\(\int \sec ^7(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [273]

Optimal result

Integrand size = 25, antiderivative size = 195 \[ \int \sec ^7(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {a^2 (5 a+6 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{16 (a+b)^{3/2} f}+\frac {a (5 a+6 b) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{16 (a+b) f}+\frac {(5 a+6 b) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{24 (a+b) f}+\frac {\sec ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2} \tan (e+f x)}{6 (a+b) f} \] Output:

1/16*a^2*(5*a+6*b)*arctanh((a+b)^(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2) 
)/(a+b)^(3/2)/f+1/16*a*(5*a+6*b)*sec(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)*tan(f 
*x+e)/(a+b)/f+1/24*(5*a+6*b)*sec(f*x+e)^3*(a+b*sin(f*x+e)^2)^(3/2)*tan(f*x 
+e)/(a+b)/f+1/6*sec(f*x+e)^5*(a+b*sin(f*x+e)^2)^(5/2)*tan(f*x+e)/(a+b)/f
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 16.89 (sec) , antiderivative size = 938, normalized size of antiderivative = 4.81 \[ \int \sec ^7(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx =\text {Too large to display} \] Input:

Integrate[Sec[e + f*x]^7*(a + b*Sin[e + f*x]^2)^(3/2),x]
 

Output:

(a^2*Sec[e + f*x]^3*(1 + (b*Sin[e + f*x]^2)/a)^2*Tan[e + f*x]*(45*a*ArcSin 
[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]] + 30*b*ArcSin[Sqrt[-(((a + b)*Tan[e 
+ f*x]^2)/a)]]*Sin[e + f*x]^2 + 210*a*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + 
f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2) + 140*b*Sin[e + f*x]^2*S 
qrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2) 
/a))^(3/2) - 120*a*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a 
+ b)*Tan[e + f*x]^2)/a))^(5/2) + 256*a*Hypergeometric2F1[2, 5, 7/2, -(((a 
+ b)*Tan[e + f*x]^2)/a)]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*( 
-(((a + b)*Tan[e + f*x]^2)/a))^(5/2) - 80*b*Sin[e + f*x]^2*Sqrt[(Sec[e + f 
*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(5/2) + 2 
56*b*Hypergeometric2F1[2, 5, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f 
*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + 
f*x]^2)/a))^(5/2) - 512*a*Hypergeometric2F1[2, 5, 7/2, -(((a + b)*Tan[e + 
f*x]^2)/a)]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Ta 
n[e + f*x]^2)/a))^(7/2) - 512*b*Hypergeometric2F1[2, 5, 7/2, -(((a + b)*Ta 
n[e + f*x]^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^ 
2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(7/2) + 256*a*Hypergeometric2F1[2, 
5, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + 
 f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(9/2) + 256*b*Hypergeometric2 
F1[2, 5, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e...
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3669, 296, 292, 292, 291, 219}

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.

\(\displaystyle \int \sec ^7(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^{3/2}}{\cos (e+f x)^7}dx\)

\(\Big \downarrow \) 3669

\(\displaystyle \frac {\int \frac {\left (b \sin ^2(e+f x)+a\right )^{3/2}}{\left (1-\sin ^2(e+f x)\right )^4}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 296

\(\displaystyle \frac {\frac {(5 a+6 b) \int \frac {\left (b \sin ^2(e+f x)+a\right )^{3/2}}{\left (1-\sin ^2(e+f x)\right )^3}d\sin (e+f x)}{6 (a+b)}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 (a+b) \left (1-\sin ^2(e+f x)\right )^3}}{f}\)

\(\Big \downarrow \) 292

\(\displaystyle \frac {\frac {(5 a+6 b) \left (\frac {3}{4} a \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\left (1-\sin ^2(e+f x)\right )^2}d\sin (e+f x)+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 \left (1-\sin ^2(e+f x)\right )^2}\right )}{6 (a+b)}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 (a+b) \left (1-\sin ^2(e+f x)\right )^3}}{f}\)

\(\Big \downarrow \) 292

\(\displaystyle \frac {\frac {(5 a+6 b) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}\right )+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 \left (1-\sin ^2(e+f x)\right )^2}\right )}{6 (a+b)}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 (a+b) \left (1-\sin ^2(e+f x)\right )^3}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {(5 a+6 b) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {(a+b) \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}\right )+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 \left (1-\sin ^2(e+f x)\right )^2}\right )}{6 (a+b)}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 (a+b) \left (1-\sin ^2(e+f x)\right )^3}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {(5 a+6 b) \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 \sqrt {a+b}}+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 \left (1-\sin ^2(e+f x)\right )}\right )+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 \left (1-\sin ^2(e+f x)\right )^2}\right )}{6 (a+b)}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 (a+b) \left (1-\sin ^2(e+f x)\right )^3}}{f}\)

Input:

Int[Sec[e + f*x]^7*(a + b*Sin[e + f*x]^2)^(3/2),x]
 

Output:

((Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(5/2))/(6*(a + b)*(1 - Sin[e + f*x]^ 
2)^3) + ((5*a + 6*b)*((Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2))/(4*(1 - 
Sin[e + f*x]^2)^2) + (3*a*((a*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + 
b*Sin[e + f*x]^2]])/(2*Sqrt[a + b]) + (Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x 
]^2])/(2*(1 - Sin[e + f*x]^2))))/4))/(6*(a + b)))/f
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 292

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Si 
mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( 
a*(p + 1)))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ 
{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt 
Q[q, 0] && NeQ[p, -1]
 

rule 296

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d))   Int[ 
(a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N 
eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1 
]) && NeQ[p, -1]
 

rule 3042

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

rule 3669

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f   S 
ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] 
/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(692\) vs. \(2(175)=350\).

Time = 0.50 (sec) , antiderivative size = 693, normalized size of antiderivative = 3.55

method	result	size
default	\(\frac {-3 a^{2} \left (5 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{4}+21 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{3} b +33 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2} b^{2}+23 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a \,b^{3}+6 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{4}-5 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{4}-21 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{3} b -33 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2} b^{2}-23 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a \,b^{3}-6 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{4}\right ) \cos \left (f x +e \right )^{6}+2 \left (a +b \right )^{\frac {7}{2}} \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (15 a^{2}+8 a b -4 b^{2}\right ) \cos \left (f x +e \right )^{4} \sin \left (f x +e \right )+4 \left (a +b \right )^{\frac {7}{2}} \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (5 a^{2}+3 a b -2 b^{2}\right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+16 \left (a +b \right )^{\frac {7}{2}} \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (a^{2}+2 a b +b^{2}\right ) \sin \left (f x +e \right )}{96 \left (a +b \right )^{\frac {9}{2}} \cos \left (f x +e \right )^{6} f}\)	\(693\)

Input:

int(sec(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/96*(-3*a^2*(5*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2 
)-b*sin(f*x+e)+a))*a^4+21*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+ 
e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3*b+33*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+ 
b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^2+23*ln(2/(1+sin(f*x+e))*(( 
a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b^3+6*ln(2/(1+sin 
(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^4-5*ln 
(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a)) 
*a^4-21*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin( 
f*x+e)+a))*a^3*b-33*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^ 
(1/2)+b*sin(f*x+e)+a))*a^2*b^2-23*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b* 
cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b^3-6*ln(2/(sin(f*x+e)-1)*((a+b)^(1 
/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^4)*cos(f*x+e)^6+2*(a+b)^ 
(7/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*(15*a^2+8*a*b-4*b^2)*cos(f*x+e)^4*sin(f*x 
+e)+4*(a+b)^(7/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*(5*a^2+3*a*b-2*b^2)*cos(f*x+e 
)^2*sin(f*x+e)+16*(a+b)^(7/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*(a^2+2*a*b+b^2)*s 
in(f*x+e))/(a+b)^(9/2)/cos(f*x+e)^6/f
 

Fricas [A] (verification not implemented)

Time = 1.82 (sec) , antiderivative size = 545, normalized size of antiderivative = 2.79 \[ \int \sec ^7(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left (5 \, a^{3} + 6 \, a^{2} b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{6} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, {\left ({\left (15 \, a^{3} + 23 \, a^{2} b + 4 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{3} + 24 \, a^{2} b + 24 \, a b^{2} + 8 \, b^{3} + 2 \, {\left (5 \, a^{3} + 8 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{192 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{6}}, -\frac {3 \, {\left (5 \, a^{3} + 6 \, a^{2} b\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{6} - 2 \, {\left ({\left (15 \, a^{3} + 23 \, a^{2} b + 4 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{3} + 24 \, a^{2} b + 24 \, a b^{2} + 8 \, b^{3} + 2 \, {\left (5 \, a^{3} + 8 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{96 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{6}}\right ] \] Input:

integrate(sec(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/192*(3*(5*a^3 + 6*a^2*b)*sqrt(a + b)*cos(f*x + e)^6*log(((a^2 + 8*a*b + 
 8*b^2)*cos(f*x + e)^4 - 8*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^2 - 4*((a + 
2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + 
b)*sin(f*x + e) + 8*a^2 + 16*a*b + 8*b^2)/cos(f*x + e)^4) + 4*((15*a^3 + 2 
3*a^2*b + 4*a*b^2 - 4*b^3)*cos(f*x + e)^4 + 8*a^3 + 24*a^2*b + 24*a*b^2 + 
8*b^3 + 2*(5*a^3 + 8*a^2*b + a*b^2 - 2*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f* 
x + e)^2 + a + b)*sin(f*x + e))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^6), -1 
/96*(3*(5*a^3 + 6*a^2*b)*sqrt(-a - b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 
 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(((a*b + b^2)*c 
os(f*x + e)^2 - a^2 - 2*a*b - b^2)*sin(f*x + e)))*cos(f*x + e)^6 - 2*((15* 
a^3 + 23*a^2*b + 4*a*b^2 - 4*b^3)*cos(f*x + e)^4 + 8*a^3 + 24*a^2*b + 24*a 
*b^2 + 8*b^3 + 2*(5*a^3 + 8*a^2*b + a*b^2 - 2*b^3)*cos(f*x + e)^2)*sqrt(-b 
*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e) 
^6)]
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^7(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)**7*(a+b*sin(f*x+e)**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^7(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{7} \,d x } \] Input:

integrate(sec(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*sin(f*x + e)^2 + a)^(3/2)*sec(f*x + e)^7, x)
 

Giac [F(-1)]

Timed out. \[ \int \sec ^7(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^7(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\cos \left (e+f\,x\right )}^7} \,d x \] Input:

int((a + b*sin(e + f*x)^2)^(3/2)/cos(e + f*x)^7,x)
 

Output:

int((a + b*sin(e + f*x)^2)^(3/2)/cos(e + f*x)^7, x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \sec ^7(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left (\int \sqrt {\sin \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{7} \sin \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{7}d x \right ) a \] Input:

int(sec(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x)
 

Output:

int(sqrt(sin(e + f*x)**2*b + a)*sec(e + f*x)**7*sin(e + f*x)**2,x)*b + int 
(sqrt(sin(e + f*x)**2*b + a)*sec(e + f*x)**7,x)*a