\(\int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx\) [708]

Optimal result

Integrand size = 23, antiderivative size = 378 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=-\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}-\frac {a \left (9 a^2-7 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{2 \sqrt {2} b^3 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}+\frac {\left (9 a^4-10 a^2 b^2-b^4\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{2 \sqrt {2} b^3 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \]

[Out]

-3/2*a^2*sec(d*x+c)*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(2/3)+3/4*(3*a^2-b^2)*(a+b*sec(d*x+c))^(1/3)*tan
(d*x+c)/b^2/(a^2-b^2)/d-1/4*a*(9*a^2-7*b^2)*AppellF1(1/2,-1/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c
))*(a+b*sec(d*x+c))^(1/3)*tan(d*x+c)/b^3/(a^2-b^2)/d/((a+b*sec(d*x+c))/(a+b))^(1/3)*2^(1/2)/(1+sec(d*x+c))^(1/
2)+1/4*(9*a^4-10*a^2*b^2-b^4)*AppellF1(1/2,2/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*((a+b*sec(d*
x+c))/(a+b))^(2/3)*tan(d*x+c)/b^3/(a^2-b^2)/d/(a+b*sec(d*x+c))^(2/3)*2^(1/2)/(1+sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3930, 4167, 4092, 3919, 144, 143} \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=-\frac {a \left (9 a^2-7 b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{2 \sqrt {2} b^3 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}+\frac {3 \left (3 a^2-b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)}}{4 b^2 d \left (a^2-b^2\right )}+\frac {\left (9 a^4-10 a^2 b^2-b^4\right ) \tan (c+d x) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{2 \sqrt {2} b^3 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}} \]

[In]

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

[Out]

(-3*a^2*Sec[c + d*x]*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(2/3)) + (3*(3*a^2 - b^2)*(a + b*Se
c[c + d*x])^(1/3)*Tan[c + d*x])/(4*b^2*(a^2 - b^2)*d) - (a*(9*a^2 - 7*b^2)*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])^(1/3)*Tan[c + d*x])/(2*Sqrt[2]*b^3*(a^2
- b^2)*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(1/3)) + ((9*a^4 - 10*a^2*b^2 - 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])/(a + b))^(2/3)*
Tan[c + d*x])/(2*Sqrt[2]*b^3*(a^2 - b^2)*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(2/3))

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])

Rule 144

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 3919

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 3930

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

Rule 4092

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 4167

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac {3 \int \frac {\sec (c+d x) \left (a^2-\frac {2}{3} a b \sec (c+d x)-\frac {2}{3} \left (3 a^2-b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{2/3}} \, dx}{2 b \left (a^2-b^2\right )} \\ & = -\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}-\frac {9 \int \frac {\sec (c+d x) \left (\frac {2}{9} b \left (3 a^2+b^2\right )+\frac {2}{9} a \left (9 a^2-7 b^2\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^{2/3}} \, dx}{8 b^2 \left (a^2-b^2\right )} \\ & = -\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}-\frac {\left (a \left (9 a^2-7 b^2\right )\right ) \int \sec (c+d x) \sqrt [3]{a+b \sec (c+d x)} \, dx}{4 b^3 \left (a^2-b^2\right )}+\frac {\left (9 a^4-10 a^2 b^2-b^4\right ) \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx}{4 b^3 \left (a^2-b^2\right )} \\ & = -\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}+\frac {\left (a \left (9 a^2-7 b^2\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{4 b^3 \left (a^2-b^2\right ) d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {\left (\left (9 a^4-10 a^2 b^2-b^4\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} (a+b x)^{2/3}} \, dx,x,\sec (c+d x)\right )}{4 b^3 \left (a^2-b^2\right ) d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \\ & = -\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}+\frac {\left (a \left (9 a^2-7 b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{4 b^3 \left (a^2-b^2\right ) d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{-\frac {a+b \sec (c+d x)}{-a-b}}}-\frac {\left (\left (9 a^4-10 a^2 b^2-b^4\right ) \left (-\frac {a+b \sec (c+d x)}{-a-b}\right )^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}} \, dx,x,\sec (c+d x)\right )}{4 b^3 \left (a^2-b^2\right ) d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \\ & = -\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}-\frac {a \left (9 a^2-7 b^2\right ) \operatorname {AppellF1}\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]{a+b \sec (c+d x)} \tan (c+d x)}{2 \sqrt {2} b^3 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}+\frac {\left (9 a^4-10 a^2 b^2-b^4\right ) \operatorname {AppellF1}\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 ) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{2 \sqrt {2} b^3 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(21987\) vs. \(2(378)=756\).

Time = 44.23 (sec) , antiderivative size = 21987, normalized size of antiderivative = 58.17 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

Result too large to show

Maple [F]

\[\int \frac {\sec \left (d x +c \right )^{4}}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {5}{3}}}d x\]

[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)

Fricas [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]

[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) + a)^(1/3)*sec(d*x + c)^4/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)

Sympy [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{3}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/3}} \,d x \]

[In]

int(1/(cos(c + d*x)^4*(a + b/cos(c + d*x))^(5/3)),x)

[Out]

int(1/(cos(c + d*x)^4*(a + b/cos(c + d*x))^(5/3)), x)