3.903 \(\int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=85 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{3/4} \sqrt [4]{d}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{3/4} \sqrt [4]{d}} \]

[Out]

2*arctan(d^(1/4)*(b*x+a)^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(3/4)/d^(1/4)+2*arctanh(d^(1/4)*(b*x+a)^(1/4)/b^(1/4)/
(d*x+c)^(1/4))/b^(3/4)/d^(1/4)

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Rubi [A]  time = 0.05, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {63, 240, 212, 208, 205} \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{3/4} \sqrt [4]{d}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{3/4} \sqrt [4]{d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTan[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(b^(3/4)*d^(1/4)) + (2*ArcTanh[(d^(1/4)*(a + b
*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(b^(3/4)*d^(1/4))

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 {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx &=\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 )}{b}\\ &=\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 )}{b}\\ &=\frac {2 \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 )}{\sqrt {b}}+\frac {2 \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 )}{\sqrt {b}}\\ &=\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{3/4} \sqrt [4]{d}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{3/4} \sqrt [4]{d}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 71, normalized size = 0.84 \[ \frac {4 \sqrt [4]{a+b x} \sqrt [4]{\frac {b (c+d x)}{b c-a d}} \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\frac {d (a+b x)}{a d-b c}\right )}{b \sqrt [4]{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.82, size = 234, normalized size = 2.75 \[ -4 \, \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} b^{2} d \left (\frac {1}{b^{3} d}\right )^{\frac {3}{4}} - {\left (b^{2} d^{2} x + b^{2} c d\right )} \sqrt {\frac {{\left (b^{2} d x + b^{2} c\right )} \sqrt {\frac {1}{b^{3} d}} + \sqrt {b x + a} \sqrt {d x + c}}{d x + c}} \left (\frac {1}{b^{3} d}\right )^{\frac {3}{4}}}{d x + c}\right ) + \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b d x + b c\right )} \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} + {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) - \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (b d x + b c\right )} \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} - {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,x\right )}^{3/4}\,{\left (c+d\,x\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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