3.39 \(\int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=52 \[ \frac{a \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a \tan (c+d x) \sec (c+d x)}{2 d}+\frac{b \sec ^3(c+d x)}{3 d} \]

[Out]

(a*ArcTanh[Sin[c + d*x]])/(2*d) + (b*Sec[c + d*x]^3)/(3*d) + (a*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.0666957, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3090, 3768, 3770, 2606, 30} \[ \frac{a \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a \tan (c+d x) \sec (c+d x)}{2 d}+\frac{b \sec ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTanh[Sin[c + d*x]])/(2*d) + (b*Sec[c + d*x]^3)/(3*d) + (a*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0165684, size = 52, normalized size = 1. \[ \frac{a \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a \tan (c+d x) \sec (c+d x)}{2 d}+\frac{b \sec ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTanh[Sin[c + d*x]])/(2*d) + (b*Sec[c + d*x]^3)/(3*d) + (a*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Maple [A]  time = 0.075, size = 54, normalized size = 1. \begin{align*}{\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}}+{\frac{b}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

1/2*a*sec(d*x+c)*tan(d*x+c)/d+1/2/d*a*ln(sec(d*x+c)+tan(d*x+c))+1/3/d*b/cos(d*x+c)^3

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Maxima [A]  time = 1.09578, size = 82, normalized size = 1.58 \begin{align*} -\frac{3 \, a{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac{4 \, b}{\cos \left (d x + c\right )^{3}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*b/cos(d*x
+ c)^3)/d

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Fricas [A]  time = 0.497882, size = 203, normalized size = 3.9 \begin{align*} \frac{3 \, a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, b}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*a*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 6*a*cos(d*x + c)*
sin(d*x + c) + 4*b)/(d*cos(d*x + c)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.19135, size = 134, normalized size = 2.58 \begin{align*} \frac{3 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(3*a*tan(1/2*d*x + 1/
2*c)^5 - 6*b*tan(1/2*d*x + 1/2*c)^4 - 3*a*tan(1/2*d*x + 1/2*c) - 2*b)/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d