\(\int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx\) [73]

Optimal result

Integrand size = 25, antiderivative size = 240 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\frac {\left (5 a^3-15 a^2 b-5 a b^2-b^3\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{5/2} f}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {(a-b) (5 a+b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a^2 f}-\frac {(5 a-b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 a f}-\frac {\cos (e+f x) \sin ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 f} \]

[Out]

1/16*(5*a^3-15*a^2*b-5*a*b^2-b^3)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(5/2)/f+arctanh(b^(1
/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f-1/16*(a-b)*(5*a+b)*cos(f*x+e)*sin(f*x+e)*(a+b+b*tan(f*x+e
)^2)^(1/2)/a^2/f-1/24*(5*a-b)*cos(f*x+e)*sin(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/a/f-1/6*cos(f*x+e)*sin(f*x+e)
^5*(a+b+b*tan(f*x+e)^2)^(1/2)/f

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4217, 478, 592, 537, 223, 212, 385, 209} \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=-\frac {(a-b) (5 a+b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{16 a^2 f}+\frac {\left (5 a^3-15 a^2 b-5 a b^2-b^3\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{16 a^{5/2} f}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac {\sin ^5(e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{6 f}-\frac {(5 a-b) \sin ^3(e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{24 a f} \]

[In]

Int[Sqrt[a + b*Sec[e + f*x]^2]*Sin[e + f*x]^6,x]

[Out]

((5*a^3 - 15*a^2*b - 5*a*b^2 - b^3)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*a^(5/2)
*f) + (Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/f - ((a - b)*(5*a + b)*Cos[e +
f*x]*Sin[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(16*a^2*f) - ((5*a - b)*Cos[e + f*x]*Sin[e + f*x]^3*Sqrt[a +
b + b*Tan[e + f*x]^2])/(24*a*f) - (Cos[e + f*x]*Sin[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(6*f)

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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

Rule 385

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

Rule 478

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*n*(p + 1))), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 592

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c -
a*d)*(p + 1))), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 4217

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1
+ ff^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^6 \sqrt {a+b+b x^2}}{\left (1+x^2\right )^4} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cos (e+f x) \sin ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 f}+\frac {\text {Subst}\left (\int \frac {x^4 \left (5 (a+b)+6 b x^2\right )}{\left (1+x^2\right )^3 \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{6 f} \\ & = -\frac {(5 a-b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 a f}-\frac {\cos (e+f x) \sin ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 f}+\frac {\text {Subst}\left (\int \frac {x^2 \left (3 (5 a-b) (a+b)+24 a b x^2\right )}{\left (1+x^2\right )^2 \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{24 a f} \\ & = -\frac {(a-b) (5 a+b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a^2 f}-\frac {(5 a-b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 a f}-\frac {\cos (e+f x) \sin ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 f}+\frac {\text {Subst}\left (\int \frac {3 (5 a+b) \left (a^2-b^2\right )+48 a^2 b x^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{48 a^2 f} \\ & = -\frac {(a-b) (5 a+b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a^2 f}-\frac {(5 a-b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 a f}-\frac {\cos (e+f x) \sin ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 f}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}+\frac {\left (-48 a^2 b+3 (5 a+b) \left (a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{48 a^2 f} \\ & = -\frac {(a-b) (5 a+b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a^2 f}-\frac {(5 a-b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 a f}-\frac {\cos (e+f x) \sin ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 f}+\frac {b \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}+\frac {\left (-48 a^2 b+3 (5 a+b) \left (a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{48 a^2 f} \\ & = -\frac {\left (16 a^2 b-(5 a+b) \left (a^2-b^2\right )\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{5/2} f}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {(a-b) (5 a+b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a^2 f}-\frac {(5 a-b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 a f}-\frac {\cos (e+f x) \sin ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 f} \\ \end{align*}

Mathematica [F]

\[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx \]

[In]

Integrate[Sqrt[a + b*Sec[e + f*x]^2]*Sin[e + f*x]^6,x]

[Out]

Integrate[Sqrt[a + b*Sec[e + f*x]^2]*Sin[e + f*x]^6, x]

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1179\) vs. \(2(214)=428\).

Time = 12.64 (sec) , antiderivative size = 1180, normalized size of antiderivative = 4.92

method	result	size
default	\(\text {Expression too large to display}\)	\(1180\)

[In]

int(sin(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/48/f/a^2/(-a)^(1/2)*(8*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*cos(f*x+e)^5*sin(f*x+e)+8
*(-a)^(1/2)*sin(f*x+e)*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2-26*(-a)^(1/2)*((b+a*cos(f*
x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*cos(f*x+e)^3*sin(f*x+e)+2*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)
^(1/2)*a*b*cos(f*x+e)^3*sin(f*x+e)-26*(-a)^(1/2)*sin(f*x+e)*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)
^(1/2)*a^2+2*(-a)^(1/2)*sin(f*x+e)*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b+33*(-a)^(1/2)*
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*cos(f*x+e)*sin(f*x+e)-14*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos
(f*x+e))^2)^(1/2)*a*b*cos(f*x+e)*sin(f*x+e)-3*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2*cos(f
*x+e)*sin(f*x+e)+33*(-a)^(1/2)*sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2-14*(-a)^(1/2)*sin(f*
x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-3*(-a)^(1/2)*sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)
)^2)^(1/2)*b^2-24*(-a)^(1/2)*b^(1/2)*ln(4*(-((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+sin
(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-a-b)/(sin(f*x+e)-1))*a^2-24*(-a)^(1/2)*b^(1/2)*l
n(-4*(-((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)
/(1+cos(f*x+e))^2)^(1/2)+a+b)/(sin(f*x+e)+1))*a^2-15*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/
2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a^3+45*ln(4*(-a)^(1/2)*
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2
)-4*sin(f*x+e)*a)*a^2*b+15*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)
*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a*b^2+3*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+co
s(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*b^3)*(a
+b*sec(f*x+e)^2)^(1/2)*cos(f*x+e)/(1+cos(f*x+e))/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 2.89 (sec) , antiderivative size = 1715, normalized size of antiderivative = 7.15 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\text {Too large to display} \]

[In]

integrate(sin(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/384*(96*a^3*sqrt(b)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos
(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(
f*x + e)^4) + 3*(5*a^3 - 15*a^2*b - 5*a*b^2 - b^3)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos
(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4
- 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x
+ e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a
)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - 8*(8*a^3*cos(f*x + e)^5 - 2*(13*a^3 - a^2*b)*cos
(f*x + e)^3 + (33*a^3 - 14*a^2*b - 3*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x
+ e))/(a^3*f), 1/384*(192*a^3*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((
a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e))) + 3*(5*a^3 - 15*a^2*b - 5*a*b
^2 - b^3)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^
2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*
cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*c
os(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e
)^2)*sin(f*x + e)) - 8*(8*a^3*cos(f*x + e)^5 - 2*(13*a^3 - a^2*b)*cos(f*x + e)^3 + (33*a^3 - 14*a^2*b - 3*a*b^
2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^3*f), 1/192*(48*a^3*sqrt(b)*log(
((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x +
e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) - 3*(5*a^3 - 15
*a^2*b - 5*a*b^2 - b^3)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b
+ b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b
^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) - 4*(8*a^3*cos(f*x + e)^5 - 2*(13*a^3 - a^2*b)*cos(f*x + e
)^3 + (33*a^3 - 14*a^2*b - 3*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a
^3*f), 1/192*(96*a^3*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x
+ e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e))) - 3*(5*a^3 - 15*a^2*b - 5*a*b^2 - b^3)
*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*s
qrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*co
s(f*x + e)^2)*sin(f*x + e))) - 4*(8*a^3*cos(f*x + e)^5 - 2*(13*a^3 - a^2*b)*cos(f*x + e)^3 + (33*a^3 - 14*a^2*
b - 3*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^3*f)]

Sympy [F(-1)]

Timed out. \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)**6*(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{6} \,d x } \]

[In]

integrate(sin(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(f*x + e)^2 + a)*sin(f*x + e)^6, x)

Giac [F]

\[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{6} \,d x } \]

[In]

integrate(sin(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(f*x + e)^2 + a)*sin(f*x + e)^6, x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^6\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}} \,d x \]

[In]

int(sin(e + f*x)^6*(a + b/cos(e + f*x)^2)^(1/2),x)

[Out]

int(sin(e + f*x)^6*(a + b/cos(e + f*x)^2)^(1/2), x)