3.8.9 \(\int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx\) [709]

3.8.9.1 Optimal result

Integrand size = 23, antiderivative size = 307 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=-\frac {3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {\left (3 a^2-2 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{\sqrt {2} b^2 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}-\frac {a \left (3 a^2-4 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{\sqrt {2} b^2 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \]

output

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

3.8.9.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(19126\) vs. \(2(307)=614\).

Time = 44.10 (sec) , antiderivative size = 19126, normalized size of antiderivative = 62.30 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\text {Result too large to show} \]

input

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

output

Result too large to show
 

3.8.9.3 Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 304, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4326, 27, 3042, 4495, 3042, 4321, 156, 155}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}}dx\)

\(\Big \downarrow \) 4326

\(\displaystyle -\frac {3 \int -\frac {\sec (c+d x) \left (2 a b+\left (3 a^2-2 b^2\right ) \sec (c+d x)\right )}{3 (a+b \sec (c+d x))^{2/3}}dx}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec (c+d x) \left (2 a b+\left (3 a^2-2 b^2\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^{2/3}}dx}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 a b+\left (3 a^2-2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\)

\(\Big \downarrow \) 4495

\(\displaystyle \frac {\frac {\left (3 a^2-2 b^2\right ) \int \sec (c+d x) \sqrt [3]{a+b \sec (c+d x)}dx}{b}-\frac {a \left (3 a^2-4 b^2\right ) \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{2/3}}dx}{b}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (3 a^2-2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {a \left (3 a^2-4 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx}{b}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\)

\(\Big \downarrow \) 4321

\(\displaystyle \frac {\frac {a \left (3 a^2-4 b^2\right ) \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}-\frac {\left (3 a^2-2 b^2\right ) \tan (c+d x) \int \frac {\sqrt [3]{a+b \sec (c+d x)}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {\frac {a \left (3 a^2-4 b^2\right ) \tan (c+d x) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \left (\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}\right )^{2/3}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}-\frac {\left (3 a^2-2 b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} \int \frac {\sqrt [3]{\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {\frac {\sqrt {2} \left (3 a^2-2 b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}-\frac {\sqrt {2} a \left (3 a^2-4 b^2\right ) \tan (c+d x) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}}{2 b \left (a^2-b^2\right )}-\frac {3 a^2 \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}\)

input

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

output

(-3*a^2*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(2/3)) + ((S 
qrt[2]*(3*a^2 - 2*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])/ 
(b*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(1/3)) - (Sqrt[ 
2]*a*(3*a^2 - 4*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])/(a + b))^(2/3)*Tan[c + 
 d*x])/(b*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(2/3)))/(2*b*(a^2 
- b^2))
 

3.8.9.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[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] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4321

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[Cot[e + f*x]/(f*Sqrt[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 4326

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

rule 4495

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)/b   Int[ 
Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] + Simp[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] && N 
eQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
 

3.8.9.4 Maple [F]

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

input

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

output

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

3.8.9.5 Fricas [F]

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

input

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

output

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

3.8.9.6 Sympy [F]

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

input

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

output

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

3.8.9.7 Maxima [F]

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

input

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

output

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

3.8.9.8 Giac [F]

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

input

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

output

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

3.8.9.9 Mupad [F(-1)]

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

input

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

output

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