3.143 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{2/3} \, dx\)

Optimal. Leaf size=383 \[ \frac {3 \tan (c+d x) (a \sec (c+d x)+a)^{5/3}}{8 a d}-\frac {9 \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{40 d}+\frac {57 \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{80 d (\sec (c+d x)+1)}-\frac {19\ 3^{3/4} \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{80 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}} \]

[Out]

-9/40*(a+a*sec(d*x+c))^(2/3)*tan(d*x+c)/d+57/80*(a+a*sec(d*x+c))^(2/3)*tan(d*x+c)/d/(1+sec(d*x+c))+3/8*(a+a*se
c(d*x+c))^(5/3)*tan(d*x+c)/a/d-19/160*3^(3/4)*((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+sec(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 = 0.68, antiderivative size = 383, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3800, 4001, 3828, 3827, 50, 63, 225} \[ \frac {3 \tan (c+d x) (a \sec (c+d x)+a)^{5/3}}{8 a d}-\frac {9 \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{40 d}+\frac {57 \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{80 d (\sec (c+d x)+1)}-\frac {19\ 3^{3/4} \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{80 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(2/3),x]

[Out]

(-9*(a + a*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(40*d) + (57*(a + a*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(80*d*(1 +
Sec[c + d*x])) + (3*(a + a*Sec[c + d*x])^(5/3)*Tan[c + d*x])/(8*a*d) - (19*3^(3/4)*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 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(80*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 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 225

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]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(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]), x]] /; FreeQ[{a, b}, x]

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 3827

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 3828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
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]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^{2/3} \, dx &=\frac {3 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{8 a d}+\frac {3 \int \sec (c+d x) \left (\frac {5 a}{3}-a \sec (c+d x)\right ) (a+a \sec (c+d x))^{2/3} \, dx}{8 a}\\ &=-\frac {9 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 d}+\frac {3 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{8 a d}+\frac {19}{40} \int \sec (c+d x) (a+a \sec (c+d x))^{2/3} \, dx\\ &=-\frac {9 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 d}+\frac {3 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{8 a d}+\frac {\left (19 (a+a \sec (c+d x))^{2/3}\right ) \int \sec (c+d x) (1+\sec (c+d x))^{2/3} \, dx}{40 (1+\sec (c+d x))^{2/3}}\\ &=-\frac {9 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 d}+\frac {3 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{8 a d}-\frac {\left (19 (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt [6]{1+x}}{\sqrt {1-x}} \, dx,x,\sec (c+d x)\right )}{40 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=-\frac {9 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 d}+\frac {57 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{80 d (1+\sec (c+d x))}+\frac {3 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{8 a d}-\frac {\left (19 (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{80 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=-\frac {9 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 d}+\frac {57 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{80 d (1+\sec (c+d x))}+\frac {3 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{8 a d}-\frac {\left (57 (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{40 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=-\frac {9 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 d}+\frac {57 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{80 d (1+\sec (c+d x))}+\frac {3 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{8 a d}-\frac {19\ 3^{3/4} 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)}{80 \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]  time = 0.34, size = 105, normalized size = 0.27 \[ \frac {\tan (c+d x) (a (\sec (c+d x)+1))^{2/3} \left (38 \sqrt [6]{2} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x))\right )+3 \sqrt [6]{\sec (c+d x)+1} \left (5 \sec ^2(c+d x)+7 \sec (c+d x)+2\right )\right )}{40 d (\sec (c+d x)+1)^{7/6}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(2/3),x]

[Out]

((a*(1 + Sec[c + d*x]))^(2/3)*(38*2^(1/6)*Hypergeometric2F1[-1/6, 1/2, 3/2, (1 - Sec[c + d*x])/2] + 3*(1 + Sec
[c + d*x])^(1/6)*(2 + 7*Sec[c + d*x] + 5*Sec[c + d*x]^2))*Tan[c + d*x])/(40*d*(1 + Sec[c + d*x])^(7/6))

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fricas [F]  time = 1.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(2/3),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.72, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{3}\left (d x +c \right )\right ) \left (a +a \sec \left (d x +c \right )\right )^{\frac {2}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{2/3}}{{\cos \left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(2/3)/cos(c + d*x)^3,x)

[Out]

int((a + a/cos(c + d*x))^(2/3)/cos(c + d*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}} \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**(2/3)*sec(c + d*x)**3, x)

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