3.13 \(\int \csc ^4(e+f x) (a+b \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=46 \[ -\frac {(a+b) \cot ^3(e+f x)}{3 f}-\frac {(a+2 b) \cot (e+f x)}{f}+\frac {b \tan (e+f x)}{f} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 46, 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 = {4132, 448} \[ -\frac {(a+b) \cot ^3(e+f x)}{3 f}-\frac {(a+2 b) \cot (e+f x)}{f}+\frac {b \tan (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 448

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

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

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Mathematica [A]  time = 0.04, size = 84, normalized size = 1.83 \[ -\frac {2 a \cot (e+f x)}{3 f}-\frac {a \cot (e+f x) \csc ^2(e+f x)}{3 f}+\frac {b \tan (e+f x)}{f}-\frac {5 b \cot (e+f x)}{3 f}-\frac {b \cot (e+f x) \csc ^2(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*Cot[e + f*x])/(3*f) - (5*b*Cot[e + f*x])/(3*f) - (a*Cot[e + f*x]*Csc[e + f*x]^2)/(3*f) - (b*Cot[e + f*x]
*Csc[e + f*x]^2)/(3*f) + (b*Tan[e + f*x])/f

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fricas [A]  time = 0.72, size = 66, normalized size = 1.43 \[ -\frac {2 \, {\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{4} - 3 \, {\left (a + 4 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b}{3 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*(a + 4*b)*cos(f*x + e)^4 - 3*(a + 4*b)*cos(f*x + e)^2 + 3*b)/((f*cos(f*x + e)^3 - f*cos(f*x + e))*sin(
f*x + e))

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giac [A]  time = 0.30, size = 54, normalized size = 1.17 \[ \frac {3 \, b \tan \left (f x + e\right ) - \frac {3 \, a \tan \left (f x + e\right )^{2} + 6 \, b \tan \left (f x + e\right )^{2} + a + b}{\tan \left (f x + e\right )^{3}}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.12, size = 73, normalized size = 1.59 \[ \frac {a \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (f x +e \right )\right )}{3}\right ) \cot \left (f x +e \right )+b \left (-\frac {1}{3 \sin \left (f x +e \right )^{3} \cos \left (f x +e \right )}+\frac {4}{3 \sin \left (f x +e \right ) \cos \left (f x +e \right )}-\frac {8 \cot \left (f x +e \right )}{3}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(a*(-2/3-1/3*csc(f*x+e)^2)*cot(f*x+e)+b*(-1/3/sin(f*x+e)^3/cos(f*x+e)+4/3/sin(f*x+e)/cos(f*x+e)-8/3*cot(f*
x+e)))

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maxima [A]  time = 0.33, size = 43, normalized size = 0.93 \[ \frac {3 \, b \tan \left (f x + e\right ) - \frac {3 \, {\left (a + 2 \, b\right )} \tan \left (f x + e\right )^{2} + a + b}{\tan \left (f x + e\right )^{3}}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.31, size = 46, normalized size = 1.00 \[ \frac {b\,\mathrm {tan}\left (e+f\,x\right )}{f}-\frac {\left (a+2\,b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+\frac {a}{3}+\frac {b}{3}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \csc ^{4}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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