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

Optimal. Leaf size=108 \[ \frac {2 \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}}+\frac {2 \sqrt [4]{b} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}}-\frac {4 \sqrt [4]{a+b x}}{d \sqrt [4]{c+d x}} \]

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Rubi [A]  time = 0.07, antiderivative size = 108, 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 = {47, 63, 240, 212, 208, 205} \begin {gather*} \frac {2 \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}}+\frac {2 \sqrt [4]{b} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}}-\frac {4 \sqrt [4]{a+b x}}{d \sqrt [4]{c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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 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 {\sqrt [4]{a+b x}}{(c+d x)^{5/4}} \, dx &=-\frac {4 \sqrt [4]{a+b x}}{d \sqrt [4]{c+d x}}+\frac {b \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{d}\\ &=-\frac {4 \sqrt [4]{a+b x}}{d \sqrt [4]{c+d x}}+\frac {4 \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 )}{d}\\ &=-\frac {4 \sqrt [4]{a+b x}}{d \sqrt [4]{c+d x}}+\frac {4 \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 )}{d}\\ &=-\frac {4 \sqrt [4]{a+b x}}{d \sqrt [4]{c+d x}}+\frac {\left (2 \sqrt {b}\right ) \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 )}{d}+\frac {\left (2 \sqrt {b}\right ) \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 )}{d}\\ &=-\frac {4 \sqrt [4]{a+b x}}{d \sqrt [4]{c+d x}}+\frac {2 \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}}+\frac {2 \sqrt [4]{b} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 73, normalized size = 0.68 \begin {gather*} \frac {4 (a+b x)^{5/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};\frac {d (a+b x)}{a d-b c}\right )}{5 b (c+d x)^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.13, size = 108, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}}+\frac {2 \sqrt [4]{b} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}}-\frac {4 \sqrt [4]{a+b x}}{d \sqrt [4]{c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.72, size = 273, normalized size = 2.53 \begin {gather*} -\frac {4 \, {\left (d^{2} x + c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} d^{4} \left (\frac {b}{d^{5}}\right )^{\frac {3}{4}} - {\left (d^{5} x + c d^{4}\right )} \sqrt {\frac {{\left (d^{3} x + c d^{2}\right )} \sqrt {\frac {b}{d^{5}}} + \sqrt {b x + a} \sqrt {d x + c}}{d x + c}} \left (\frac {b}{d^{5}}\right )^{\frac {3}{4}}}{b d x + b c}\right ) - {\left (d^{2} x + c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (d^{2} x + c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} + {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) + {\left (d^{2} x + c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (d^{2} x + c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} - {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) + 4 \, {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d^{2} x + c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*(d^2*x + c*d)*(b/d^5)^(1/4)*arctan(-((b*x + a)^(1/4)*(d*x + c)^(3/4)*d^4*(b/d^5)^(3/4) - (d^5*x + c*d^4)*s
qrt(((d^3*x + c*d^2)*sqrt(b/d^5) + sqrt(b*x + a)*sqrt(d*x + c))/(d*x + c))*(b/d^5)^(3/4))/(b*d*x + b*c)) - (d^
2*x + c*d)*(b/d^5)^(1/4)*log(((d^2*x + c*d)*(b/d^5)^(1/4) + (b*x + a)^(1/4)*(d*x + c)^(3/4))/(d*x + c)) + (d^2
*x + c*d)*(b/d^5)^(1/4)*log(-((d^2*x + c*d)*(b/d^5)^(1/4) - (b*x + a)^(1/4)*(d*x + c)^(3/4))/(d*x + c)) + 4*(b
*x + a)^(1/4)*(d*x + c)^(3/4))/(d^2*x + c*d)

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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}}}{{\left (d x + c\right )}^{\frac {5}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x +a \right )^{\frac {1}{4}}}{\left (d x +c \right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\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((a + b*x)^(1/4)/(c + d*x)^(5/4),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\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((b*x+a)**(1/4)/(d*x+c)**(5/4),x)

[Out]

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

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