3.432 \(\int \sec ^4(c+d x) (a+b \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=46 \[ \frac{(a+b) \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{b \tan ^5(c+d x)}{5 d} \]

[Out]

(a*Tan[c + d*x])/d + ((a + b)*Tan[c + d*x]^3)/(3*d) + (b*Tan[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.0405622, antiderivative size = 46, normalized size of antiderivative = 1., 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 = {3675, 373} \[ \frac{(a+b) \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{b \tan ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Tan[c + d*x])/d + ((a + b)*Tan[c + d*x]^3)/(3*d) + (b*Tan[c + d*x]^5)/(5*d)

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 373

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

Rubi steps

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

Mathematica [A]  time = 0.138067, size = 53, normalized size = 1.15 \[ \frac{\tan (c+d x) \left (5 a \tan ^2(c+d x)+15 a+3 b \sec ^4(c+d x)-b \sec ^2(c+d x)-2 b\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Tan[c + d*x]*(15*a - 2*b - b*Sec[c + d*x]^2 + 3*b*Sec[c + d*x]^4 + 5*a*Tan[c + d*x]^2))/(15*d)

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Maple [A]  time = 0.043, size = 66, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( b \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) -a \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+b*tan(d*x+c)^2),x)

[Out]

1/d*(b*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)-a*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.00245, size = 53, normalized size = 1.15 \begin{align*} \frac{3 \, b \tan \left (d x + c\right )^{5} + 5 \,{\left (a + b\right )} \tan \left (d x + c\right )^{3} + 15 \, a \tan \left (d x + c\right )}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b*tan(d*x + c)^5 + 5*(a + b)*tan(d*x + c)^3 + 15*a*tan(d*x + c))/d

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Fricas [A]  time = 1.31229, size = 135, normalized size = 2.93 \begin{align*} \frac{{\left (2 \,{\left (5 \, a - b\right )} \cos \left (d x + c\right )^{4} +{\left (5 \, a - b\right )} \cos \left (d x + c\right )^{2} + 3 \, b\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(2*(5*a - b)*cos(d*x + c)^4 + (5*a - b)*cos(d*x + c)^2 + 3*b)*sin(d*x + c)/(d*cos(d*x + c)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+b*tan(d*x+c)**2),x)

[Out]

Integral((a + b*tan(c + d*x)**2)*sec(c + d*x)**4, x)

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Giac [A]  time = 1.53058, size = 65, normalized size = 1.41 \begin{align*} \frac{3 \, b \tan \left (d x + c\right )^{5} + 5 \, a \tan \left (d x + c\right )^{3} + 5 \, b \tan \left (d x + c\right )^{3} + 15 \, a \tan \left (d x + c\right )}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b*tan(d*x + c)^5 + 5*a*tan(d*x + c)^3 + 5*b*tan(d*x + c)^3 + 15*a*tan(d*x + c))/d