3.15.98 \(\int \frac {(a+b x)^{3/4}}{(c+d x)^{3/4}} \, dx\)

Optimal. Leaf size=127 \[ \frac {3 (b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 \sqrt [4]{b} d^{7/4}}-\frac {3 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 \sqrt [4]{b} d^{7/4}}+\frac {(a+b x)^{3/4} \sqrt [4]{c+d x}}{d} \]

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Rubi [A]  time = 0.08, antiderivative size = 127, 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 = {50, 63, 331, 298, 205, 208} \begin {gather*} \frac {3 (b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 \sqrt [4]{b} d^{7/4}}-\frac {3 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 \sqrt [4]{b} d^{7/4}}+\frac {(a+b x)^{3/4} \sqrt [4]{c+d x}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 298

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

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 73, normalized size = 0.57 \begin {gather*} \frac {4 (a+b x)^{7/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \, _2F_1\left (\frac {3}{4},\frac {7}{4};\frac {11}{4};\frac {d (a+b x)}{a d-b c}\right )}{7 b (c+d x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*x)^(7/4)*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[3/4, 7/4, 11/4, (d*(a + b*x))/(-(b*c) +
a*d)])/(7*b*(c + d*x)^(3/4))

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IntegrateAlgebraic [A]  time = 8.42, size = 176, normalized size = 1.39 \begin {gather*} \frac {d^{3/4} (a+b x)^{3/4} \left (\frac {\sqrt [4]{c+d x} (a d+b (c+d x)-b c)^{3/4}}{d^{7/4}}-\frac {3 (b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{a d+b (c+d x)-b c}}\right )}{2 \sqrt [4]{b} d^{7/4}}-\frac {3 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{a d+b (c+d x)-b c}}\right )}{2 \sqrt [4]{b} d^{7/4}}\right )}{(a d+b d x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.42, size = 808, normalized size = 6.36 \begin {gather*} -\frac {12 \, d \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{b d^{7}}\right )^{\frac {1}{4}} \arctan \left (\frac {{\left (b^{2} c d^{5} - a b d^{6}\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{b d^{7}}\right )^{\frac {3}{4}} + {\left (b^{2} d^{5} x + a b d^{5}\right )} \sqrt {\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} + {\left (b d^{4} x + a d^{4}\right )} \sqrt {\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{b d^{7}}}}{b x + a}} \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{b d^{7}}\right )^{\frac {3}{4}}}{a b^{4} c^{4} - 4 \, a^{2} b^{3} c^{3} d + 6 \, a^{3} b^{2} c^{2} d^{2} - 4 \, a^{4} b c d^{3} + a^{5} d^{4} + {\left (b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}\right )} x}\right ) + 3 \, d \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{b d^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left ({\left (b c - a d\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} + {\left (b d^{2} x + a d^{2}\right )} \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{b d^{7}}\right )^{\frac {1}{4}}\right )}}{b x + a}\right ) - 3 \, d \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{b d^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left ({\left (b c - a d\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} - {\left (b d^{2} x + a d^{2}\right )} \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{b d^{7}}\right )^{\frac {1}{4}}\right )}}{b x + a}\right ) - 4 \, {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(12*d*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(b*d^7))^(1/4)*arctan(((b^
2*c*d^5 - a*b*d^6)*(b*x + a)^(3/4)*(d*x + c)^(1/4)*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d
^3 + a^4*d^4)/(b*d^7))^(3/4) + (b^2*d^5*x + a*b*d^5)*sqrt(((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(b*x + a)*sqrt(
d*x + c) + (b*d^4*x + a*d^4)*sqrt((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(b*d
^7)))/(b*x + a))*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(b*d^7))^(3/4))/(a*b
^4*c^4 - 4*a^2*b^3*c^3*d + 6*a^3*b^2*c^2*d^2 - 4*a^4*b*c*d^3 + a^5*d^4 + (b^5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^3*
c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4)*x)) + 3*d*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3
+ a^4*d^4)/(b*d^7))^(1/4)*log(-3*((b*c - a*d)*(b*x + a)^(3/4)*(d*x + c)^(1/4) + (b*d^2*x + a*d^2)*((b^4*c^4 -
4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(b*d^7))^(1/4))/(b*x + a)) - 3*d*((b^4*c^4 - 4*a
*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(b*d^7))^(1/4)*log(-3*((b*c - a*d)*(b*x + a)^(3/4)*(
d*x + c)^(1/4) - (b*d^2*x + a*d^2)*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(b
*d^7))^(1/4))/(b*x + a)) - 4*(b*x + a)^(3/4)*(d*x + c)^(1/4))/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 {3}{4}}}{{\left (d x + c\right )}^{\frac {3}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)^(3/4)/(d*x + c)^(3/4), 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 {3}{4}}}{\left (d x +c \right )^{\frac {3}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((b*x+a)^(3/4)/(d*x+c)^(3/4),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 {3}{4}}}{{\left (d x + c\right )}^{\frac {3}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)^(3/4)/(d*x + c)^(3/4), 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 )}^{3/4}}{{\left (c+d\,x\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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