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

Optimal. Leaf size=24 \[ \frac {b \tan (e+f x)}{f}-\frac {a \cot (e+f x)}{f} \]

[Out]

-a*cot(f*x+e)/f+b*tan(f*x+e)/f

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Rubi [A]  time = 0.03, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3663, 14} \[ \frac {b \tan (e+f x)}{f}-\frac {a \cot (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Cot[e + f*x])/f) + (b*Tan[e + f*x])/f

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 3663

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 ^2(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b x^2}{x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (b+\frac {a}{x^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {a \cot (e+f x)}{f}+\frac {b \tan (e+f x)}{f}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.00 \[ \frac {b \tan (e+f x)}{f}-\frac {a \cot (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Cot[e + f*x])/f) + (b*Tan[e + f*x])/f

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fricas [A]  time = 0.44, size = 37, normalized size = 1.54 \[ -\frac {{\left (a + b\right )} \cos \left (f x + e\right )^{2} - b}{f \cos \left (f x + e\right ) \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a + b)*cos(f*x + e)^2 - b)/(f*cos(f*x + e)*sin(f*x + e))

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giac [A]  time = 1.36, size = 26, normalized size = 1.08 \[ \frac {b \tan \left (f x + e\right ) - \frac {a}{\tan \left (f x + e\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*tan(f*x + e) - a/tan(f*x + e))/f

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maple [A]  time = 0.57, size = 23, normalized size = 0.96 \[ \frac {-a \cot \left (f x +e \right )+b \tan \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-a*cot(f*x+e)+b*tan(f*x+e))

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maxima [A]  time = 0.57, size = 24, normalized size = 1.00 \[ \frac {b \tan \left (f x + e\right ) - \frac {a}{\tan \left (f x + e\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*tan(f*x + e) - a/tan(f*x + e))/f

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mupad [B]  time = 11.27, size = 26, normalized size = 1.08 \[ \frac {b\,\mathrm {tan}\left (e+f\,x\right )}{f}-\frac {a}{f\,\mathrm {tan}\left (e+f\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*tan(e + f*x))/f - a/(f*tan(e + f*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right ) \csc ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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