3.687 \(\int \sec ^4(c+d x) (a+b \sec (c+d x))^{2/3} \, dx\)

Optimal. Leaf size=362 \[ \frac{a \left (18 a^2+49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{110 \sqrt{2} b^3 d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac{\left (23 a^2 b^2+9 a^4-32 b^4\right ) \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{55 \sqrt{2} b^3 d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac{3 \left (9 a^2+32 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{220 b^2 d}-\frac{9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{44 b^2 d}+\frac{3 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/3}}{11 b d} \]

[Out]

(3*(9*a^2 + 32*b^2)*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(220*b^2*d) - (9*a*(a + b*Sec[c + d*x])^(5/3)*Tan
[c + d*x])/(44*b^2*d) + (3*Sec[c + d*x]*(a + b*Sec[c + d*x])^(5/3)*Tan[c + d*x])/(11*b*d) + (a*(18*a^2 + 49*b^
2)*AppellF1[1/2, 1/2, -2/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*(a + b*Sec[c + d*x])^(2
/3)*Tan[c + d*x])/(110*Sqrt[2]*b^3*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3)) - ((9*a^4 +
23*a^2*b^2 - 32*b^4)*AppellF1[1/2, 1/2, 1/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*((a +
b*Sec[c + d*x])/(a + b))^(1/3)*Tan[c + d*x])/(55*Sqrt[2]*b^3*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(1/
3))

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Rubi [A]  time = 0.66823, antiderivative size = 362, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3865, 4082, 4002, 4007, 3834, 139, 138} \[ \frac{a \left (18 a^2+49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{110 \sqrt{2} b^3 d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac{\left (23 a^2 b^2+9 a^4-32 b^4\right ) \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{55 \sqrt{2} b^3 d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac{3 \left (9 a^2+32 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{220 b^2 d}-\frac{9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{44 b^2 d}+\frac{3 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/3}}{11 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Sec[c + d*x])^(2/3),x]

[Out]

(3*(9*a^2 + 32*b^2)*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(220*b^2*d) - (9*a*(a + b*Sec[c + d*x])^(5/3)*Tan
[c + d*x])/(44*b^2*d) + (3*Sec[c + d*x]*(a + b*Sec[c + d*x])^(5/3)*Tan[c + d*x])/(11*b*d) + (a*(18*a^2 + 49*b^
2)*AppellF1[1/2, 1/2, -2/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*(a + b*Sec[c + d*x])^(2
/3)*Tan[c + d*x])/(110*Sqrt[2]*b^3*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3)) - ((9*a^4 +
23*a^2*b^2 - 32*b^4)*AppellF1[1/2, 1/2, 1/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*((a +
b*Sec[c + d*x])/(a + b))^(1/3)*Tan[c + d*x])/(55*Sqrt[2]*b^3*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(1/
3))

Rule 3865

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(d^3*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3))/(b*f*(m + n - 1)), x] + Dist[d^3/(b*(m + n
- 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 3)*Simp[a*(n - 3) + b*(m + n - 2)*Csc[e + f*x] - a*(n
- 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 3] && (Integer
Q[n] || IntegersQ[2*m, 2*n]) &&  !IGtQ[m, 2]

Rule 4082

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

Rule 4002

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[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4007

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

Rule 3834

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

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+b \sec (c+d x))^{2/3} \, dx &=\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}+\frac{3 \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \left (a+\frac{8}{3} b \sec (c+d x)-2 a \sec ^2(c+d x)\right ) \, dx}{11 b}\\ &=-\frac{9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}+\frac{9 \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \left (-\frac{2 a b}{3}+\frac{2}{9} \left (9 a^2+32 b^2\right ) \sec (c+d x)\right ) \, dx}{88 b^2}\\ &=\frac{3 \left (9 a^2+32 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 b^2 d}-\frac{9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}+\frac{27 \int \frac{\sec (c+d x) \left (\frac{2}{27} b \left (3 a^2+64 b^2\right )+\frac{2}{27} a \left (18 a^2+49 b^2\right ) \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{440 b^2}\\ &=\frac{3 \left (9 a^2+32 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 b^2 d}-\frac{9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}+\frac{\left (a \left (18 a^2+49 b^2\right )\right ) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx}{220 b^3}-\frac{\left (9 a^4+23 a^2 b^2-32 b^4\right ) \int \frac{\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{110 b^3}\\ &=\frac{3 \left (9 a^2+32 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 b^2 d}-\frac{9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}-\frac{\left (a \left (18 a^2+49 b^2\right ) \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{220 b^3 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}+\frac{\left (\left (9 a^4+23 a^2 b^2-32 b^4\right ) \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{110 b^3 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}\\ &=\frac{3 \left (9 a^2+32 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 b^2 d}-\frac{9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}-\frac{\left (a \left (18 a^2+49 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{2/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{220 b^3 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \left (-\frac{a+b \sec (c+d x)}{-a-b}\right )^{2/3}}+\frac{\left (\left (9 a^4+23 a^2 b^2-32 b^4\right ) \sqrt [3]{-\frac{a+b \sec (c+d x)}{-a-b}} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} \sqrt [3]{-\frac{a}{-a-b}-\frac{b x}{-a-b}}} \, dx,x,\sec (c+d x)\right )}{110 b^3 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=\frac{3 \left (9 a^2+32 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 b^2 d}-\frac{9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{44 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{11 b d}+\frac{a \left (18 a^2+49 b^2\right ) F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{110 \sqrt{2} b^3 d \sqrt{1+\sec (c+d x)} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac{\left (9 a^4+23 a^2 b^2-32 b^4\right ) F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{55 \sqrt{2} b^3 d \sqrt{1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 26.6812, size = 21877, normalized size = 60.43 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sec[c + d*x])^(2/3),x]

[Out]

Result too large to show

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Maple [F]  time = 0.12, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{4} \left ( a+b\sec \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+b*sec(d*x+c))^(2/3),x)

[Out]

int(sec(d*x+c)^4*(a+b*sec(d*x+c))^(2/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sec \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sec \left (d x + c\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^4, x)

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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(sec(d*x+c)**4*(a+b*sec(d*x+c))**(2/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sec \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^(2/3),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^4, x)