3.52 \(\int \frac {\csc ^4(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=123 \[ -\frac {\sqrt {b} (3 a-2 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 f (a+b)^{7/2}}-\frac {a b \tan (e+f x)}{2 f (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\cot ^3(e+f x)}{3 f (a+b)^2}-\frac {(a-b) \cot (e+f x)}{f (a+b)^3} \]

[Out]

-(a-b)*cot(f*x+e)/(a+b)^3/f-1/3*cot(f*x+e)^3/(a+b)^2/f-1/2*(3*a-2*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*b^
(1/2)/(a+b)^(7/2)/f-1/2*a*b*tan(f*x+e)/(a+b)^3/f/(a+b+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.17, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4132, 456, 1261, 205} \[ -\frac {\sqrt {b} (3 a-2 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 f (a+b)^{7/2}}-\frac {a b \tan (e+f x)}{2 f (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\cot ^3(e+f x)}{3 f (a+b)^2}-\frac {(a-b) \cot (e+f x)}{f (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

-((3*a - 2*b)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*(a + b)^(7/2)*f) - ((a - b)*Cot[e + f*x])
/((a + b)^3*f) - Cot[e + f*x]^3/(3*(a + b)^2*f) - (a*b*Tan[e + f*x])/(2*(a + b)^3*f*(a + b + b*Tan[e + f*x]^2)
)

Rule 205

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

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 4132

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*} \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1+x^2}{x^4 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {a b \tan (e+f x)}{2 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {b \operatorname {Subst}\left (\int \frac {-\frac {2}{b (a+b)}-\frac {2 a x^2}{b (a+b)^2}+\frac {a x^4}{(a+b)^3}}{x^4 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {a b \tan (e+f x)}{2 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {b \operatorname {Subst}\left (\int \left (-\frac {2}{b (a+b)^2 x^4}-\frac {2 (a-b)}{b (a+b)^3 x^2}+\frac {3 a-2 b}{(a+b)^3 \left (a+b+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {(a-b) \cot (e+f x)}{(a+b)^3 f}-\frac {\cot ^3(e+f x)}{3 (a+b)^2 f}-\frac {a b \tan (e+f x)}{2 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {((3 a-2 b) b) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 (a+b)^3 f}\\ &=-\frac {(3 a-2 b) \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 (a+b)^{7/2} f}-\frac {(a-b) \cot (e+f x)}{(a+b)^3 f}-\frac {\cot ^3(e+f x)}{3 (a+b)^2 f}-\frac {a b \tan (e+f x)}{2 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [C]  time = 6.18, size = 303, normalized size = 2.46 \[ \frac {\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-3 a b \sec (2 e) \sin (2 f x)-2 (a+b) \cot (e) \csc ^2(e+f x) (a \cos (2 (e+f x))+a+2 b)+2 (a+b) \csc (e) \sin (f x) \csc ^3(e+f x) (a \cos (2 (e+f x))+a+2 b)+4 (a-2 b) \csc (e) \sin (f x) \csc (e+f x) (a \cos (2 (e+f x))+a+2 b)+\frac {3 b (3 a-2 b) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}+3 b (a+2 b) \tan (2 e)\right )}{24 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*(-2*(a + b)*(a + 2*b + a*Cos[2*(e + f*x)])*Cot[e]*Csc[e + f*x]^
2 + (3*(3*a - 2*b)*b*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*S
qrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b
]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + 4*(a - 2*b)*(a + 2*b + a*Cos[2*(e + f*x)])*Csc[e]*Csc[e + f*x]*Sin[f*x] + 2
*(a + b)*(a + 2*b + a*Cos[2*(e + f*x)])*Csc[e]*Csc[e + f*x]^3*Sin[f*x] - 3*a*b*Sec[2*e]*Sin[2*f*x] + 3*b*(a +
2*b)*Tan[2*e]))/(24*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^2)

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fricas [B]  time = 0.63, size = 663, normalized size = 5.39 \[ \left [-\frac {4 \, {\left (4 \, a^{2} - 11 \, a b\right )} \cos \left (f x + e\right )^{5} - 8 \, {\left (3 \, a^{2} - 8 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (3 \, a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 5 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 2 \, b^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) - 12 \, {\left (3 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )}{24 \, {\left ({\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 2 \, a^{3} b - 2 \, a b^{3} - b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (4 \, a^{2} - 11 \, a b\right )} \cos \left (f x + e\right )^{5} - 4 \, {\left (3 \, a^{2} - 8 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (3 \, a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 5 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 2 \, b^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 6 \, {\left (3 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )}{12 \, {\left ({\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 2 \, a^{3} b - 2 \, a b^{3} - b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[-1/24*(4*(4*a^2 - 11*a*b)*cos(f*x + e)^5 - 8*(3*a^2 - 8*a*b + 4*b^2)*cos(f*x + e)^3 + 3*((3*a^2 - 2*a*b)*cos(
f*x + e)^4 - (3*a^2 - 5*a*b + 2*b^2)*cos(f*x + e)^2 - 3*a*b + 2*b^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^
2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 - 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*c
os(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2))*sin(f*x +
e) - 12*(3*a*b - 2*b^2)*cos(f*x + e))/(((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^4 - (a^4 + 2*a^3*b
- 2*a*b^3 - b^4)*f*cos(f*x + e)^2 - (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*f)*sin(f*x + e)), -1/12*(2*(4*a^2 - 1
1*a*b)*cos(f*x + e)^5 - 4*(3*a^2 - 8*a*b + 4*b^2)*cos(f*x + e)^3 - 3*((3*a^2 - 2*a*b)*cos(f*x + e)^4 - (3*a^2
- 5*a*b + 2*b^2)*cos(f*x + e)^2 - 3*a*b + 2*b^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqr
t(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) - 6*(3*a*b - 2*b^2)*cos(f*x + e))/(((a^4 + 3*a^3*b +
3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^4 - (a^4 + 2*a^3*b - 2*a*b^3 - b^4)*f*cos(f*x + e)^2 - (a^3*b + 3*a^2*b^2 +
3*a*b^3 + b^4)*f)*sin(f*x + e))]

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giac [A]  time = 0.46, size = 192, normalized size = 1.56 \[ -\frac {\frac {3 \, a b \tan \left (f x + e\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (3 \, a b - 2 \, b^{2}\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b + b^{2}}} + \frac {2 \, {\left (3 \, a \tan \left (f x + e\right )^{2} - 3 \, b \tan \left (f x + e\right )^{2} + a + b\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{3}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/6*(3*a*b*tan(f*x + e)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(b*tan(f*x + e)^2 + a + b)) + 3*(pi*floor((f*x + e)/
pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*(3*a*b - 2*b^2)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sq
rt(a*b + b^2)) + 2*(3*a*tan(f*x + e)^2 - 3*b*tan(f*x + e)^2 + a + b)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(f*x
+ e)^3))/f

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maple [A]  time = 1.42, size = 160, normalized size = 1.30 \[ -\frac {a b \tan \left (f x +e \right )}{2 \left (a +b \right )^{3} f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) a}{2 f \left (a +b \right )^{3} \sqrt {\left (a +b \right ) b}}+\frac {b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \left (a +b \right )^{3} \sqrt {\left (a +b \right ) b}}-\frac {1}{3 f \left (a +b \right )^{2} \tan \left (f x +e \right )^{3}}-\frac {a}{f \left (a +b \right )^{3} \tan \left (f x +e \right )}+\frac {b}{f \left (a +b \right )^{3} \tan \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^2,x)

[Out]

-1/2*a*b*tan(f*x+e)/(a+b)^3/f/(a+b+b*tan(f*x+e)^2)-3/2/f/(a+b)^3*b/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*
b)^(1/2))*a+1/f/(a+b)^3*b^2/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-1/3/f/(a+b)^2/tan(f*x+e)^3-1/
f/(a+b)^3/tan(f*x+e)*a+1/f/(a+b)^3/tan(f*x+e)*b

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maxima [A]  time = 0.55, size = 193, normalized size = 1.57 \[ -\frac {\frac {3 \, {\left (3 \, a b - 2 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {3 \, {\left (3 \, a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}}{{\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{5} + {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{3}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

-1/6*(3*(3*a*b - 2*b^2)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt((a + b)*b
)) + (3*(3*a*b - 2*b^2)*tan(f*x + e)^4 + 2*(3*a^2 + a*b - 2*b^2)*tan(f*x + e)^2 + 2*a^2 + 4*a*b + 2*b^2)/((a^3
*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*tan(f*x + e)^5 + (a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*tan(f*x + e)^3))/
f

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mupad [B]  time = 5.49, size = 141, normalized size = 1.15 \[ -\frac {\frac {1}{3\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (3\,a-2\,b\right )}{3\,{\left (a+b\right )}^2}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (3\,a-2\,b\right )}{2\,{\left (a+b\right )}^3}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (a+b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{{\left (a+b\right )}^{7/2}}\right )\,\left (3\,a-2\,b\right )}{2\,f\,{\left (a+b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^4*(a + b/cos(e + f*x)^2)^2),x)

[Out]

- (1/(3*(a + b)) + (tan(e + f*x)^2*(3*a - 2*b))/(3*(a + b)^2) + (b*tan(e + f*x)^4*(3*a - 2*b))/(2*(a + b)^3))/
(f*(tan(e + f*x)^3*(a + b) + b*tan(e + f*x)^5)) - (b^(1/2)*atan((b^(1/2)*tan(e + f*x)*(3*a*b^2 + 3*a^2*b + a^3
+ b^3))/(a + b)^(7/2))*(3*a - 2*b))/(2*f*(a + b)^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**4/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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