*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(1/6))/(1
6*d^2) + (3*(a + b*x)^(5/2)*(c + d*x)^(1/6))/(8*d) - (81*3^(3/4)*(b*c - a*d)^(8/3)*(c + d*x)^(1/6)*((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))/((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))^2]*EllipticF[ArcCos[((b*c - a*
d)^(1/3) - (1 - Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))/((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))]
, (2 + Sqrt[3])/4])/(128*d^4*Sqrt[a + b*x]*Sqrt[-((b^(1/3)*(c + d*x)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d
*x)^(1/3)))/((b*c - a*d)^(1/3) - (1 + Sqrt[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 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{5/2}}{(c+d x)^{5/6}} \, dx &=\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 d}-\frac {(15 (b c-a d)) \int \frac {(a+b x)^{3/2}}{(c+d x)^{5/6}} \, dx}{16 d}\\ &=-\frac {9 (b c-a d) (a+b x)^{3/2} \sqrt [6]{c+d x}}{16 d^2}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 d}+\frac {\left (27 (b c-a d)^2\right ) \int \frac {\sqrt {a+b x}}{(c+d x)^{5/6}} \, dx}{32 d^2}\\ &=\frac {81 (b c-a d)^2 \sqrt {a+b x} \sqrt [6]{c+d x}}{64 d^3}-\frac {9 (b c-a d) (a+b x)^{3/2} \sqrt [6]{c+d x}}{16 d^2}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 d}-\frac {\left (81 (b c-a d)^3\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}} \, dx}{128 d^3}\\ &=\frac {81 (b c-a d)^2 \sqrt {a+b x} \sqrt [6]{c+d x}}{64 d^3}-\frac {9 (b c-a d) (a+b x)^{3/2} \sqrt [6]{c+d x}}{16 d^2}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 d}-\frac {\left (243 (b c-a d)^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-\frac {b c}{d}+\frac {b x^6}{d}}} \, dx,x,\sqrt [6]{c+d x}\right )}{64 d^4}\\ &=\frac {81 (b c-a d)^2 \sqrt {a+b x} \sqrt [6]{c+d x}}{64 d^3}-\frac {9 (b c-a d) (a+b x)^{3/2} \sqrt [6]{c+d x}}{16 d^2}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 d}-\frac {81\ 3^{3/4} (b c-a d)^{8/3} \sqrt [6]{c+d x} \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 (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{128 d^4 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 73, normalized size = 0.17 \[ \frac {2 (a+b x)^{7/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/6} \, _2F_1\left (\frac {5}{6},\frac {7}{2};\frac {9}{2};\frac {d (a+b x)}{a d-b c}\right )}{7 b (c+d x)^{5/6}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/(c + d*x)^(5/6),x]

[Out]

(2*(a + b*x)^(7/2)*((b*(c + d*x))/(b*c - a*d))^(5/6)*Hypergeometric2F1[5/6, 7/2, 9/2, (d*(a + b*x))/(-(b*c) +
a*d)])/(7*b*(c + d*x)^(5/6))

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b x + a}}{{\left (d x + c\right )}^{\frac {5}{6}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/(d*x+c)^(5/6),x, algorithm="fricas")

[Out]

integral((b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*x + a)/(d*x + c)^(5/6), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{\frac {5}{2}}}{{\left (d x + c\right )}^{\frac {5}{6}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/(d*x+c)^(5/6),x, algorithm="giac")

[Out]

integrate((b*x + a)^(5/2)/(d*x + c)^(5/6), x)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{\frac {5}{2}}}{\left (d x +c \right )^{\frac {5}{6}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)/(d*x+c)^(5/6),x)

[Out]

int((b*x+a)^(5/2)/(d*x+c)^(5/6),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{\frac {5}{2}}}{{\left (d x + c\right )}^{\frac {5}{6}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/(d*x+c)^(5/6),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(5/2)/(d*x + c)^(5/6), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,x\right )}^{5/2}}{{\left (c+d\,x\right )}^{5/6}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(5/2)/(c + d*x)^(5/6),x)

[Out]

int((a + b*x)^(5/2)/(c + d*x)^(5/6), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right )^{\frac {5}{2}}}{\left (c + d x\right )^{\frac {5}{6}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)/(d*x+c)**(5/6),x)

[Out]

Integral((a + b*x)**(5/2)/(c + d*x)**(5/6), x)

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