3.1680 \(\int \frac{(c+d x)^{5/4}}{(a+b x)^{5/4}} \, dx\)

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

[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))

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Rubi [A]  time = 0.0975301, antiderivative size = 152, normalized size of antiderivative = 1., 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 = {47, 50, 63, 331, 298, 205, 208} \[ \frac{5 d (a+b x)^{3/4} \sqrt [4]{c+d x}}{b^2}-\frac{5 \sqrt [4]{d} (b c-a d) \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 b^{9/4}}+\frac{5 \sqrt [4]{d} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 b^{9/4}}-\frac{4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}} \]

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 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 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 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]

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 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]

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)) \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^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)) \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^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 ) \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 b^2}-\frac{\left (5 \sqrt{d} (b c-a d)\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 )}{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*}

Mathematica [C]  time = 0.0438857, size = 71, normalized size = 0.47 \[ -\frac{4 (c+d x)^{5/4} \, _2F_1\left (-\frac{5}{4},-\frac{1}{4};\frac{3}{4};\frac{d (a+b x)}{a d-b c}\right )}{b \sqrt [4]{a+b x} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/4}} \]

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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Maple [F]  time = 0.042, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx+c \right ) ^{{\frac{5}{4}}} \left ( bx+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{5}{4}}}{{\left (b x + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

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]  time = 3.10835, size = 1817, normalized size = 11.95 \begin{align*} \text{result too large to display} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x\right )^{\frac{5}{4}}}{\left (a + b x\right )^{\frac{5}{4}}}\, dx \end{align*}

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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{5}{4}}}{{\left (b x + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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