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

Optimal. Leaf size=48 \[ \frac{a \tan ^3(e+f x)}{3 f}-\frac{a \tan (e+f x)}{f}+a x+\frac{b \tan ^5(e+f x)}{5 f} \]

[Out]

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

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Rubi [A]  time = 0.0565195, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4141, 1802, 203} \[ \frac{a \tan ^3(e+f x)}{3 f}-\frac{a \tan (e+f x)}{f}+a x+\frac{b \tan ^5(e+f x)}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.0206226, size = 57, normalized size = 1.19 \[ \frac{a \tan ^3(e+f x)}{3 f}+\frac{a \tan ^{-1}(\tan (e+f x))}{f}-\frac{a \tan (e+f x)}{f}+\frac{b \tan ^5(e+f x)}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.048, size = 50, normalized size = 1. \begin{align*}{\frac{1}{f} \left ( a \left ({\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{3}}-\tan \left ( fx+e \right ) +fx+e \right ) +{\frac{b \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{5\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(a*(1/3*tan(f*x+e)^3-tan(f*x+e)+f*x+e)+1/5*b*sin(f*x+e)^5/cos(f*x+e)^5)

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Maxima [A]  time = 1.47521, size = 61, normalized size = 1.27 \begin{align*} \frac{3 \, b \tan \left (f x + e\right )^{5} + 5 \, a \tan \left (f x + e\right )^{3} + 15 \,{\left (f x + e\right )} a - 15 \, a \tan \left (f x + e\right )}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b*tan(f*x + e)^5 + 5*a*tan(f*x + e)^3 + 15*(f*x + e)*a - 15*a*tan(f*x + e))/f

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Fricas [A]  time = 0.495778, size = 177, normalized size = 3.69 \begin{align*} \frac{15 \, a f x \cos \left (f x + e\right )^{5} -{\left ({\left (20 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{4} -{\left (5 \, a - 6 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, b\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*a*f*x*cos(f*x + e)^5 - ((20*a - 3*b)*cos(f*x + e)^4 - (5*a - 6*b)*cos(f*x + e)^2 - 3*b)*sin(f*x + e))
/(f*cos(f*x + e)^5)

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Sympy [A]  time = 2.9966, size = 54, normalized size = 1.12 \begin{align*} a \left (\begin{cases} x + \frac{\tan ^{3}{\left (e + f x \right )}}{3 f} - \frac{\tan{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \tan ^{4}{\left (e \right )} & \text{otherwise} \end{cases}\right ) + b \left (\begin{cases} x \tan ^{4}{\left (e \right )} \sec ^{2}{\left (e \right )} & \text{for}\: f = 0 \\\frac{\tan ^{5}{\left (e + f x \right )}}{5 f} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((x + tan(e + f*x)**3/(3*f) - tan(e + f*x)/f, Ne(f, 0)), (x*tan(e)**4, True)) + b*Piecewise((x*tan(
e)**4*sec(e)**2, Eq(f, 0)), (tan(e + f*x)**5/(5*f), True))

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Giac [A]  time = 2.26045, size = 66, normalized size = 1.38 \begin{align*} \frac{3 \, b \tan \left (f x + e\right )^{5} + 5 \, a \tan \left (f x + e\right )^{3} + 15 \,{\left (f x + e\right )} a - 15 \, a \tan \left (f x + e\right )}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b*tan(f*x + e)^5 + 5*a*tan(f*x + e)^3 + 15*(f*x + e)*a - 15*a*tan(f*x + e))/f