3.1759 \(\int \frac {1}{(a+b x)^{5/2} (c+d x)^{5/6}} \, dx\)

Optimal. Leaf size=410 \[ \frac {16 d \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\cos ^{-1}\left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{9 \sqrt [4]{3} \sqrt {a+b x} (b c-a d)^{7/3} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}+\frac {16 d \sqrt [6]{c+d x}}{9 \sqrt {a+b x} (b c-a d)^2}-\frac {2 \sqrt [6]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)} \]

[Out]

-2/3*(d*x+c)^(1/6)/(-a*d+b*c)/(b*x+a)^(3/2)+16/9*d*(d*x+c)^(1/6)/(-a*d+b*c)^2/(b*x+a)^(1/2)+16/27*d*(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)*3^(3/4)/(-a*d+b*c)^(7/3)/(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 = 0.27, antiderivative size = 410, 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 = {51, 63, 225} \[ \frac {16 d \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{9 \sqrt [4]{3} \sqrt {a+b x} (b c-a d)^{7/3} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}+\frac {16 d \sqrt [6]{c+d x}}{9 \sqrt {a+b x} (b c-a d)^2}-\frac {2 \sqrt [6]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(5/2)*(c + d*x)^(5/6)),x]

[Out]

(-2*(c + d*x)^(1/6))/(3*(b*c - a*d)*(a + b*x)^(3/2)) + (16*d*(c + d*x)^(1/6))/(9*(b*c - a*d)^2*Sqrt[a + b*x])
+ (16*d*(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])/(9*3^(1/4)*(b*c - a*d)^(7/3)*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 51

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 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 {1}{(a+b x)^{5/2} (c+d x)^{5/6}} \, dx &=-\frac {2 \sqrt [6]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}-\frac {(8 d) \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/6}} \, dx}{9 (b c-a d)}\\ &=-\frac {2 \sqrt [6]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}+\frac {16 d \sqrt [6]{c+d x}}{9 (b c-a d)^2 \sqrt {a+b x}}+\frac {\left (16 d^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}} \, dx}{27 (b c-a d)^2}\\ &=-\frac {2 \sqrt [6]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}+\frac {16 d \sqrt [6]{c+d x}}{9 (b c-a d)^2 \sqrt {a+b x}}+\frac {(32 d) \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 )}{9 (b c-a d)^2}\\ &=-\frac {2 \sqrt [6]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}+\frac {16 d \sqrt [6]{c+d x}}{9 (b c-a d)^2 \sqrt {a+b x}}+\frac {16 d \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 )}{9 \sqrt [4]{3} (b c-a d)^{7/3} \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 = 0.03, size = 73, normalized size = 0.18 \[ -\frac {2 \left (\frac {b (c+d x)}{b c-a d}\right )^{5/6} \, _2F_1\left (-\frac {3}{2},\frac {5}{6};-\frac {1}{2};\frac {d (a+b x)}{a d-b c}\right )}{3 b (a+b x)^{3/2} (c+d x)^{5/6}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(5/2)*(c + d*x)^(5/6)),x]

[Out]

(-2*((b*(c + d*x))/(b*c - a*d))^(5/6)*Hypergeometric2F1[-3/2, 5/6, -1/2, (d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(
a + b*x)^(3/2)*(c + d*x)^(5/6))

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x + a} {\left (d x + c\right )}^{\frac {1}{6}}}{b^{3} d x^{4} + a^{3} c + {\left (b^{3} c + 3 \, a b^{2} d\right )} x^{3} + 3 \, {\left (a b^{2} c + a^{2} b d\right )} x^{2} + {\left (3 \, a^{2} b c + a^{3} d\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(5/6),x, algorithm="fricas")

[Out]

integral(sqrt(b*x + a)*(d*x + c)^(1/6)/(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {5}{6}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(5/6),x, algorithm="giac")

[Out]

integrate(1/((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 \[ \int \frac {1}{\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {5}{6}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(5/2)/(d*x+c)^(5/6),x)

[Out]

int(1/(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 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {5}{6}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(5/6),x, algorithm="maxima")

[Out]

integrate(1/((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 \[ \int \frac {1}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{5/6}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(5/2)*(c + d*x)^(5/6)),x)

[Out]

int(1/((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 \[ \int \frac {1}{\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {5}{6}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(5/2)/(d*x+c)**(5/6),x)

[Out]

Integral(1/((a + b*x)**(5/2)*(c + d*x)**(5/6)), x)

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