3.3045 \(\int \frac {1}{(a+b x) \sqrt {c+d x} (e+f x)^{3/4}} \, dx\)

Optimal. Leaf size=252 \[ -\frac {2 \sqrt [4]{d e-c f} \sqrt {-\frac {f (c+d x)}{d e-c f}} \Pi \left (-\frac {\sqrt {b} \sqrt {d e-c f}}{\sqrt {d} \sqrt {b e-a f}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{e+f x}}{\sqrt [4]{d e-c f}}\right )\right |-1\right )}{\sqrt [4]{d} \sqrt {c+d x} (b e-a f)}-\frac {2 \sqrt [4]{d e-c f} \sqrt {-\frac {f (c+d x)}{d e-c f}} \Pi \left (\frac {\sqrt {b} \sqrt {d e-c f}}{\sqrt {d} \sqrt {b e-a f}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{e+f x}}{\sqrt [4]{d e-c f}}\right )\right |-1\right )}{\sqrt [4]{d} \sqrt {c+d x} (b e-a f)} \]

[Out]

-2*(-c*f+d*e)^(1/4)*EllipticPi(d^(1/4)*(f*x+e)^(1/4)/(-c*f+d*e)^(1/4),-b^(1/2)*(-c*f+d*e)^(1/2)/d^(1/2)/(-a*f+
b*e)^(1/2),I)*(-f*(d*x+c)/(-c*f+d*e))^(1/2)/d^(1/4)/(-a*f+b*e)/(d*x+c)^(1/2)-2*(-c*f+d*e)^(1/4)*EllipticPi(d^(
1/4)*(f*x+e)^(1/4)/(-c*f+d*e)^(1/4),b^(1/2)*(-c*f+d*e)^(1/2)/d^(1/2)/(-a*f+b*e)^(1/2),I)*(-f*(d*x+c)/(-c*f+d*e
))^(1/2)/d^(1/4)/(-a*f+b*e)/(d*x+c)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {109, 108, 409, 1218} \[ -\frac {2 \sqrt [4]{d e-c f} \sqrt {-\frac {f (c+d x)}{d e-c f}} \Pi \left (-\frac {\sqrt {b} \sqrt {d e-c f}}{\sqrt {d} \sqrt {b e-a f}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{e+f x}}{\sqrt [4]{d e-c f}}\right )\right |-1\right )}{\sqrt [4]{d} \sqrt {c+d x} (b e-a f)}-\frac {2 \sqrt [4]{d e-c f} \sqrt {-\frac {f (c+d x)}{d e-c f}} \Pi \left (\frac {\sqrt {b} \sqrt {d e-c f}}{\sqrt {d} \sqrt {b e-a f}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{e+f x}}{\sqrt [4]{d e-c f}}\right )\right |-1\right )}{\sqrt [4]{d} \sqrt {c+d x} (b e-a f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d*e - c*f)^(1/4)*Sqrt[-((f*(c + d*x))/(d*e - c*f))]*EllipticPi[-((Sqrt[b]*Sqrt[d*e - c*f])/(Sqrt[d]*Sqrt[
b*e - a*f])), ArcSin[(d^(1/4)*(e + f*x)^(1/4))/(d*e - c*f)^(1/4)], -1])/(d^(1/4)*(b*e - a*f)*Sqrt[c + d*x]) -
(2*(d*e - c*f)^(1/4)*Sqrt[-((f*(c + d*x))/(d*e - c*f))]*EllipticPi[(Sqrt[b]*Sqrt[d*e - c*f])/(Sqrt[d]*Sqrt[b*e
 - a*f]), ArcSin[(d^(1/4)*(e + f*x)^(1/4))/(d*e - c*f)^(1/4)], -1])/(d^(1/4)*(b*e - a*f)*Sqrt[c + d*x])

Rule 108

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(3/4)), x_Symbol] :> Dist[-4, Subst[
Int[1/((b*e - a*f - b*x^4)*Sqrt[c - (d*e)/f + (d*x^4)/f]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e
, f}, x] && GtQ[-(f/(d*e - c*f)), 0]

Rule 109

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(3/4)), x_Symbol] :> Dist[Sqrt[-((f*
(c + d*x))/(d*e - c*f))]/Sqrt[c + d*x], Int[1/((a + b*x)*Sqrt[-((c*f)/(d*e - c*f)) - (d*f*x)/(d*e - c*f)]*(e +
 f*x)^(3/4)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&  !GtQ[-(f/(d*e - c*f)), 0]

Rule 409

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

Rule 1218

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rubi steps

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

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Mathematica [C]  time = 0.10, size = 118, normalized size = 0.47 \[ -\frac {4 \sqrt {\frac {b (c+d x)}{d (a+b x)}} \left (\frac {b (e+f x)}{f (a+b x)}\right )^{3/4} F_1\left (\frac {5}{4};\frac {1}{2},\frac {3}{4};\frac {9}{4};\frac {a d-b c}{d (a+b x)},\frac {a f-b e}{f (a+b x)}\right )}{5 b \sqrt {c+d x} (e+f x)^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*Sqrt[(b*(c + d*x))/(d*(a + b*x))]*((b*(e + f*x))/(f*(a + b*x)))^(3/4)*AppellF1[5/4, 1/2, 3/4, 9/4, (-(b*c)
 + a*d)/(d*(a + b*x)), (-(b*e) + a*f)/(f*(a + b*x))])/(5*b*Sqrt[c + d*x]*(e + f*x)^(3/4))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(f*x+e)^(3/4)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(f*x+e)^(3/4)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x +a \right ) \left (f x +e \right )^{\frac {3}{4}} \sqrt {d x +c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)/(f*x+e)^(3/4)/(d*x+c)^(1/2),x)

[Out]

int(1/(b*x+a)/(f*x+e)^(3/4)/(d*x+c)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(f*x+e)^(3/4)/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e + f*x)^(3/4)*(a + b*x)*(c + d*x)^(1/2)),x)

[Out]

int(1/((e + f*x)^(3/4)*(a + b*x)*(c + d*x)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(f*x+e)**(3/4)/(d*x+c)**(1/2),x)

[Out]

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

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