3.2.3 \(\int \csc ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) [103]

Optimal. Leaf size=141 \[ \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {2 (5 a-b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a f} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3744, 473, 462, 283, 223, 212} \begin {gather*} -\frac {2 (5 a-b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{15 a^2 f}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a f}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f - (Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]
^2])/f - (2*(5*a - b)*Cot[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(3/2))/(15*a^2*f) - (Cot[e + f*x]^5*(a + b*Tan[e +
f*x]^2)^(3/2))/(5*a*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 3744

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

Rubi steps

\begin {align*} \int \csc ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2 \sqrt {a+b x^2}}{x^6} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a f}+\frac {\text {Subst}\left (\int \frac {\left (2 (5 a-b)+5 a x^2\right ) \sqrt {a+b x^2}}{x^4} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac {2 (5 a-b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a f}+\frac {\text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {2 (5 a-b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a f}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {2 (5 a-b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a f}+\frac {b \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}\\ &=\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {2 (5 a-b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 3.57, size = 287, normalized size = 2.04 \begin {gather*} -\frac {\left (\left (80 a^3+198 a^2 b+98 a b^2-20 b^3+\left (40 a^3-241 a^2 b-149 a b^2+30 b^3\right ) \cos (2 (e+f x))+\left (-32 a^3+42 a^2 b+62 a b^2-12 b^3\right ) \cos (4 (e+f x))+8 a^3 \cos (6 (e+f x))+a^2 b \cos (6 (e+f x))-11 a b^2 \cos (6 (e+f x))+2 b^3 \cos (6 (e+f x))\right ) \csc ^6(e+f x)-240 \sqrt {2} a^2 b \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} F\left (\left .\text {ArcSin}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )\right ) \tan (e+f x)}{240 \sqrt {2} a^2 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/240*(((80*a^3 + 198*a^2*b + 98*a*b^2 - 20*b^3 + (40*a^3 - 241*a^2*b - 149*a*b^2 + 30*b^3)*Cos[2*(e + f*x)]
+ (-32*a^3 + 42*a^2*b + 62*a*b^2 - 12*b^3)*Cos[4*(e + f*x)] + 8*a^3*Cos[6*(e + f*x)] + a^2*b*Cos[6*(e + f*x)]
- 11*a*b^2*Cos[6*(e + f*x)] + 2*b^3*Cos[6*(e + f*x)])*Csc[e + f*x]^6 - 240*Sqrt[2]*a^2*b*Sqrt[((a + b + (a - b
)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^
2)/b]/Sqrt[2]], 1])*Tan[e + f*x])/(Sqrt[2]*a^2*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.37, size = 3769, normalized size = 26.73

method	result	size
default	\(\text {Expression too large to display}\)	\(3769\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15/f*(32*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b^2-25*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(
1/2)+a-2*b)/a)^(1/2)*a^2*b-31*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b^2+cos(f*x+e)^6*((2*I*
b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b-11*cos(f*x+e)^6*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b^2+9*co
s(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b+30*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(
1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*
b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/
2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b
)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*a^2*b*sin(f*x+e)-15*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)
^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a
-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticF((cos(f*x+e)-1)*
((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8
*a*b+8*b^2)/a^2)^(1/2))*a^2*b*sin(f*x+e)-2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^3+8*cos(f*x+e)^6*((2*I*
b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^3-20*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^3+15*cos(f
*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^3+2*cos(f*x+e)^6*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)
*b^3-6*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^3+6*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2
*b)/a)^(1/2)*b^3+15*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b+10*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/
2)*a*b^2+30*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos
(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(
cos(f*x+e)+1)/a)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),1/(2*I*b
^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/
2))*sin(f*x+e)*cos(f*x+e)^4*a^2*b-15*sin(f*x+e)*cos(f*x+e)^4*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1
/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b
^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)
*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a
^2)^(1/2))*a^2*b-60*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)
+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x
+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),
1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b
)/a)^(1/2))*sin(f*x+e)*cos(f*x+e)^3*a^2*b+30*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+
cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/
2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2
*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*sin(f
*x+e)*cos(f*x+e)^3*a^2*b-60*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*co
s(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b
*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin
(f*x+e),1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/
2)+a-2*b)/a)^(1/2))*sin(f*x+e)*cos(f*x+e)^2*a^2*b+30*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b
)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(
a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(
1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2
))*sin(f*x+e)*cos(f*x+e)^2*a^2*b+30*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e
)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*
x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(
1/2)/sin(f*x+e),1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(
a-b)^(1/2)+a-2*b)/a)^(1/2))*sin(f*x+e)*cos(f*x+e)*a^2*b-15*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2
)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos...

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Maxima [A]
time = 0.29, size = 140, normalized size = 0.99 \begin {gather*} \frac {15 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right ) - \frac {15 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{\tan \left (f x + e\right )} - \frac {10 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \tan \left (f x + e\right )^{3}} + \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b}{a^{2} \tan \left (f x + e\right )^{3}} - \frac {3 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \tan \left (f x + e\right )^{5}}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 297 vs. \(2 (133) = 266\).
time = 2.97, size = 623, normalized size = 4.42 \begin {gather*} \left [\frac {15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (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 )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \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} + 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (20 \, a^{2} + 19 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 10 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )} \sin \left (f x + e\right )}, -\frac {15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {{\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a b - b^{2}\right )} \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} + 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (20 \, a^{2} + 19 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 10 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{30 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )} \sin \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(15*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 + a^2)*sqrt(b)*log(((a^2 - 8*a*b + 8*b^2)*cos(f*x + e)^4
+ 8*(a*b - 2*b^2)*cos(f*x + e)^2 + 4*((a - 2*b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt(((a - b)*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 + 9*a*b - 2*b^2)
*cos(f*x + e)^5 - (20*a^2 + 19*a*b - 4*b^2)*cos(f*x + e)^3 + (15*a^2 + 10*a*b - 2*b^2)*cos(f*x + e))*sqrt(((a
- b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^2*f*cos(f*x + e)^4 - 2*a^2*f*cos(f*x + e)^2 + a^2*f)*sin(f*x + e
)), -1/30*(15*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 + a^2)*sqrt(-b)*arctan(1/2*((a - 2*b)*cos(f*x + e)^3
+ 2*b*cos(f*x + e))*sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(((a*b - b^2)*cos(f*x + e)^2 +
b^2)*sin(f*x + e)))*sin(f*x + e) + 2*((8*a^2 + 9*a*b - 2*b^2)*cos(f*x + e)^5 - (20*a^2 + 19*a*b - 4*b^2)*cos(f
*x + e)^3 + (15*a^2 + 10*a*b - 2*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^2*f
*cos(f*x + e)^4 - 2*a^2*f*cos(f*x + e)^2 + a^2*f)*sin(f*x + e))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \csc ^{6}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*tan(e + f*x)**2)*csc(e + f*x)**6, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*csc(f*x + e)^6, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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