\(\int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [79]

Optimal result

Integrand size = 25, antiderivative size = 149 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{f}-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f} \]

[Out]

arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f-cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/f-2/15*
(5*a+4*b)*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(3/2)/(a+b)^2/f-1/5*cot(f*x+e)^5*(a+b+b*tan(f*x+e)^2)^(3/2)/(a+b)/
f

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4217, 473, 462, 283, 223, 212} \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 f (a+b)}-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{15 f (a+b)^2}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f} \]

[In]

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

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/f - (Cot[e + f*x]*Sqrt[a + b + b*Tan[
e + f*x]^2])/f - (2*(5*a + 4*b)*Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(3/2))/(15*(a + b)^2*f) - (Cot[e + f
*x]^5*(a + b + b*Tan[e + f*x]^2)^(3/2))/(5*(a + b)*f)

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 283

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

Rule 462

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

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 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 {\left (1+x^2\right )^2 \sqrt {a+b+b x^2}}{x^6} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}+\frac {\text {Subst}\left (\int \frac {\sqrt {a+b+b x^2} \left (2 (5 a+4 b)+5 (a+b) x^2\right )}{x^4} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f} \\ & = -\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}+\frac {\text {Subst}\left (\int \frac {\sqrt {a+b+b x^2}}{x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{f}-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{f}-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) 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 {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{f}-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 8.73 (sec) , antiderivative size = 941, normalized size of antiderivative = 6.32 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {2} \csc ^5(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \left (-3 (a+b) \cos ^2(e+f x)-4 a \cos ^2(e+f x) \sin ^2(e+f x)-16 b \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x)+8 b \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x)-\frac {8 a^2 \cos ^2(e+f x) \sin ^4(e+f x)}{a+b}-\frac {8 a b \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x)}{a+b}-\frac {16 a b \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x)}{a+b}+\frac {24 a^2 b \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x)}{(a+b)^2}+\frac {8 a^2 b \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x)}{(a+b)^2}-\frac {16 b^2 \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{a+b}+\frac {8 b^2 \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{a+b}-\frac {8 a b^2 \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x) \tan ^2(e+f x)}{(a+b)^2}-\frac {16 a b^2 \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x) \tan ^2(e+f x)}{(a+b)^2}+\frac {24 a^2 b^2 \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x) \tan ^2(e+f x)}{(a+b)^3}+\frac {8 a^2 b^2 \, _3F_2\left (2,2,2;1,\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x) \tan ^2(e+f x)}{(a+b)^3}-\frac {4 a \arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \cos ^2(e+f x) \sin ^2(e+f x) \sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}}{\sqrt {\frac {a+b \sec ^2(e+f x)}{a+b}}}+\frac {3 b \arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \sin ^2(e+f x)}{\sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}}+\frac {8 a^2 b \arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \sin ^6(e+f x)}{(a+b)^2 \sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}}\right )}{15 f \sqrt {a+2 b+a \cos (2 e+2 f x)} \sqrt {a+b-a \sin ^2(e+f x)}} \]

[In]

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

[Out]

(Sqrt[2]*Csc[e + f*x]^5*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*(1 - (a*Sin[e + f*x]^2)/(a + b))*(-3*(a + b)*C
os[e + f*x]^2 - 4*a*Cos[e + f*x]^2*Sin[e + f*x]^2 - 16*b*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a
+ b))]*Sin[e + f*x]^2 + 8*b*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]
^2 - (8*a^2*Cos[e + f*x]^2*Sin[e + f*x]^4)/(a + b) - (8*a*b*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/
(a + b))]*Sin[e + f*x]^4)/(a + b) - (16*a*b*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b
))]*Sin[e + f*x]^4)/(a + b) + (24*a^2*b*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*
x]^6)/(a + b)^2 + (8*a^2*b*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^
6)/(a + b)^2 - (16*b^2*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^2*Tan[e + f*x]
^2)/(a + b) + (8*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^2*Tan[
e + f*x]^2)/(a + b) - (8*a*b^2*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^4*Tan[
e + f*x]^2)/(a + b)^2 - (16*a*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e
+ f*x]^4*Tan[e + f*x]^2)/(a + b)^2 + (24*a^2*b^2*Hypergeometric2F1[2, 2, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*S
in[e + f*x]^6*Tan[e + f*x]^2)/(a + b)^3 + (8*a^2*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, -((b*Tan[e + f*x]^
2)/(a + b))]*Sin[e + f*x]^6*Tan[e + f*x]^2)/(a + b)^3 - (4*a*ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]]*Cos[e
+ f*x]^2*Sin[e + f*x]^2*Sqrt[-((b*Tan[e + f*x]^2)/(a + b))])/Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)] + (3*b*ArcS
in[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]]*Sin[e + f*x]^2)/Sqrt[-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Ta
n[e + f*x]^2)/(a + b)^2)] + (8*a^2*b*ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]]*Sin[e + f*x]^6)/((a + b)^2*Sq
rt[-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2)])))/(15*f*Sqrt[a + 2*b + a*Cos[2*
e + 2*f*x]]*Sqrt[a + b - a*Sin[e + f*x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1679\) vs. \(2(133)=266\).

Time = 7.18 (sec) , antiderivative size = 1680, normalized size of antiderivative = 11.28

method	result	size
default	\(\text {Expression too large to display}\)	\(1680\)

[In]

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

[Out]

-1/30/f/(a+b)^2*(15*sin(f*x+e)^3*b^(5/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+
b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*cos(f*x+e)+15*sin(f*x+e)
^3*b^(5/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(
1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*cos(f*x+e)+30*sin(f*x+e)^3*b^(3/2)*a*ln(4*(((b+a*cos(
f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f
*x+e)*a-a-b)/(sin(f*x+e)+1))*cos(f*x+e)+30*sin(f*x+e)^3*b^(3/2)*a*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)
^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1)
)*cos(f*x+e)-15*sin(f*x+e)^3*b^(5/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1
/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))-15*sin(f*x+e)^3*b^(5/2)*ln(-
4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2
)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))+15*sin(f*x+e)^3*b^(1/2)*a^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)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)
+1))*cos(f*x+e)+15*sin(f*x+e)^3*b^(1/2)*a^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)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*cos(f*x+e)-30*sin(f*
x+e)^3*b^(3/2)*a*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)
^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))-30*sin(f*x+e)^3*b^(3/2)*a*ln(-4*(((b+a*cos(f*x+e
)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)
*a+a+b)/(sin(f*x+e)-1))+16*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+50*cos(f*x+e)^4*((b+a*
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b+30*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2-15*s
in(f*x+e)^3*b^(1/2)*a^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)+b^(1/2)*((b+a*cos
(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))-15*sin(f*x+e)^3*b^(1/2)*a^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)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-s
in(f*x+e)*a+a+b)/(sin(f*x+e)-1))-40*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2-118*cos(f*x+e
)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-70*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2
)*b^2+30*a^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+80*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b+46
*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2)*(a+b*sec(f*x+e)^2)^(1/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^
2)^(1/2)*cot(f*x+e)*csc(f*x+e)^4

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (133) = 266\).

Time = 1.39 (sec) , antiderivative size = 656, normalized size of antiderivative = 4.40 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [\frac {15 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (8 \, a^{2} + 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (20 \, a^{2} + 59 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )} \sin \left (f x + e\right )}, \frac {15 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left ({\left (8 \, a^{2} + 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (20 \, a^{2} + 59 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{30 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]

[In]

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

[Out]

[1/60*(15*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*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*c
os(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)*sin(f*x
+ e) - 4*((8*a^2 + 25*a*b + 15*b^2)*cos(f*x + e)^5 - (20*a^2 + 59*a*b + 35*b^2)*cos(f*x + e)^3 + (15*a^2 + 40
*a*b + 23*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^2 + 2*a*b + b^2)*f*cos(f*x + e)
^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f)*sin(f*x + e)), 1/30*(15*((a^2 + 2*a*b + b
^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b)*arctan(-1/2*((a - b)*c
os(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)))*sin(f*x + e) - 2*((8*a^2 + 25*a*b + 15*b^2)*cos(f*x + e)^5 - (20*a^2 + 59*a*b + 35*b^2)*co
s(f*x + e)^3 + (15*a^2 + 40*a*b + 23*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^2 +
2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f)*sin(f*x + e))]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.96 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {15 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) - \frac {15 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b}}{\tan \left (f x + e\right )} - \frac {10 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{{\left (a + b\right )} \tan \left (f x + e\right )^{3}} + \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b}{{\left (a + b\right )}^{2} \tan \left (f x + e\right )^{3}} - \frac {3 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{{\left (a + b\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \]

[In]

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

[Out]

1/15*(15*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) - 15*sqrt(b*tan(f*x + e)^2 + a + b)/tan(f*x + e) - 10
*(b*tan(f*x + e)^2 + a + b)^(3/2)/((a + b)*tan(f*x + e)^3) + 2*(b*tan(f*x + e)^2 + a + b)^(3/2)*b/((a + b)^2*t
an(f*x + e)^3) - 3*(b*tan(f*x + e)^2 + a + b)^(3/2)/((a + b)*tan(f*x + e)^5))/f

Giac [F]

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

[In]

integrate(csc(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)*csc(f*x + e)^6, x)

Mupad [F(-1)]

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

[In]

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

[Out]

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