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

Optimal. Leaf size=188 \[ \frac {(b c-a d) (3 a d+5 b c) \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{7/4} d^{9/4}}+\frac {(b c-a d) (3 a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{7/4} d^{9/4}}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (3 a d+5 b c)}{8 b d^2}+\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 b d} \]

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Rubi [A]  time = 0.11, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {80, 50, 63, 240, 212, 208, 205} \begin {gather*} \frac {(b c-a d) (3 a d+5 b c) \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{7/4} d^{9/4}}+\frac {(b c-a d) (3 a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{7/4} d^{9/4}}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (3 a d+5 b c)}{8 b d^2}+\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 80

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

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]

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 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 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rubi steps

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

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Mathematica [C]  time = 0.09, size = 96, normalized size = 0.51 \begin {gather*} \frac {(a+b x)^{5/4} \left (5 b (c+d x)-(3 a d+5 b c) \sqrt [4]{\frac {b (c+d x)}{b c-a d}} \, _2F_1\left (\frac {1}{4},\frac {5}{4};\frac {9}{4};\frac {d (a+b x)}{a d-b c}\right )\right )}{10 b^2 d \sqrt [4]{c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(5/4)*(5*b*(c + d*x) - (5*b*c + 3*a*d)*((b*(c + d*x))/(b*c - a*d))^(1/4)*Hypergeometric2F1[1/4, 5/4
, 9/4, (d*(a + b*x))/(-(b*c) + a*d)]))/(10*b^2*d*(c + d*x)^(1/4))

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IntegrateAlgebraic [A]  time = 8.70, size = 241, normalized size = 1.28 \begin {gather*} \frac {\sqrt [4]{d} \sqrt [4]{a+b x} \left (\frac {\left (3 a^2 d^2+2 a b c d-5 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{a d+b (c+d x)-b c}}\right )}{16 b^{7/4} d^{9/4}}+\frac {\left (-3 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{a d+b (c+d x)-b c}}\right )}{16 b^{7/4} d^{9/4}}+\frac {\sqrt [4]{a d+b (c+d x)-b c} \left (a d (c+d x)^{3/4}+4 b (c+d x)^{7/4}-9 b c (c+d x)^{3/4}\right )}{8 b d^{9/4}}\right )}{\sqrt [4]{a d+b d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.72, size = 1506, normalized size = 8.01

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x +a \right )^{\frac {1}{4}} x}{\left (d x +c \right )^{\frac {1}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,{\left (a+b\,x\right )}^{1/4}}{{\left (c+d\,x\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \sqrt [4]{a + b x}}{\sqrt [4]{c + d x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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