3.694 \(\int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/3} \, dx\)

Optimal. Leaf size=412 \[ \frac{\left (164 a^2 b^2+36 a^4+605 b^4\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{616 \sqrt{2} b^3 d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac{a \left (79 a^2 b^2+18 a^4-97 b^4\right ) \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{308 \sqrt{2} b^3 d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac{3 \left (18 a^2+121 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{1232 b^2 d}+\frac{3 a \left (18 a^2+97 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{1232 b^2 d}-\frac{9 a \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{77 b^2 d}+\frac{3 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{8/3}}{14 b d} \]

[Out]

(3*a*(18*a^2 + 97*b^2)*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(1232*b^2*d) + (3*(18*a^2 + 121*b^2)*(a + b*Se
c[c + d*x])^(5/3)*Tan[c + d*x])/(1232*b^2*d) - (9*a*(a + b*Sec[c + d*x])^(8/3)*Tan[c + d*x])/(77*b^2*d) + (3*S
ec[c + d*x]*(a + b*Sec[c + d*x])^(8/3)*Tan[c + d*x])/(14*b*d) + ((36*a^4 + 164*a^2*b^2 + 605*b^4)*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
])/(616*Sqrt[2]*b^3*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3)) - (a*(18*a^4 + 79*a^2*b^2 -
97*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])/(308*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 = 0.825241, antiderivative size = 412, normalized size of antiderivative = 1., number of steps used = 11, 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} \[ \frac{\left (164 a^2 b^2+36 a^4+605 b^4\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{616 \sqrt{2} b^3 d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac{a \left (79 a^2 b^2+18 a^4-97 b^4\right ) \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{308 \sqrt{2} b^3 d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac{3 \left (18 a^2+121 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{1232 b^2 d}+\frac{3 a \left (18 a^2+97 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{1232 b^2 d}-\frac{9 a \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{77 b^2 d}+\frac{3 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{8/3}}{14 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Sec[c + d*x])^(5/3),x]

[Out]

(3*a*(18*a^2 + 97*b^2)*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(1232*b^2*d) + (3*(18*a^2 + 121*b^2)*(a + b*Se
c[c + d*x])^(5/3)*Tan[c + d*x])/(1232*b^2*d) - (9*a*(a + b*Sec[c + d*x])^(8/3)*Tan[c + d*x])/(77*b^2*d) + (3*S
ec[c + d*x]*(a + b*Sec[c + d*x])^(8/3)*Tan[c + d*x])/(14*b*d) + ((36*a^4 + 164*a^2*b^2 + 605*b^4)*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
])/(616*Sqrt[2]*b^3*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3)) - (a*(18*a^4 + 79*a^2*b^2 -
97*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])/(308*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))^{5/3} \, dx &=\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac{3 \int \sec (c+d x) (a+b \sec (c+d x))^{5/3} \left (a+\frac{11}{3} b \sec (c+d x)-2 a \sec ^2(c+d x)\right ) \, dx}{14 b}\\ &=-\frac{9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac{9 \int \sec (c+d x) (a+b \sec (c+d x))^{5/3} \left (-\frac{5 a b}{3}+\frac{1}{9} \left (18 a^2+121 b^2\right ) \sec (c+d x)\right ) \, dx}{154 b^2}\\ &=\frac{3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac{9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac{27 \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \left (-\frac{5}{27} b \left (6 a^2-121 b^2\right )+\frac{5}{27} a \left (18 a^2+97 b^2\right ) \sec (c+d x)\right ) \, dx}{1232 b^2}\\ &=\frac{3 a \left (18 a^2+97 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{1232 b^2 d}+\frac{3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac{9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac{81 \int \frac{\sec (c+d x) \left (\frac{5}{81} a b \left (6 a^2+799 b^2\right )+\frac{5}{81} \left (36 a^4+164 a^2 b^2+605 b^4\right ) \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{6160 b^2}\\ &=\frac{3 a \left (18 a^2+97 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{1232 b^2 d}+\frac{3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac{9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}-\frac{\left (a \left (18 a^4+79 a^2 b^2-97 b^4\right )\right ) \int \frac{\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{616 b^3}+\frac{\left (36 a^4+164 a^2 b^2+605 b^4\right ) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx}{1232 b^3}\\ &=\frac{3 a \left (18 a^2+97 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{1232 b^2 d}+\frac{3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac{9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac{\left (a \left (18 a^4+79 a^2 b^2-97 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 )}{616 b^3 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}-\frac{\left (\left (36 a^4+164 a^2 b^2+605 b^4\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 )}{1232 b^3 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}\\ &=\frac{3 a \left (18 a^2+97 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{1232 b^2 d}+\frac{3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac{9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}-\frac{\left (\left (36 a^4+164 a^2 b^2+605 b^4\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 )}{1232 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 (a \left (18 a^4+79 a^2 b^2-97 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 )}{616 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 a \left (18 a^2+97 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{1232 b^2 d}+\frac{3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac{9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac{3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac{\left (36 a^4+164 a^2 b^2+605 b^4\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)}{616 \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{a \left (18 a^4+79 a^2 b^2-97 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)}{308 \sqrt{2} b^3 d \sqrt{1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 26.9911, size = 28057, normalized size = 68.1 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sec[c + d*x])^(5/3),x]

[Out]

Result too large to show

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Maple [F]  time = 0.122, 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{5}{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))^(5/3),x)

[Out]

int(sec(d*x+c)^4*(a+b*sec(d*x+c))^(5/3),x)

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Maxima [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))^(5/3),x, algorithm="maxima")

[Out]

Timed out

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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 )^{5} + a \sec \left (d x + c\right )^{4}\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}}, 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))^(5/3),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c)^5 + a*sec(d*x + c)^4)*(b*sec(d*x + c) + a)^(2/3), 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))**(5/3),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{5}{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))^(5/3),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^(5/3)*sec(d*x + c)^4, x)