3.1714 \(\int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx\)

Optimal. Leaf size=270 \[ \frac {\sqrt {2} \sqrt {b c-a d} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt [4]{d} (a+b x)^{3/4} (c+d x)^{3/4} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]

[Out]

((b*x+a)*(d*x+c))^(3/4)*(cos(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))^2)^(1
/2)/cos(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))*EllipticF(sin(2*arctan(b^(
1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2))),1/2*2^(1/2))*2^(1/2)*(-a*d+b*c)^(1/2)*(1+2*b^(
1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d+b*(2*d*x+c))^2/(-a*d+b*c)^2
/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))^2)^(1/2)/b^(1/4)/d^(1/4)/(b*x+a)^(3/4)/(d*x+c)^(3/4)
/(2*b*d*x+a*d+b*c)/((a*d+b*(2*d*x+c))^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {62, 623, 220} \[ \frac {\sqrt {2} \sqrt {b c-a d} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt [4]{d} (a+b x)^{3/4} (c+d x)^{3/4} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(3/4)*(c + d*x)^(3/4)),x]

[Out]

(Sqrt[2]*Sqrt[b*c - a*d]*((a + b*x)*(c + d*x))^(3/4)*Sqrt[(b*c + a*d + 2*b*d*x)^2]*(1 + (2*Sqrt[b]*Sqrt[d]*Sqr
t[(a + b*x)*(c + d*x)])/(b*c - a*d))*Sqrt[(a*d + b*(c + 2*d*x))^2/((b*c - a*d)^2*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[
(a + b*x)*(c + d*x)])/(b*c - a*d))^2)]*EllipticF[2*ArcTan[(Sqrt[2]*b^(1/4)*d^(1/4)*((a + b*x)*(c + d*x))^(1/4)
)/Sqrt[b*c - a*d]], 1/2])/(b^(1/4)*d^(1/4)*(a + b*x)^(3/4)*(c + d*x)^(3/4)*(b*c + a*d + 2*b*d*x)*Sqrt[(a*d + b
*(c + 2*d*x))^2])

Rule 62

Int[((a_.) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Dist[((a + b*x)^m*(c + d*x)^m)/((a + b*x)
*(c + d*x))^m, Int[(a*c + (b*c + a*d)*x + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] &&
LtQ[-1, m, 0] && LeQ[3, Denominator[m], 4]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 623

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[(d*Sqrt[(b + 2*c*x)
^2])/(b + 2*c*x), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rubi steps

\begin {align*} \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx &=\frac {((a+b x) (c+d x))^{3/4} \int \frac {1}{\left (a c+(b c+a d) x+b d x^2\right )^{3/4}} \, dx}{(a+b x)^{3/4} (c+d x)^{3/4}}\\ &=\frac {\left (4 ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{(a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x)}\\ &=\frac {\sqrt {2} \sqrt {b c-a d} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt [4]{d} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 71, normalized size = 0.26 \[ \frac {4 \sqrt [4]{a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};\frac {d (a+b x)}{a d-b c}\right )}{b (c+d x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(3/4)*(c + d*x)^(3/4)),x]

[Out]

(4*(a + b*x)^(1/4)*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[1/4, 3/4, 5/4, (d*(a + b*x))/(-(b*c) +
a*d)])/(b*(c + d*x)^(3/4))

________________________________________________________________________________________

fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{b d x^{2} + a c + {\left (b c + a d\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(3/4)/(d*x+c)^(3/4),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(3/4)/(d*x+c)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(3/4)*(d*x + c)^(3/4)), x)

________________________________________________________________________________________

maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x +a \right )^{\frac {3}{4}} \left (d x +c \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(3/4)/(d*x+c)^(3/4),x)

[Out]

int(1/(b*x+a)^(3/4)/(d*x+c)^(3/4),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(3/4)/(d*x+c)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(3/4)*(d*x + c)^(3/4)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+b\,x\right )}^{3/4}\,{\left (c+d\,x\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(3/4)*(c + d*x)^(3/4)),x)

[Out]

int(1/((a + b*x)^(3/4)*(c + d*x)^(3/4)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x\right )^{\frac {3}{4}} \left (c + d x\right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(3/4)/(d*x+c)**(3/4),x)

[Out]

Integral(1/((a + b*x)**(3/4)*(c + d*x)**(3/4)), x)

________________________________________________________________________________________