3.60 \(\int \frac{\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=130 \[ -\frac{\left (3 a^2-12 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^3 f}-\frac{(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac{\sqrt{b} (a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{a^3 f}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f} \]

[Out]

-(((a - b)^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(a^3*f)) - ((3*a^2 - 12*a*b + 8*b^2)*ArcT
anh[Cos[e + f*x]])/(8*a^3*f) - ((5*a - 4*b)*Cot[e + f*x]*Csc[e + f*x])/(8*a^2*f) - (Cot[e + f*x]^3*Csc[e + f*x
])/(4*a*f)

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Rubi [A]  time = 0.175632, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3664, 470, 527, 522, 207, 205} \[ -\frac{\left (3 a^2-12 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^3 f}-\frac{(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac{\sqrt{b} (a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{a^3 f}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2),x]

[Out]

-(((a - b)^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(a^3*f)) - ((3*a^2 - 12*a*b + 8*b^2)*ArcT
anh[Cos[e + f*x]])/(8*a^3*f) - ((5*a - 4*b)*Cot[e + f*x]*Csc[e + f*x])/(8*a^2*f) - (Cot[e + f*x]^3*Csc[e + f*x
])/(4*a*f)

Rule 3664

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

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

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

Rule 207

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

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]

Rubi steps

\begin{align*} \int \frac{\csc ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^3 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f}-\frac{\operatorname{Subst}\left (\int \frac{-a+b+(-4 a+3 b) x^2}{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{4 a f}\\ &=-\frac{(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f}-\frac{\operatorname{Subst}\left (\int \frac{-(3 a-4 b) (a-b)+(5 a-4 b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{8 a^2 f}\\ &=-\frac{(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f}-\frac{\left ((a-b)^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{a^3 f}+\frac{\left (3 a^2-12 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{8 a^3 f}\\ &=-\frac{(a-b)^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{a^3 f}-\frac{\left (3 a^2-12 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^3 f}-\frac{(5 a-4 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f}\\ \end{align*}

Mathematica [B]  time = 6.24972, size = 326, normalized size = 2.51 \[ \frac{\left (3 a^2-12 a b+8 b^2\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{8 a^3 f}+\frac{\left (-3 a^2+12 a b-8 b^2\right ) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{8 a^3 f}+\frac{(4 b-3 a) \csc ^2\left (\frac{1}{2} (e+f x)\right )}{32 a^2 f}+\frac{(3 a-4 b) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{32 a^2 f}+\frac{\sqrt{b} (a-b)^{3/2} \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (e+f x)\right ) \left (\sqrt{a-b} \cos \left (\frac{1}{2} (e+f x)\right )-\sqrt{a} \sin \left (\frac{1}{2} (e+f x)\right )\right )}{\sqrt{b}}\right )}{a^3 f}+\frac{\sqrt{b} (a-b)^{3/2} \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (e+f x)\right ) \left (\sqrt{a-b} \cos \left (\frac{1}{2} (e+f x)\right )+\sqrt{a} \sin \left (\frac{1}{2} (e+f x)\right )\right )}{\sqrt{b}}\right )}{a^3 f}-\frac{\csc ^4\left (\frac{1}{2} (e+f x)\right )}{64 a f}+\frac{\sec ^4\left (\frac{1}{2} (e+f x)\right )}{64 a f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2),x]

[Out]

((a - b)^(3/2)*Sqrt[b]*ArcTan[(Sec[(e + f*x)/2]*(Sqrt[a - b]*Cos[(e + f*x)/2] - Sqrt[a]*Sin[(e + f*x)/2]))/Sqr
t[b]])/(a^3*f) + ((a - b)^(3/2)*Sqrt[b]*ArcTan[(Sec[(e + f*x)/2]*(Sqrt[a - b]*Cos[(e + f*x)/2] + Sqrt[a]*Sin[(
e + f*x)/2]))/Sqrt[b]])/(a^3*f) + ((-3*a + 4*b)*Csc[(e + f*x)/2]^2)/(32*a^2*f) - Csc[(e + f*x)/2]^4/(64*a*f) +
((-3*a^2 + 12*a*b - 8*b^2)*Log[Cos[(e + f*x)/2]])/(8*a^3*f) + ((3*a^2 - 12*a*b + 8*b^2)*Log[Sin[(e + f*x)/2]]
)/(8*a^3*f) + ((3*a - 4*b)*Sec[(e + f*x)/2]^2)/(32*a^2*f) + Sec[(e + f*x)/2]^4/(64*a*f)

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Maple [B]  time = 0.081, size = 344, normalized size = 2.7 \begin{align*}{\frac{1}{16\,fa \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}+{\frac{3}{16\,fa \left ( \cos \left ( fx+e \right ) +1 \right ) }}-{\frac{b}{4\,f{a}^{2} \left ( \cos \left ( fx+e \right ) +1 \right ) }}-{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) +1 \right ) }{16\,fa}}+{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) +1 \right ) b}{4\,f{a}^{2}}}-{\frac{\ln \left ( \cos \left ( fx+e \right ) +1 \right ){b}^{2}}{2\,f{a}^{3}}}+{\frac{b}{fa}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}}-2\,{\frac{{b}^{2}}{f{a}^{2}\sqrt{b \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \cos \left ( fx+e \right ) }{\sqrt{b \left ( a-b \right ) }}} \right ) }+{\frac{{b}^{3}}{f{a}^{3}}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}}-{\frac{1}{16\,fa \left ( \cos \left ( fx+e \right ) -1 \right ) ^{2}}}+{\frac{3}{16\,fa \left ( \cos \left ( fx+e \right ) -1 \right ) }}-{\frac{b}{4\,f{a}^{2} \left ( \cos \left ( fx+e \right ) -1 \right ) }}+{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) -1 \right ) }{16\,fa}}-{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) -1 \right ) b}{4\,f{a}^{2}}}+{\frac{\ln \left ( \cos \left ( fx+e \right ) -1 \right ){b}^{2}}{2\,f{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^5/(a+b*tan(f*x+e)^2),x)

[Out]

1/16/f/a/(cos(f*x+e)+1)^2+3/16/f/a/(cos(f*x+e)+1)-1/4/f/a^2/(cos(f*x+e)+1)*b-3/16/f/a*ln(cos(f*x+e)+1)+3/4/f/a
^2*ln(cos(f*x+e)+1)*b-1/2/f/a^3*ln(cos(f*x+e)+1)*b^2+1/f*b/a/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))
^(1/2))-2/f*b^2/a^2/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))+1/f*b^3/a^3/(b*(a-b))^(1/2)*arcta
n((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))-1/16/f/a/(cos(f*x+e)-1)^2+3/16/f/a/(cos(f*x+e)-1)-1/4/f/a^2/(cos(f*x+e)-1)
*b+3/16/f/a*ln(cos(f*x+e)-1)-3/4/f/a^2*ln(cos(f*x+e)-1)*b+1/2/f/a^3*ln(cos(f*x+e)-1)*b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.52071, size = 1523, normalized size = 11.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/16*(2*(3*a^2 - 4*a*b)*cos(f*x + e)^3 - 8*((a - b)*cos(f*x + e)^4 - 2*(a - b)*cos(f*x + e)^2 + a - b)*sqrt(-
a*b + b^2)*log(((a - b)*cos(f*x + e)^2 - 2*sqrt(-a*b + b^2)*cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) -
2*(5*a^2 - 4*a*b)*cos(f*x + e) - ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 12*a*b + 8*b^2)*cos(f*x
+ e)^2 + 3*a^2 - 12*a*b + 8*b^2)*log(1/2*cos(f*x + e) + 1/2) + ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(
3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^2 + 3*a^2 - 12*a*b + 8*b^2)*log(-1/2*cos(f*x + e) + 1/2))/(a^3*f*cos(f*x
+ e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f), 1/16*(2*(3*a^2 - 4*a*b)*cos(f*x + e)^3 + 16*((a - b)*cos(f*x + e)^4
- 2*(a - b)*cos(f*x + e)^2 + a - b)*sqrt(a*b - b^2)*arctan(sqrt(a*b - b^2)*cos(f*x + e)/b) - 2*(5*a^2 - 4*a*b)
*cos(f*x + e) - ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^2 + 3*a^2 -
12*a*b + 8*b^2)*log(1/2*cos(f*x + e) + 1/2) + ((3*a^2 - 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 - 12*a*b +
8*b^2)*cos(f*x + e)^2 + 3*a^2 - 12*a*b + 8*b^2)*log(-1/2*cos(f*x + e) + 1/2))/(a^3*f*cos(f*x + e)^4 - 2*a^3*f*
cos(f*x + e)^2 + a^3*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**5/(a+b*tan(f*x+e)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.47657, size = 505, normalized size = 3.88 \begin{align*} -\frac{\frac{\frac{8 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{8 \, b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{a{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{a^{2}} - \frac{4 \,{\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}{a^{3}} + \frac{64 \,{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac{a \cos \left (f x + e\right ) - b \cos \left (f x + e\right ) - b}{\sqrt{a b - b^{2}} \cos \left (f x + e\right ) + \sqrt{a b - b^{2}}}\right )}{\sqrt{a b - b^{2}} a^{3}} + \frac{{\left (a^{2} - \frac{8 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{8 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{18 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{72 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{48 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{a^{3}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}}{64 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

-1/64*((8*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 8*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - a*(cos(f*x + e
) - 1)^2/(cos(f*x + e) + 1)^2)/a^2 - 4*(3*a^2 - 12*a*b + 8*b^2)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/a^
3 + 64*(a^2*b - 2*a*b^2 + b^3)*arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b)/(sqrt(a*b - b^2)*cos(f*x + e) + s
qrt(a*b - b^2)))/(sqrt(a*b - b^2)*a^3) + (a^2 - 8*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 8*a*b*(cos(f*x +
e) - 1)/(cos(f*x + e) + 1) + 18*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 72*a*b*(cos(f*x + e) - 1)^2/(
cos(f*x + e) + 1)^2 + 48*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)^2/(a^3*(cos(f*x + e
) - 1)^2))/f