3.18.26 \(\int \frac {1}{\sqrt [4]{a+b x} (c+d x)^{5/4}} \, dx\) [1726]

Optimal. Leaf size=719 \[ \frac {4 (a+b x)^{3/4}}{(b c-a d) \sqrt [4]{c+d x}}-\frac {4 \sqrt {b} \sqrt {(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \sqrt {(a d+b (c+2 d x))^2}}{\sqrt {d} (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )}+\frac {2 \sqrt {2} \sqrt [4]{b} \sqrt {b c-a d} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{d^{3/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {b c-a d} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{d^{3/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]

[Out]

4*(b*x+a)^(3/4)/(-a*d+b*c)/(d*x+c)^(1/4)-4*b^(1/2)*((b*x+a)*(d*x+c))^(1/2)*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d+b
*(2*d*x+c))^2)^(1/2)/(-a*d+b*c)^2/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*d*x+a*d+b*c)/d^(1/2)/(1+2*b^(1/2)*d^(1/2)*(
(b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))+2*b^(1/4)*((b*x+a)*(d*x+c))^(1/4)*(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)))*EllipticE(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
)/d^(3/4)/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*d*x+a*d+b*c)/((a*d+b*(2*d*x+c))^2)^(1/2)-b^(1/4)*((b*x+a)*(d*x+c))^
(1/4)*(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)/d^(3/4)/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*d*x+a*d+b*c)/((a*d+b
*(2*d*x+c))^2)^(1/2)

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Rubi [A]
time = 0.44, antiderivative size = 719, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {53, 64, 637, 311, 226, 1210} \begin {gather*} -\frac {\sqrt {2} \sqrt [4]{b} \sqrt {b c-a d} \sqrt [4]{(a+b x) (c+d x)} \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 \text {ArcTan}\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 )}{d^{3/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {2 \sqrt {2} \sqrt [4]{b} \sqrt {b c-a d} \sqrt [4]{(a+b x) (c+d x)} \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}} E\left (2 \text {ArcTan}\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 )}{d^{3/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {4 \sqrt {b} \sqrt {(a+b x) (c+d x)} \sqrt {(a d+b c+2 b d x)^2} \sqrt {(a d+b (c+2 d x))^2}}{\sqrt {d} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c-a d)^2 (a d+b c+2 b d x) \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )}+\frac {4 (a+b x)^{3/4}}{\sqrt [4]{c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*x)^(3/4))/((b*c - a*d)*(c + d*x)^(1/4)) - (4*Sqrt[b]*Sqrt[(a + b*x)*(c + d*x)]*Sqrt[(b*c + a*d + 2*b
*d*x)^2]*Sqrt[(a*d + b*(c + 2*d*x))^2])/(Sqrt[d]*(b*c - a*d)^2*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(b*c + a*d + 2*
b*d*x)*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))) + (2*Sqrt[2]*b^(1/4)*Sqrt[b*c - a*d]*(
(a + b*x)*(c + d*x))^(1/4)*Sqrt[(b*c + a*d + 2*b*d*x)^2]*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(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)]*EllipticE[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])/
(d^(3/4)*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(b*c + a*d + 2*b*d*x)*Sqrt[(a*d + b*(c + 2*d*x))^2]) - (Sqrt[2]*b^(1/
4)*Sqrt[b*c - a*d]*((a + b*x)*(c + d*x))^(1/4)*Sqrt[(b*c + a*d + 2*b*d*x)^2]*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(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])/(d^(3/4)*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(b*c + a*d + 2*b*d*x)*Sqrt[(a*d + b*(c + 2*d*x))^
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 64

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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

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

Rule 637

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]

Rule 1210

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

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [4]{a+b x} (c+d x)^{5/4}} \, dx &=\frac {4 (a+b x)^{3/4}}{(b c-a d) \sqrt [4]{c+d x}}-\frac {(2 b) \int \frac {1}{\sqrt [4]{a+b x} \sqrt [4]{c+d x}} \, dx}{b c-a d}\\ &=\frac {4 (a+b x)^{3/4}}{(b c-a d) \sqrt [4]{c+d x}}-\frac {\left (2 b \sqrt [4]{(a+b x) (c+d x)}\right ) \int \frac {1}{\sqrt [4]{a c+(b c+a d) x+b d x^2}} \, dx}{(b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}\\ &=\frac {4 (a+b x)^{3/4}}{(b c-a d) \sqrt [4]{c+d x}}-\frac {\left (8 b \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2}\right ) \text {Subst}\left (\int \frac {x^2}{\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 )}{(b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}\\ &=\frac {4 (a+b x)^{3/4}}{(b c-a d) \sqrt [4]{c+d x}}-\frac {\left (4 \sqrt {b} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2}\right ) \text {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 )}{\sqrt {d} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}+\frac {\left (4 \sqrt {b} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2}\right ) \text {Subst}\left (\int \frac {1-\frac {2 \sqrt {b} \sqrt {d} x^2}{b c-a d}}{\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 )}{\sqrt {d} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}\\ &=\frac {4 (a+b x)^{3/4}}{(b c-a d) \sqrt [4]{c+d x}}-\frac {4 \sqrt {b} \sqrt {(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \sqrt {(a d+b (c+2 d x))^2}}{\sqrt {d} (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )}+\frac {2 \sqrt {2} \sqrt [4]{b} \sqrt {b c-a d} \sqrt [4]{(a+b x) (c+d x)} \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}} E\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 )}{d^{3/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {b c-a d} \sqrt [4]{(a+b x) (c+d x)} \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 )}{d^{3/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^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 = 10.03, size = 73, normalized size = 0.10 \begin {gather*} \frac {4 (a+b x)^{3/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4} \, _2F_1\left (\frac {3}{4},\frac {5}{4};\frac {7}{4};\frac {d (a+b x)}{-b c+a d}\right )}{3 b (c+d x)^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*x)^(3/4)*((b*(c + d*x))/(b*c - a*d))^(5/4)*Hypergeometric2F1[3/4, 5/4, 7/4, (d*(a + b*x))/(-(b*c) +
a*d)])/(3*b*(c + d*x)^(5/4))

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b x +a \right )^{\frac {1}{4}} \left (d x +c \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(1/4)/(d*x+c)^(5/4),x)

[Out]

int(1/(b*x+a)^(1/4)/(d*x+c)^(5/4),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)^(1/4)/(d*x+c)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(1/4)*(d*x + c)^(5/4)), 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)^(1/4)/(d*x+c)^(5/4),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{a + b x} \left (c + d x\right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(1/4)/(d*x+c)**(5/4),x)

[Out]

Integral(1/((a + b*x)**(1/4)*(c + d*x)**(5/4)), 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)^(1/4)/(d*x+c)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(1/4)*(d*x + c)^(5/4)), 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 )}^{1/4}\,{\left (c+d\,x\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(1/4)*(c + d*x)^(5/4)),x)

[Out]

int(1/((a + b*x)^(1/4)*(c + d*x)^(5/4)), x)

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