3.1781 \(\int \frac{(a+b x)^{5/6}}{(c+d x)^{11/6}} \, dx\)

Optimal. Leaf size=334 \[ -\frac{b^{5/6} \log \left (-\frac{\sqrt [6]{b} \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}+\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}+\sqrt [3]{b}\right )}{2 d^{11/6}}+\frac{b^{5/6} \log \left (\frac{\sqrt [6]{b} \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}+\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}+\sqrt [3]{b}\right )}{2 d^{11/6}}+\frac{\sqrt{3} b^{5/6} \tan ^{-1}\left (\frac{1}{\sqrt{3}}-\frac{2 \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt{3} \sqrt [6]{b} \sqrt [6]{c+d x}}\right )}{d^{11/6}}-\frac{\sqrt{3} b^{5/6} \tan ^{-1}\left (\frac{2 \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt{3} \sqrt [6]{b} \sqrt [6]{c+d x}}+\frac{1}{\sqrt{3}}\right )}{d^{11/6}}+\frac{2 b^{5/6} \tanh ^{-1}\left (\frac{\sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{b} \sqrt [6]{c+d x}}\right )}{d^{11/6}}-\frac{6 (a+b x)^{5/6}}{5 d (c+d x)^{5/6}} \]

[Out]

(-6*(a + b*x)^(5/6))/(5*d*(c + d*x)^(5/6)) + (Sqrt[3]*b^(5/6)*ArcTan[1/Sqrt[3] - (2*d^(1/6)*(a + b*x)^(1/6))/(
Sqrt[3]*b^(1/6)*(c + d*x)^(1/6))])/d^(11/6) - (Sqrt[3]*b^(5/6)*ArcTan[1/Sqrt[3] + (2*d^(1/6)*(a + b*x)^(1/6))/
(Sqrt[3]*b^(1/6)*(c + d*x)^(1/6))])/d^(11/6) + (2*b^(5/6)*ArcTanh[(d^(1/6)*(a + b*x)^(1/6))/(b^(1/6)*(c + d*x)
^(1/6))])/d^(11/6) - (b^(5/6)*Log[b^(1/3) + (d^(1/3)*(a + b*x)^(1/3))/(c + d*x)^(1/3) - (b^(1/6)*d^(1/6)*(a +
b*x)^(1/6))/(c + d*x)^(1/6)])/(2*d^(11/6)) + (b^(5/6)*Log[b^(1/3) + (d^(1/3)*(a + b*x)^(1/3))/(c + d*x)^(1/3)
+ (b^(1/6)*d^(1/6)*(a + b*x)^(1/6))/(c + d*x)^(1/6)])/(2*d^(11/6))

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Rubi [A]  time = 0.560891, antiderivative size = 334, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {47, 63, 331, 296, 634, 618, 204, 628, 208} \[ -\frac{b^{5/6} \log \left (-\frac{\sqrt [6]{b} \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}+\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}+\sqrt [3]{b}\right )}{2 d^{11/6}}+\frac{b^{5/6} \log \left (\frac{\sqrt [6]{b} \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}+\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}+\sqrt [3]{b}\right )}{2 d^{11/6}}+\frac{\sqrt{3} b^{5/6} \tan ^{-1}\left (\frac{1}{\sqrt{3}}-\frac{2 \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt{3} \sqrt [6]{b} \sqrt [6]{c+d x}}\right )}{d^{11/6}}-\frac{\sqrt{3} b^{5/6} \tan ^{-1}\left (\frac{2 \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt{3} \sqrt [6]{b} \sqrt [6]{c+d x}}+\frac{1}{\sqrt{3}}\right )}{d^{11/6}}+\frac{2 b^{5/6} \tanh ^{-1}\left (\frac{\sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{b} \sqrt [6]{c+d x}}\right )}{d^{11/6}}-\frac{6 (a+b x)^{5/6}}{5 d (c+d x)^{5/6}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/6)/(c + d*x)^(11/6),x]

[Out]

(-6*(a + b*x)^(5/6))/(5*d*(c + d*x)^(5/6)) + (Sqrt[3]*b^(5/6)*ArcTan[1/Sqrt[3] - (2*d^(1/6)*(a + b*x)^(1/6))/(
Sqrt[3]*b^(1/6)*(c + d*x)^(1/6))])/d^(11/6) - (Sqrt[3]*b^(5/6)*ArcTan[1/Sqrt[3] + (2*d^(1/6)*(a + b*x)^(1/6))/
(Sqrt[3]*b^(1/6)*(c + d*x)^(1/6))])/d^(11/6) + (2*b^(5/6)*ArcTanh[(d^(1/6)*(a + b*x)^(1/6))/(b^(1/6)*(c + d*x)
^(1/6))])/d^(11/6) - (b^(5/6)*Log[b^(1/3) + (d^(1/3)*(a + b*x)^(1/3))/(c + d*x)^(1/3) - (b^(1/6)*d^(1/6)*(a +
b*x)^(1/6))/(c + d*x)^(1/6)])/(2*d^(11/6)) + (b^(5/6)*Log[b^(1/3) + (d^(1/3)*(a + b*x)^(1/3))/(c + d*x)^(1/3)
+ (b^(1/6)*d^(1/6)*(a + b*x)^(1/6))/(c + d*x)^(1/6)])/(2*d^(11/6))

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 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 296

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt
[-(a/b), n]], k, u}, Simp[u = Int[(r*Cos[(2*k*m*Pi)/n] - s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi
)/n]*x + s^2*x^2), x] + Int[(r*Cos[(2*k*m*Pi)/n] + s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x
+ s^2*x^2), x]; (2*r^(m + 2)*Int[1/(r^2 - s^2*x^2), x])/(a*n*s^m) + Dist[(2*r^(m + 1))/(a*n*s^m), Sum[u, {k,
1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && NegQ[a/b]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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{(a+b x)^{5/6}}{(c+d x)^{11/6}} \, dx &=-\frac{6 (a+b x)^{5/6}}{5 d (c+d x)^{5/6}}+\frac{b \int \frac{1}{\sqrt [6]{a+b x} (c+d x)^{5/6}} \, dx}{d}\\ &=-\frac{6 (a+b x)^{5/6}}{5 d (c+d x)^{5/6}}+\frac{6 \operatorname{Subst}\left (\int \frac{x^4}{\left (c-\frac{a d}{b}+\frac{d x^6}{b}\right )^{5/6}} \, dx,x,\sqrt [6]{a+b x}\right )}{d}\\ &=-\frac{6 (a+b x)^{5/6}}{5 d (c+d x)^{5/6}}+\frac{6 \operatorname{Subst}\left (\int \frac{x^4}{1-\frac{d x^6}{b}} \, dx,x,\frac{\sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{d}\\ &=-\frac{6 (a+b x)^{5/6}}{5 d (c+d x)^{5/6}}+\frac{\left (2 b^{5/6}\right ) \operatorname{Subst}\left (\int \frac{-\frac{\sqrt [6]{b}}{2}-\frac{\sqrt [6]{d} x}{2}}{\sqrt [3]{b}-\sqrt [6]{b} \sqrt [6]{d} x+\sqrt [3]{d} x^2} \, dx,x,\frac{\sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{d^{5/3}}+\frac{\left (2 b^{5/6}\right ) \operatorname{Subst}\left (\int \frac{-\frac{\sqrt [6]{b}}{2}+\frac{\sqrt [6]{d} x}{2}}{\sqrt [3]{b}+\sqrt [6]{b} \sqrt [6]{d} x+\sqrt [3]{d} x^2} \, dx,x,\frac{\sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{d^{5/3}}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{b}-\sqrt [3]{d} x^2} \, dx,x,\frac{\sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{d^{5/3}}\\ &=-\frac{6 (a+b x)^{5/6}}{5 d (c+d x)^{5/6}}+\frac{2 b^{5/6} \tanh ^{-1}\left (\frac{\sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{b} \sqrt [6]{c+d x}}\right )}{d^{11/6}}-\frac{b^{5/6} \operatorname{Subst}\left (\int \frac{-\sqrt [6]{b} \sqrt [6]{d}+2 \sqrt [3]{d} x}{\sqrt [3]{b}-\sqrt [6]{b} \sqrt [6]{d} x+\sqrt [3]{d} x^2} \, dx,x,\frac{\sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{2 d^{11/6}}+\frac{b^{5/6} \operatorname{Subst}\left (\int \frac{\sqrt [6]{b} \sqrt [6]{d}+2 \sqrt [3]{d} x}{\sqrt [3]{b}+\sqrt [6]{b} \sqrt [6]{d} x+\sqrt [3]{d} x^2} \, dx,x,\frac{\sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{2 d^{11/6}}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{b}-\sqrt [6]{b} \sqrt [6]{d} x+\sqrt [3]{d} x^2} \, dx,x,\frac{\sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{2 d^{5/3}}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{b}+\sqrt [6]{b} \sqrt [6]{d} x+\sqrt [3]{d} x^2} \, dx,x,\frac{\sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{2 d^{5/3}}\\ &=-\frac{6 (a+b x)^{5/6}}{5 d (c+d x)^{5/6}}+\frac{2 b^{5/6} \tanh ^{-1}\left (\frac{\sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{b} \sqrt [6]{c+d x}}\right )}{d^{11/6}}-\frac{b^{5/6} \log \left (\sqrt [3]{b}+\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}-\frac{\sqrt [6]{b} \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{2 d^{11/6}}+\frac{b^{5/6} \log \left (\sqrt [3]{b}+\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}+\frac{\sqrt [6]{b} \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{2 d^{11/6}}-\frac{\left (3 b^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{b} \sqrt [6]{c+d x}}\right )}{d^{11/6}}+\frac{\left (3 b^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{b} \sqrt [6]{c+d x}}\right )}{d^{11/6}}\\ &=-\frac{6 (a+b x)^{5/6}}{5 d (c+d x)^{5/6}}+\frac{\sqrt{3} b^{5/6} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{b} \sqrt [6]{c+d x}}}{\sqrt{3}}\right )}{d^{11/6}}-\frac{\sqrt{3} b^{5/6} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{b} \sqrt [6]{c+d x}}}{\sqrt{3}}\right )}{d^{11/6}}+\frac{2 b^{5/6} \tanh ^{-1}\left (\frac{\sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{b} \sqrt [6]{c+d x}}\right )}{d^{11/6}}-\frac{b^{5/6} \log \left (\sqrt [3]{b}+\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}-\frac{\sqrt [6]{b} \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{2 d^{11/6}}+\frac{b^{5/6} \log \left (\sqrt [3]{b}+\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}+\frac{\sqrt [6]{b} \sqrt [6]{d} \sqrt [6]{a+b x}}{\sqrt [6]{c+d x}}\right )}{2 d^{11/6}}\\ \end{align*}

Mathematica [C]  time = 0.0505406, size = 73, normalized size = 0.22 \[ \frac{6 (a+b x)^{11/6} \left (\frac{b (c+d x)}{b c-a d}\right )^{11/6} \, _2F_1\left (\frac{11}{6},\frac{11}{6};\frac{17}{6};\frac{d (a+b x)}{a d-b c}\right )}{11 b (c+d x)^{11/6}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/6)/(c + d*x)^(11/6),x]

[Out]

(6*(a + b*x)^(11/6)*((b*(c + d*x))/(b*c - a*d))^(11/6)*Hypergeometric2F1[11/6, 11/6, 17/6, (d*(a + b*x))/(-(b*
c) + a*d)])/(11*b*(c + d*x)^(11/6))

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Maple [F]  time = 0.037, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{{\frac{5}{6}}} \left ( dx+c \right ) ^{-{\frac{11}{6}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/6)/(d*x+c)^(11/6),x)

[Out]

int((b*x+a)^(5/6)/(d*x+c)^(11/6),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{5}{6}}}{{\left (d x + c\right )}^{\frac{11}{6}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/6)/(d*x+c)^(11/6),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(5/6)/(d*x + c)^(11/6), x)

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Fricas [B]  time = 2.18966, size = 1887, normalized size = 5.65 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/6)/(d*x+c)^(11/6),x, algorithm="fricas")

[Out]

-1/10*(20*sqrt(3)*(d^2*x + c*d)*(b^5/d^11)^(1/6)*arctan(-1/3*(2*sqrt(3)*(b*x + a)^(5/6)*(d*x + c)^(1/6)*b*d^9*
(b^5/d^11)^(5/6) - 2*sqrt(3)*(b*d^9*x + a*d^9)*sqrt(((b*x + a)^(5/6)*(d*x + c)^(1/6)*b*d^2*(b^5/d^11)^(1/6) +
(b*x + a)^(2/3)*(d*x + c)^(1/3)*b^2 + (b*d^4*x + a*d^4)*(b^5/d^11)^(1/3))/(b*x + a))*(b^5/d^11)^(5/6) + sqrt(3
)*(b^6*x + a*b^5))/(b^6*x + a*b^5)) + 20*sqrt(3)*(d^2*x + c*d)*(b^5/d^11)^(1/6)*arctan(-1/3*(2*sqrt(3)*(b*x +
a)^(5/6)*(d*x + c)^(1/6)*b*d^9*(b^5/d^11)^(5/6) - 2*sqrt(3)*(b*d^9*x + a*d^9)*sqrt(-((b*x + a)^(5/6)*(d*x + c)
^(1/6)*b*d^2*(b^5/d^11)^(1/6) - (b*x + a)^(2/3)*(d*x + c)^(1/3)*b^2 - (b*d^4*x + a*d^4)*(b^5/d^11)^(1/3))/(b*x
+ a))*(b^5/d^11)^(5/6) - sqrt(3)*(b^6*x + a*b^5))/(b^6*x + a*b^5)) - 5*(d^2*x + c*d)*(b^5/d^11)^(1/6)*log(4*(
(b*x + a)^(5/6)*(d*x + c)^(1/6)*b*d^2*(b^5/d^11)^(1/6) + (b*x + a)^(2/3)*(d*x + c)^(1/3)*b^2 + (b*d^4*x + a*d^
4)*(b^5/d^11)^(1/3))/(b*x + a)) + 5*(d^2*x + c*d)*(b^5/d^11)^(1/6)*log(-4*((b*x + a)^(5/6)*(d*x + c)^(1/6)*b*d
^2*(b^5/d^11)^(1/6) - (b*x + a)^(2/3)*(d*x + c)^(1/3)*b^2 - (b*d^4*x + a*d^4)*(b^5/d^11)^(1/3))/(b*x + a)) - 1
0*(d^2*x + c*d)*(b^5/d^11)^(1/6)*log(((b*x + a)^(5/6)*(d*x + c)^(1/6)*b + (b*d^2*x + a*d^2)*(b^5/d^11)^(1/6))/
(b*x + a)) + 10*(d^2*x + c*d)*(b^5/d^11)^(1/6)*log(((b*x + a)^(5/6)*(d*x + c)^(1/6)*b - (b*d^2*x + a*d^2)*(b^5
/d^11)^(1/6))/(b*x + a)) + 12*(b*x + a)^(5/6)*(d*x + c)^(1/6))/(d^2*x + c*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/6)/(d*x+c)**(11/6),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{5}{6}}}{{\left (d x + c\right )}^{\frac{11}{6}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/6)/(d*x+c)^(11/6),x, algorithm="giac")

[Out]

integrate((b*x + a)^(5/6)/(d*x + c)^(11/6), x)