3.47 \(\int \csc ^3(e+f x) (a+b \tan ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=82 \[ -\frac{a^2 \csc ^2(e+f x) \sec (e+f x)}{2 f}+\frac{a (a+4 b) \sec (e+f x)}{2 f}-\frac{a (a+4 b) \tanh ^{-1}(\cos (e+f x))}{2 f}+\frac{b^2 \sec ^3(e+f x)}{3 f} \]

[Out]

-(a*(a + 4*b)*ArcTanh[Cos[e + f*x]])/(2*f) + (a*(a + 4*b)*Sec[e + f*x])/(2*f) - (a^2*Csc[e + f*x]^2*Sec[e + f*
x])/(2*f) + (b^2*Sec[e + f*x]^3)/(3*f)

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Rubi [A]  time = 0.109035, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3664, 463, 459, 321, 207} \[ -\frac{a^2 \csc ^2(e+f x) \sec (e+f x)}{2 f}+\frac{a (a+4 b) \sec (e+f x)}{2 f}-\frac{a (a+4 b) \tanh ^{-1}(\cos (e+f x))}{2 f}+\frac{b^2 \sec ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3*(a + b*Tan[e + f*x]^2)^2,x]

[Out]

-(a*(a + 4*b)*ArcTanh[Cos[e + f*x]])/(2*f) + (a*(a + 4*b)*Sec[e + f*x])/(2*f) - (a^2*Csc[e + f*x]^2*Sec[e + f*
x])/(2*f) + (b^2*Sec[e + f*x]^3)/(3*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 463

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

Rule 459

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

Rule 321

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

Rubi steps

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

Mathematica [B]  time = 6.12979, size = 376, normalized size = 4.59 \[ \frac{\left (a^2+4 a b\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{2 f}+\frac{\left (-a^2-4 a b\right ) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{2 f}-\frac{a^2 \csc ^2\left (\frac{1}{2} (e+f x)\right )}{8 f}+\frac{a^2 \sec ^2\left (\frac{1}{2} (e+f x)\right )}{8 f}+\frac{b^2 \left (-\sin \left (\frac{1}{2} (e+f x)\right )\right )-12 a b \sin \left (\frac{1}{2} (e+f x)\right )}{6 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{12 a b \sin \left (\frac{1}{2} (e+f x)\right )+b^2 \sin \left (\frac{1}{2} (e+f x)\right )}{6 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{b^2 \sin \left (\frac{1}{2} (e+f x)\right )}{6 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3}+\frac{b^2}{12 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}+\frac{b^2}{12 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{b^2 \sin \left (\frac{1}{2} (e+f x)\right )}{6 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^3*(a + b*Tan[e + f*x]^2)^2,x]

[Out]

-(a^2*Csc[(e + f*x)/2]^2)/(8*f) + ((-a^2 - 4*a*b)*Log[Cos[(e + f*x)/2]])/(2*f) + ((a^2 + 4*a*b)*Log[Sin[(e + f
*x)/2]])/(2*f) + (a^2*Sec[(e + f*x)/2]^2)/(8*f) + b^2/(12*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2) + (b^2*Si
n[(e + f*x)/2])/(6*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3) - (b^2*Sin[(e + f*x)/2])/(6*f*(Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2])^3) + b^2/(12*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2) + (-12*a*b*Sin[(e + f*x)/2] - b^2*
Sin[(e + f*x)/2])/(6*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])) + (12*a*b*Sin[(e + f*x)/2] + b^2*Sin[(e + f*x)/2
])/(6*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

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Maple [A]  time = 0.061, size = 100, normalized size = 1.2 \begin{align*}{\frac{{b}^{2}}{3\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}}}+2\,{\frac{ab}{f\cos \left ( fx+e \right ) }}+2\,{\frac{ab\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{f}}-{\frac{{a}^{2}\csc \left ( fx+e \right ) \cot \left ( fx+e \right ) }{2\,f}}+{\frac{{a}^{2}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{2\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^2,x)

[Out]

1/3/f*b^2/cos(f*x+e)^3+2/f*a*b/cos(f*x+e)+2/f*a*b*ln(csc(f*x+e)-cot(f*x+e))-1/2/f*a^2*csc(f*x+e)*cot(f*x+e)+1/
2/f*a^2*ln(csc(f*x+e)-cot(f*x+e))

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Maxima [A]  time = 1.14151, size = 150, normalized size = 1.83 \begin{align*} -\frac{3 \,{\left (a^{2} + 4 \, a b\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) - 3 \,{\left (a^{2} + 4 \, a b\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) - \frac{2 \,{\left (3 \,{\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, b^{2}\right )}}{\cos \left (f x + e\right )^{5} - \cos \left (f x + e\right )^{3}}}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

-1/12*(3*(a^2 + 4*a*b)*log(cos(f*x + e) + 1) - 3*(a^2 + 4*a*b)*log(cos(f*x + e) - 1) - 2*(3*(a^2 + 4*a*b)*cos(
f*x + e)^4 - 2*(6*a*b - b^2)*cos(f*x + e)^2 - 2*b^2)/(cos(f*x + e)^5 - cos(f*x + e)^3))/f

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Fricas [B]  time = 2.02787, size = 414, normalized size = 5.05 \begin{align*} \frac{6 \,{\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{4} - 4 \,{\left (6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, b^{2} - 3 \,{\left ({\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{5} -{\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) + 3 \,{\left ({\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{5} -{\left (a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right )}{12 \,{\left (f \cos \left (f x + e\right )^{5} - f \cos \left (f x + e\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/12*(6*(a^2 + 4*a*b)*cos(f*x + e)^4 - 4*(6*a*b - b^2)*cos(f*x + e)^2 - 4*b^2 - 3*((a^2 + 4*a*b)*cos(f*x + e)^
5 - (a^2 + 4*a*b)*cos(f*x + e)^3)*log(1/2*cos(f*x + e) + 1/2) + 3*((a^2 + 4*a*b)*cos(f*x + e)^5 - (a^2 + 4*a*b
)*cos(f*x + e)^3)*log(-1/2*cos(f*x + e) + 1/2))/(f*cos(f*x + e)^5 - f*cos(f*x + e)^3)

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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)**3*(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.64372, size = 339, normalized size = 4.13 \begin{align*} -\frac{\frac{3 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - 6 \,{\left (a^{2} + 4 \, a b\right )} \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right ) - \frac{3 \,{\left (a^{2} - \frac{2 \, 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}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}}{\cos \left (f x + e\right ) - 1} - \frac{16 \,{\left (6 \, a b + b^{2} + \frac{12 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{6 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{3 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}^{3}}}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/24*(3*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 6*(a^2 + 4*a*b)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1
)) - 3*(a^2 - 2*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))*(cos(
f*x + e) + 1)/(cos(f*x + e) - 1) - 16*(6*a*b + b^2 + 12*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 6*a*b*(cos
(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 3*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/((cos(f*x + e) - 1)/(
cos(f*x + e) + 1) + 1)^3)/f