3.889 \(\int \frac{\sqrt [4]{a+b x}}{x^3 \sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=194 \[ \frac{(b c-a d) (5 a d+3 b c) \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{7/4} c^{9/4}}+\frac{(b c-a d) (5 a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{7/4} c^{9/4}}+\frac{\sqrt [4]{a+b x} (c+d x)^{3/4} (5 a d+3 b c)}{8 a c^2 x}-\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2} \]

[Out]

((3*b*c + 5*a*d)*(a + b*x)^(1/4)*(c + d*x)^(3/4))/(8*a*c^2*x) - ((a + b*x)^(5/4)*(c + d*x)^(3/4))/(2*a*c*x^2)
+ ((b*c - a*d)*(3*b*c + 5*a*d)*ArcTan[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/(16*a^(7/4)*c^(9/4
)) + ((b*c - a*d)*(3*b*c + 5*a*d)*ArcTanh[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/(16*a^(7/4)*c^
(9/4))

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Rubi [A]  time = 0.0987722, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {96, 94, 93, 212, 208, 205} \[ \frac{(b c-a d) (5 a d+3 b c) \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{7/4} c^{9/4}}+\frac{(b c-a d) (5 a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{7/4} c^{9/4}}+\frac{\sqrt [4]{a+b x} (c+d x)^{3/4} (5 a d+3 b c)}{8 a c^2 x}-\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b*c + 5*a*d)*(a + b*x)^(1/4)*(c + d*x)^(3/4))/(8*a*c^2*x) - ((a + b*x)^(5/4)*(c + d*x)^(3/4))/(2*a*c*x^2)
+ ((b*c - a*d)*(3*b*c + 5*a*d)*ArcTan[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/(16*a^(7/4)*c^(9/4
)) + ((b*c - a*d)*(3*b*c + 5*a*d)*ArcTanh[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/(16*a^(7/4)*c^
(9/4))

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\sqrt [4]{a+b x}}{x^3 \sqrt [4]{c+d x}} \, dx &=-\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}-\frac{\left (\frac{3 b c}{4}+\frac{5 a d}{4}\right ) \int \frac{\sqrt [4]{a+b x}}{x^2 \sqrt [4]{c+d x}} \, dx}{2 a c}\\ &=\frac{(3 b c+5 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 a c^2 x}-\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}-\frac{((b c-a d) (3 b c+5 a d)) \int \frac{1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{32 a c^2}\\ &=\frac{(3 b c+5 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 a c^2 x}-\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}-\frac{((b c-a d) (3 b c+5 a d)) \operatorname{Subst}\left (\int \frac{1}{-a+c x^4} \, dx,x,\frac{\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{8 a c^2}\\ &=\frac{(3 b c+5 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 a c^2 x}-\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}+\frac{((b c-a d) (3 b c+5 a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a}-\sqrt{c} x^2} \, dx,x,\frac{\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{16 a^{3/2} c^2}+\frac{((b c-a d) (3 b c+5 a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a}+\sqrt{c} x^2} \, dx,x,\frac{\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{16 a^{3/2} c^2}\\ &=\frac{(3 b c+5 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 a c^2 x}-\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}+\frac{(b c-a d) (3 b c+5 a d) \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{7/4} c^{9/4}}+\frac{(b c-a d) (3 b c+5 a d) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{7/4} c^{9/4}}\\ \end{align*}

Mathematica [C]  time = 0.0489809, size = 106, normalized size = 0.55 \[ \frac{\sqrt [4]{a+b x} \left (x^2 \left (-5 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \, _2F_1\left (\frac{1}{4},1;\frac{5}{4};\frac{c (a+b x)}{a (c+d x)}\right )-a (c+d x) (4 a c-5 a d x+b c x)\right )}{8 a^2 c^2 x^2 \sqrt [4]{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(1/4)*(-(a*(c + d*x)*(4*a*c + b*c*x - 5*a*d*x)) + (3*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)*x^2*Hypergeom
etric2F1[1/4, 1, 5/4, (c*(a + b*x))/(a*(c + d*x))]))/(8*a^2*c^2*x^2*(c + d*x)^(1/4))

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}}\sqrt [4]{bx+a}{\frac{1}{\sqrt [4]{dx+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{1}{4}}}{{\left (d x + c\right )}^{\frac{1}{4}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.5699, size = 3394, normalized size = 17.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(4*a*c^2*x^2*((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4
*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^(1/4)*arctan(((
3*a^5*b^2*c^9 + 2*a^6*b*c^8*d - 5*a^7*c^7*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4)*((81*b^8*c^8 + 216*a*b^7*c^7*d
- 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6
- 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^(3/4) + (a^5*c^7*d*x + a^5*c^8)*sqrt(((9*b^4*c^4 + 12*a*b^3*c^3*
d - 26*a^2*b^2*c^2*d^2 - 20*a^3*b*c*d^3 + 25*a^4*d^4)*sqrt(b*x + a)*sqrt(d*x + c) + (a^4*c^4*d*x + a^4*c^5)*sq
rt((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*
b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9)))/(d*x + c))*((81*b^8*c^8 + 216*
a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6
*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^(3/4))/(81*b^8*c^9 + 216*a*b^7*c^8*d - 324*a^2*b^6*c
^7*d^2 - 984*a^3*b^5*c^6*d^3 + 646*a^4*b^4*c^5*d^4 + 1640*a^5*b^3*c^4*d^5 - 900*a^6*b^2*c^3*d^6 - 1000*a^7*b*c
^2*d^7 + 625*a^8*c*d^8 + (81*b^8*c^8*d + 216*a*b^7*c^7*d^2 - 324*a^2*b^6*c^6*d^3 - 984*a^3*b^5*c^5*d^4 + 646*a
^4*b^4*c^4*d^5 + 1640*a^5*b^3*c^3*d^6 - 900*a^6*b^2*c^2*d^7 - 1000*a^7*b*c*d^8 + 625*a^8*d^9)*x)) + a*c^2*x^2*
((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^
3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^(1/4)*log(-((3*b^2*c^2 + 2*a*b*c*
d - 5*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (a^2*c^2*d*x + a^2*c^3)*((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*
a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 100
0*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^(1/4))/(d*x + c)) - a*c^2*x^2*((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2
*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a
^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^(1/4)*log(-((3*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)
^(3/4) - (a^2*c^2*d*x + a^2*c^3)*((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 +
646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^
(1/4))/(d*x + c)) - 4*(4*a*c + (b*c - 5*a*d)*x)*(b*x + a)^(1/4)*(d*x + c)^(3/4))/(a*c^2*x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt [4]{a + b x}}{x^{3} \sqrt [4]{c + d x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out