3.1.81 \(\int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2} \, dx\) [81]

3.1.81.1 Optimal result

Integrand size = 34, antiderivative size = 85 \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2} \, dx=-\frac {8 c^2 (a+a \sec (e+f x))^3 \tan (e+f x)}{63 f \sqrt {c-c \sec (e+f x)}}-\frac {2 c (a+a \sec (e+f x))^3 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{9 f} \]

output

-8/63*c^2*(a+a*sec(f*x+e))^3*tan(f*x+e)/f/(c-c*sec(f*x+e))^(1/2)-2/9*c*(a+ 
a*sec(f*x+e))^3*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/f
 

3.1.81.2 Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.78 \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2} \, dx=\frac {16 a^3 c \cos ^6\left (\frac {1}{2} (e+f x)\right ) (-7+11 \cos (e+f x)) \cot \left (\frac {1}{2} (e+f x)\right ) \sec ^4(e+f x) \sqrt {c-c \sec (e+f x)}}{63 f} \]

input

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

output

(16*a^3*c*Cos[(e + f*x)/2]^6*(-7 + 11*Cos[e + f*x])*Cot[(e + f*x)/2]*Sec[e 
 + f*x]^4*Sqrt[c - c*Sec[e + f*x]])/(63*f)
 

3.1.81.3 Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3042, 4443, 3042, 4441}

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.

\(\displaystyle \int \sec (e+f x) (a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}dx\)

\(\Big \downarrow \) 4443

\(\displaystyle \frac {4}{9} c \int \sec (e+f x) (\sec (e+f x) a+a)^3 \sqrt {c-c \sec (e+f x)}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a)^3 \sqrt {c-c \sec (e+f x)}}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{9} c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^3 \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a)^3 \sqrt {c-c \sec (e+f x)}}{9 f}\)

\(\Big \downarrow \) 4441

\(\displaystyle -\frac {8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^3}{63 f \sqrt {c-c \sec (e+f x)}}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a)^3 \sqrt {c-c \sec (e+f x)}}{9 f}\)

input

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

output

(-8*c^2*(a + a*Sec[e + f*x])^3*Tan[e + f*x])/(63*f*Sqrt[c - c*Sec[e + f*x] 
]) - (2*c*(a + a*Sec[e + f*x])^3*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(9 
*f)
 

3.1.81.3.1 Defintions of rubi rules used

rule 3042

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

rule 4441

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

rule 4443

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

3.1.81.4 Maple [A] (verified)

Time = 6.83 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.67

method	result	size
default	\(\frac {2 a^{3} c \left (11 \cos \left (f x +e \right )-7\right ) \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )+1\right )^{4} \sec \left (f x +e \right )^{4} \csc \left (f x +e \right )}{63 f}\)	\(57\)
parts	\(-\frac {2 a^{3} \left (\sec \left (f x +e \right )-1\right ) \left (5 \cos \left (f x +e \right )-1\right ) c \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )+1\right ) \csc \left (f x +e \right )}{3 f \left (\cos \left (f x +e \right )-1\right )}+\frac {2 a^{3} \left (\sec \left (f x +e \right )-1\right ) \left (272 \cos \left (f x +e \right )^{4}-136 \cos \left (f x +e \right )^{3}+102 \cos \left (f x +e \right )^{2}-85 \cos \left (f x +e \right )+35\right ) c \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )+1\right ) \sec \left (f x +e \right )^{3} \csc \left (f x +e \right )}{315 f \left (\cos \left (f x +e \right )-1\right )}+\frac {6 a^{3} \left (6 \cos \left (f x +e \right )^{2}-3 \cos \left (f x +e \right )+1\right ) \left (\sec \left (f x +e \right )-1\right ) c \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )+1\right ) \sec \left (f x +e \right ) \csc \left (f x +e \right )}{5 f \left (\cos \left (f x +e \right )-1\right )}-\frac {2 a^{3} \left (104 \cos \left (f x +e \right )^{3}-52 \cos \left (f x +e \right )^{2}+39 \cos \left (f x +e \right )-15\right ) \left (\sec \left (f x +e \right )-1\right ) c \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )+1\right ) \sec \left (f x +e \right )^{2} \csc \left (f x +e \right )}{35 f \left (\cos \left (f x +e \right )-1\right )}\)	\(340\)

input

int(sec(f*x+e)*(a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^(3/2),x,method=_RETURNV 
ERBOSE)
 

output

2/63*a^3*c/f*(11*cos(f*x+e)-7)*(-c*(sec(f*x+e)-1))^(1/2)*(cos(f*x+e)+1)^4* 
sec(f*x+e)^4*csc(f*x+e)
 

3.1.81.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.40 \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2} \, dx=\frac {2 \, {\left (11 \, a^{3} c \cos \left (f x + e\right )^{5} + 37 \, a^{3} c \cos \left (f x + e\right )^{4} + 38 \, a^{3} c \cos \left (f x + e\right )^{3} + 2 \, a^{3} c \cos \left (f x + e\right )^{2} - 17 \, a^{3} c \cos \left (f x + e\right ) - 7 \, a^{3} c\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{63 \, f \cos \left (f x + e\right )^{4} \sin \left (f x + e\right )} \]

input

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^(3/2),x, algorith 
m="fricas")
 

output

2/63*(11*a^3*c*cos(f*x + e)^5 + 37*a^3*c*cos(f*x + e)^4 + 38*a^3*c*cos(f*x 
 + e)^3 + 2*a^3*c*cos(f*x + e)^2 - 17*a^3*c*cos(f*x + e) - 7*a^3*c)*sqrt(( 
c*cos(f*x + e) - c)/cos(f*x + e))/(f*cos(f*x + e)^4*sin(f*x + e))
 

3.1.81.6 Sympy [F]

\[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2} \, dx=a^{3} \left (\int c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )}\, dx + \int 2 c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )}\, dx + \int \left (- 2 c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{4}{\left (e + f x \right )}\right )\, dx + \int \left (- c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{5}{\left (e + f x \right )}\right )\, dx\right ) \]

input

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**3*(c-c*sec(f*x+e))**(3/2),x)
 

output

a**3*(Integral(c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x), x) + Integral(2*c 
*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**2, x) + Integral(-2*c*sqrt(-c*sec 
(e + f*x) + c)*sec(e + f*x)**4, x) + Integral(-c*sqrt(-c*sec(e + f*x) + c) 
*sec(e + f*x)**5, x))
 

3.1.81.7 Maxima [F(-1)]

Timed out. \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2} \, dx=\text {Timed out} \]

input

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^(3/2),x, algorith 
m="maxima")
 

output

Timed out
 

3.1.81.8 Giac [A] (verification not implemented)

Time = 0.86 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68 \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2} \, dx=\frac {32 \, \sqrt {2} {\left (9 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c^{5} + 7 \, c^{6}\right )} a^{3}}{63 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {9}{2}} f} \]

input

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3*(c-c*sec(f*x+e))^(3/2),x, algorith 
m="giac")
 

output

32/63*sqrt(2)*(9*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^5 + 7*c^6)*a^3/((c*tan(1 
/2*f*x + 1/2*e)^2 - c)^(9/2)*f)
 

3.1.81.9 Mupad [B] (verification not implemented)

Time = 21.87 (sec) , antiderivative size = 471, normalized size of antiderivative = 5.54 \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2} \, dx=\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^3\,c\,2{}\mathrm {i}}{f}+\frac {a^3\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,22{}\mathrm {i}}{63\,f}\right )}{{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1}-\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^3\,c\,32{}\mathrm {i}}{9\,f}+\frac {a^3\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,32{}\mathrm {i}}{9\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^3\,c\,8{}\mathrm {i}}{3\,f}-\frac {a^3\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,200{}\mathrm {i}}{63\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}+\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^3\,c\,32{}\mathrm {i}}{7\,f}+\frac {a^3\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,608{}\mathrm {i}}{63\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {a^3\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,160{}\mathrm {i}}{21\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2} \]

input

int(((a + a/cos(e + f*x))^3*(c - c/cos(e + f*x))^(3/2))/cos(e + f*x),x)
 

output

((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a^3*c*2i) 
/f + (a^3*c*exp(e*1i + f*x*1i)*22i)/(63*f)))/(exp(e*1i + f*x*1i) - 1) - (( 
c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a^3*c*32i)/ 
(9*f) + (a^3*c*exp(e*1i + f*x*1i)*32i)/(9*f)))/((exp(e*1i + f*x*1i) - 1)*( 
exp(e*2i + f*x*2i) + 1)^4) - ((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + 
f*x*1i)/2))^(1/2)*((a^3*c*8i)/(3*f) - (a^3*c*exp(e*1i + f*x*1i)*200i)/(63* 
f)))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)) + ((c - c/(exp(- 
e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a^3*c*32i)/(7*f) + (a^3* 
c*exp(e*1i + f*x*1i)*608i)/(63*f)))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + 
f*x*2i) + 1)^3) - (a^3*c*exp(e*1i + f*x*1i)*(c - c/(exp(- e*1i - f*x*1i)/2 
 + exp(e*1i + f*x*1i)/2))^(1/2)*160i)/(21*f*(exp(e*1i + f*x*1i) - 1)*(exp( 
e*2i + f*x*2i) + 1)^2)