3.2.50 \(\int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx\) [150]

Optimal. Leaf size=356 \[ \frac {3 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {21 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {7\ 3^{3/4} a F\left (\text {ArcCos}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right ) (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{10 \sqrt [3]{2} d (1-\sec (c+d x)) (1+\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]

[Out]

3/5*a*(a+a*sec(d*x+c))^(2/3)*tan(d*x+c)/d+21/10*a*(a+a*sec(d*x+c))^(2/3)*tan(d*x+c)/d/(1+sec(d*x+c))-7/20*3^(3
/4)*a*((2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)/(2^(1/
3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))*(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))*EllipticF((1-(2^(1/3)-(1+sec(d
*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(a+a*
sec(d*x+c))^(2/3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))*((2^(2/3)+2^(1/3)*(1+sec(d*x+c))^(1/3)+(1+sec(d*x+c))^(2/3))/
(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)*tan(d*x+c)*2^(2/3)/d/(1-sec(d*x+c))/(1+sec(d*x+c))/(-(1+se
c(d*x+c))^(1/3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3913, 3912, 52, 65, 231} \begin {gather*} -\frac {7\ 3^{3/4} a \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} F\left (\text {ArcCos}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{10 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}+\frac {3 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}+\frac {21 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{10 d (\sec (c+d x)+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + a*Sec[c + d*x])^(5/3),x]

[Out]

(3*a*(a + a*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(5*d) + (21*a*(a + a*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(10*d*(1
+ Sec[c + d*x])) - (7*3^(3/4)*a*EllipticF[ArcCos[(2^(1/3) - (1 - Sqrt[3])*(1 + Sec[c + d*x])^(1/3))/(2^(1/3) -
(1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))], (2 + Sqrt[3])/4]*(a + a*Sec[c + d*x])^(2/3)*(2^(1/3) - (1 + Sec[c +
d*x])^(1/3))*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[c + d*x])^(1/3) + (1 + Sec[c + d*x])^(2/3))/(2^(1/3) - (1 + Sqr
t[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(10*2^(1/3)*d*(1 - Sec[c + d*x])*(1 + Sec[c + d*x])*Sqrt[-(((
1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3
))^2)])

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 231

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[x*(s +
r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*(
(s + r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^
2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x]

Rule 3912

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^2*d
*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])), Subst[Int[(d*x)^(n - 1)*((a + b*x)^(m -
1/2)/Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !In
tegerQ[m] && GtQ[a, 0]

Rule 3913

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^Int
Part[m]*((a + b*Csc[e + f*x])^FracPart[m]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]), Int[(1 + (b/a)*Csc[e + f*x])^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rubi steps

\begin {align*} \int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx &=\frac {\left (a (a+a \sec (c+d x))^{2/3}\right ) \int \sec (c+d x) (1+\sec (c+d x))^{5/3} \, dx}{(1+\sec (c+d x))^{2/3}}\\ &=-\frac {\left (a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(1+x)^{7/6}}{\sqrt {1-x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac {3 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}-\frac {\left (7 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt [6]{1+x}}{\sqrt {1-x}} \, dx,x,\sec (c+d x)\right )}{5 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac {3 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {21 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {\left (7 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{10 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac {3 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {21 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {\left (21 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{5 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac {3 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {21 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {7\ 3^{3/4} a F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right ) (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{10 \sqrt [3]{2} d (1-\sec (c+d x)) (1+\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.09, size = 66, normalized size = 0.19 \begin {gather*} \frac {4 \sqrt [6]{2} \, _2F_1\left (-\frac {7}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x))\right ) (a (1+\sec (c+d x)))^{5/3} \tan (c+d x)}{d (1+\sec (c+d x))^{13/6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + a*Sec[c + d*x])^(5/3),x]

[Out]

(4*2^(1/6)*Hypergeometric2F1[-7/6, 1/2, 3/2, (1 - Sec[c + d*x])/2]*(a*(1 + Sec[c + d*x]))^(5/3)*Tan[c + d*x])/
(d*(1 + Sec[c + d*x])^(13/6))

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \sec \left (d x +c \right ) \left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+a*sec(d*x+c))^(5/3),x)

[Out]

int(sec(d*x+c)*(a+a*sec(d*x+c))^(5/3),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(sec(d*x+c)*(a+a*sec(d*x+c))^(5/3),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^(5/3)*sec(d*x + c), x)

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Fricas [F]
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(sec(d*x+c)*(a+a*sec(d*x+c))^(5/3),x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c)^2 + a*sec(d*x + c))*(a*sec(d*x + c) + a)^(2/3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{3}} \sec {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))**(5/3),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**(5/3)*sec(c + d*x), x)

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Giac [F]
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(sec(d*x+c)*(a+a*sec(d*x+c))^(5/3),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^(5/3)*sec(d*x + c), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/3}}{\cos \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(5/3)/cos(c + d*x),x)

[Out]

int((a + a/cos(c + d*x))^(5/3)/cos(c + d*x), x)

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