3.17.80 \(\int \frac {(c+d x)^{5/4}}{(a+b x)^{5/4}} \, dx\) [1680]

Optimal. Leaf size=152 \[ \frac {5 d (a+b x)^{3/4} \sqrt [4]{c+d x}}{b^2}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}-\frac {5 \sqrt [4]{d} (b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 b^{9/4}}+\frac {5 \sqrt [4]{d} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 b^{9/4}} \]

[Out]

5*d*(b*x+a)^(3/4)*(d*x+c)^(1/4)/b^2-4*(d*x+c)^(5/4)/b/(b*x+a)^(1/4)-5/2*d^(1/4)*(-a*d+b*c)*arctan(d^(1/4)*(b*x
+a)^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(9/4)+5/2*d^(1/4)*(-a*d+b*c)*arctanh(d^(1/4)*(b*x+a)^(1/4)/b^(1/4)/(d*x+c)^
(1/4))/b^(9/4)

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Rubi [A]
time = 0.07, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {49, 52, 65, 338, 304, 211, 214} \begin {gather*} -\frac {5 \sqrt [4]{d} (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 b^{9/4}}+\frac {5 \sqrt [4]{d} (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 b^{9/4}}+\frac {5 d (a+b x)^{3/4} \sqrt [4]{c+d x}}{b^2}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*d*(a + b*x)^(3/4)*(c + d*x)^(1/4))/b^2 - (4*(c + d*x)^(5/4))/(b*(a + b*x)^(1/4)) - (5*d^(1/4)*(b*c - a*d)*A
rcTan[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(2*b^(9/4)) + (5*d^(1/4)*(b*c - a*d)*ArcTanh[(d^(1
/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(2*b^(9/4))

Rule 49

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 52

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 65

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 211

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

Rule 214

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

Rule 304

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 338

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 {(c+d x)^{5/4}}{(a+b x)^{5/4}} \, dx &=-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}+\frac {(5 d) \int \frac {\sqrt [4]{c+d x}}{\sqrt [4]{a+b x}} \, dx}{b}\\ &=\frac {5 d (a+b x)^{3/4} \sqrt [4]{c+d x}}{b^2}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}+\frac {(5 d (b c-a d)) \int \frac {1}{\sqrt [4]{a+b x} (c+d x)^{3/4}} \, dx}{4 b^2}\\ &=\frac {5 d (a+b x)^{3/4} \sqrt [4]{c+d x}}{b^2}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}+\frac {(5 d (b c-a d)) \text {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^3}\\ &=\frac {5 d (a+b x)^{3/4} \sqrt [4]{c+d x}}{b^2}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}+\frac {(5 d (b c-a d)) \text {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^3}\\ &=\frac {5 d (a+b x)^{3/4} \sqrt [4]{c+d x}}{b^2}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}+\frac {\left (5 \sqrt {d} (b c-a d)\right ) \text {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 b^2}-\frac {\left (5 \sqrt {d} (b c-a d)\right ) \text {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 b^2}\\ &=\frac {5 d (a+b x)^{3/4} \sqrt [4]{c+d x}}{b^2}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}-\frac {5 \sqrt [4]{d} (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 b^{9/4}}+\frac {5 \sqrt [4]{d} (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 b^{9/4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 10.04, size = 71, normalized size = 0.47 \begin {gather*} -\frac {4 (c+d x)^{5/4} \, _2F_1\left (-\frac {5}{4},-\frac {1}{4};\frac {3}{4};\frac {d (a+b x)}{-b c+a d}\right )}{b \sqrt [4]{a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathics [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Timed out

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 857 vs. \(2 (116) = 232\).
time = 0.35, size = 857, normalized size = 5.64 \begin {gather*} \frac {20 \, {\left (b^{3} x + a b^{2}\right )} \left (\frac {b^{4} c^{4} d - 4 \, a b^{3} c^{3} d^{2} + 6 \, a^{2} b^{2} c^{2} d^{3} - 4 \, a^{3} b c d^{4} + a^{4} d^{5}}{b^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {{\left (b^{8} c - a b^{7} d\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} \left (\frac {b^{4} c^{4} d - 4 \, a b^{3} c^{3} d^{2} + 6 \, a^{2} b^{2} c^{2} d^{3} - 4 \, a^{3} b c d^{4} + a^{4} d^{5}}{b^{9}}\right )^{\frac {3}{4}} + {\left (b^{8} x + a b^{7}\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^{5} x + a b^{4}\right )} \sqrt {\frac {b^{4} c^{4} d - 4 \, a b^{3} c^{3} d^{2} + 6 \, a^{2} b^{2} c^{2} d^{3} - 4 \, a^{3} b c d^{4} + a^{4} d^{5}}{b^{9}}}}{b x + a}} \left (\frac {b^{4} c^{4} d - 4 \, a b^{3} c^{3} d^{2} + 6 \, a^{2} b^{2} c^{2} d^{3} - 4 \, a^{3} b c d^{4} + a^{4} d^{5}}{b^{9}}\right )^{\frac {3}{4}}}{a b^{4} c^{4} d - 4 \, a^{2} b^{3} c^{3} d^{2} + 6 \, a^{3} b^{2} c^{2} d^{3} - 4 \, a^{4} b c d^{4} + a^{5} d^{5} + {\left (b^{5} c^{4} d - 4 \, a b^{4} c^{3} d^{2} + 6 \, a^{2} b^{3} c^{2} d^{3} - 4 \, a^{3} b^{2} c d^{4} + a^{4} b d^{5}\right )} x}\right ) + 5 \, {\left (b^{3} x + a b^{2}\right )} \left (\frac {b^{4} c^{4} d - 4 \, a b^{3} c^{3} d^{2} + 6 \, a^{2} b^{2} c^{2} d^{3} - 4 \, a^{3} b c d^{4} + a^{4} d^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {5 \, {\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^{3} x + a b^{2}\right )} \left (\frac {b^{4} c^{4} d - 4 \, a b^{3} c^{3} d^{2} + 6 \, a^{2} b^{2} c^{2} d^{3} - 4 \, a^{3} b c d^{4} + a^{4} d^{5}}{b^{9}}\right )^{\frac {1}{4}}\right )}}{b x + a}\right ) - 5 \, {\left (b^{3} x + a b^{2}\right )} \left (\frac {b^{4} c^{4} d - 4 \, a b^{3} c^{3} d^{2} + 6 \, a^{2} b^{2} c^{2} d^{3} - 4 \, a^{3} b c d^{4} + a^{4} d^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {5 \, {\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^{3} x + a b^{2}\right )} \left (\frac {b^{4} c^{4} d - 4 \, a b^{3} c^{3} d^{2} + 6 \, a^{2} b^{2} c^{2} d^{3} - 4 \, a^{3} b c d^{4} + a^{4} d^{5}}{b^{9}}\right )^{\frac {1}{4}}\right )}}{b x + a}\right ) + 4 \, {\left (b d x - 4 \, b c + 5 \, a d\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{4 \, {\left (b^{3} x + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(20*(b^3*x + a*b^2)*((b^4*c^4*d - 4*a*b^3*c^3*d^2 + 6*a^2*b^2*c^2*d^3 - 4*a^3*b*c*d^4 + a^4*d^5)/b^9)^(1/4
)*arctan(((b^8*c - a*b^7*d)*(b*x + a)^(3/4)*(d*x + c)^(1/4)*((b^4*c^4*d - 4*a*b^3*c^3*d^2 + 6*a^2*b^2*c^2*d^3
- 4*a^3*b*c*d^4 + a^4*d^5)/b^9)^(3/4) + (b^8*x + a*b^7)*sqrt(((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(b*x + a)*sq
rt(d*x + c) + (b^5*x + a*b^4)*sqrt((b^4*c^4*d - 4*a*b^3*c^3*d^2 + 6*a^2*b^2*c^2*d^3 - 4*a^3*b*c*d^4 + a^4*d^5)
/b^9))/(b*x + a))*((b^4*c^4*d - 4*a*b^3*c^3*d^2 + 6*a^2*b^2*c^2*d^3 - 4*a^3*b*c*d^4 + a^4*d^5)/b^9)^(3/4))/(a*
b^4*c^4*d - 4*a^2*b^3*c^3*d^2 + 6*a^3*b^2*c^2*d^3 - 4*a^4*b*c*d^4 + a^5*d^5 + (b^5*c^4*d - 4*a*b^4*c^3*d^2 + 6
*a^2*b^3*c^2*d^3 - 4*a^3*b^2*c*d^4 + a^4*b*d^5)*x)) + 5*(b^3*x + a*b^2)*((b^4*c^4*d - 4*a*b^3*c^3*d^2 + 6*a^2*
b^2*c^2*d^3 - 4*a^3*b*c*d^4 + a^4*d^5)/b^9)^(1/4)*log(-5*((b*c - a*d)*(b*x + a)^(3/4)*(d*x + c)^(1/4) + (b^3*x
+ a*b^2)*((b^4*c^4*d - 4*a*b^3*c^3*d^2 + 6*a^2*b^2*c^2*d^3 - 4*a^3*b*c*d^4 + a^4*d^5)/b^9)^(1/4))/(b*x + a))
- 5*(b^3*x + a*b^2)*((b^4*c^4*d - 4*a*b^3*c^3*d^2 + 6*a^2*b^2*c^2*d^3 - 4*a^3*b*c*d^4 + a^4*d^5)/b^9)^(1/4)*lo
g(-5*((b*c - a*d)*(b*x + a)^(3/4)*(d*x + c)^(1/4) - (b^3*x + a*b^2)*((b^4*c^4*d - 4*a*b^3*c^3*d^2 + 6*a^2*b^2*
c^2*d^3 - 4*a^3*b*c*d^4 + a^4*d^5)/b^9)^(1/4))/(b*x + a)) + 4*(b*d*x - 4*b*c + 5*a*d)*(b*x + a)^(3/4)*(d*x + c
)^(1/4))/(b^3*x + a*b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Could not integrate

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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