3.38 \(\int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=28 \[ \frac {a \tan (c+d x)}{d}+\frac {b \sec ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 28, normalized size of antiderivative = 1.00, 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, 3767, 8, 2606, 30} \[ \frac {a \tan (c+d x)}{d}+\frac {b \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

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 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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.00 \[ \frac {a \tan (c+d x)}{d}+\frac {b \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.26, size = 30, normalized size = 1.07 \[ \frac {2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b}{2 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*cos(d*x + c)*sin(d*x + c) + b)/(d*cos(d*x + c)^2)

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giac [A]  time = 0.26, size = 25, normalized size = 0.89 \[ \frac {b \tan \left (d x + c\right )^{2} + 2 \, a \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*tan(d*x + c)^2 + 2*a*tan(d*x + c))/d

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maple [A]  time = 1.70, size = 25, normalized size = 0.89 \[ \frac {a \tan \left (d x +c \right )+\frac {b}{2 \cos \left (d x +c \right )^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.32, size = 30, normalized size = 1.07 \[ \frac {2 \, a \tan \left (d x + c\right ) - \frac {b}{\sin \left (d x + c\right )^{2} - 1}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*tan(d*x + c) - b/(sin(d*x + c)^2 - 1))/d

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mupad [B]  time = 0.40, size = 23, normalized size = 0.82 \[ \frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))/cos(c + d*x)^3,x)

[Out]

(tan(c + d*x)*(2*a + b*tan(c + d*x)))/(2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Integral((a*cos(c + d*x) + b*sin(c + d*x))*sec(c + d*x)**3, x)

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