3.425 \(\int \sec ^3(c+d x) (a+b \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=70 \[ \frac{(4 a-b) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(4 a-b) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[Out]

((4*a - b)*ArcTanh[Sin[c + d*x]])/(8*d) + ((4*a - b)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*Sec[c + d*x]^3*Tan[
c + d*x])/(4*d)

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Rubi [A]  time = 0.0513948, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3676, 385, 199, 206} \[ \frac{(4 a-b) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(4 a-b) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + b*Tan[c + d*x]^2),x]

[Out]

((4*a - b)*ArcTanh[Sin[c + d*x]])/(8*d) + ((4*a - b)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*Sec[c + d*x]^3*Tan[
c + d*x])/(4*d)

Rule 3676

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

Rule 385

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

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

Rubi steps

\begin{align*} \int \sec ^3(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a-(a-b) x^2}{\left (1-x^2\right )^3} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{(4 a-b) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{4 d}\\ &=\frac{(4 a-b) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{(4 a-b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{8 d}\\ &=\frac{(4 a-b) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(4 a-b) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.0670364, size = 93, normalized size = 1.33 \[ \frac{a \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a \tan (c+d x) \sec (c+d x)}{2 d}-\frac{b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac{b \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + b*Tan[c + d*x]^2),x]

[Out]

(a*ArcTanh[Sin[c + d*x]])/(2*d) - (b*ArcTanh[Sin[c + d*x]])/(8*d) + (a*Sec[c + d*x]*Tan[c + d*x])/(2*d) - (b*S
ec[c + d*x]*Tan[c + d*x])/(8*d) + (b*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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Maple [A]  time = 0.088, size = 116, normalized size = 1.7 \begin{align*}{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{\sin \left ( dx+c \right ) b}{8\,d}}-{\frac{b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{a\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*tan(d*x+c)^2),x)

[Out]

1/4/d*b*sin(d*x+c)^3/cos(d*x+c)^4+1/8/d*b*sin(d*x+c)^3/cos(d*x+c)^2+1/8/d*sin(d*x+c)*b-1/8/d*b*ln(sec(d*x+c)+t
an(d*x+c))+1/2/d*a*sec(d*x+c)*tan(d*x+c)+1/2/d*a*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.10627, size = 128, normalized size = 1.83 \begin{align*} \frac{{\left (4 \, a - b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (4 \, a - b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left ({\left (4 \, a - b\right )} \sin \left (d x + c\right )^{3} -{\left (4 \, a + b\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

1/16*((4*a - b)*log(sin(d*x + c) + 1) - (4*a - b)*log(sin(d*x + c) - 1) - 2*((4*a - b)*sin(d*x + c)^3 - (4*a +
b)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

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Fricas [A]  time = 1.56155, size = 235, normalized size = 3.36 \begin{align*} \frac{{\left (4 \, a - b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (4 \, a - b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left ({\left (4 \, a - b\right )} \cos \left (d x + c\right )^{2} + 2 \, b\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

1/16*((4*a - b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (4*a - b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*((4
*a - b)*cos(d*x + c)^2 + 2*b)*sin(d*x + c))/(d*cos(d*x + c)^4)

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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 ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+b*tan(d*x+c)**2),x)

[Out]

Integral((a + b*tan(c + d*x)**2)*sec(c + d*x)**3, x)

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Giac [A]  time = 1.70409, size = 132, normalized size = 1.89 \begin{align*} \frac{{\left (4 \, a - b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (4 \, a - b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (4 \, a \sin \left (d x + c\right )^{3} - b \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)^2),x, algorithm="giac")

[Out]

1/16*((4*a - b)*log(abs(sin(d*x + c) + 1)) - (4*a - b)*log(abs(sin(d*x + c) - 1)) - 2*(4*a*sin(d*x + c)^3 - b*
sin(d*x + c)^3 - 4*a*sin(d*x + c) - b*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d