3.287 \(\int \frac {(a B+b B \cos (c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=36 \[ \frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

1/2*B*arctanh(sin(d*x+c))/d+1/2*B*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {21, 3768, 3770} \[ \frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*ArcTanh[Sin[c + d*x]])/(2*d) + (B*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 36, normalized size = 1.00 \[ B \left (\frac {\tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

B*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d))

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fricas [A]  time = 0.65, size = 64, normalized size = 1.78 \[ \frac {B \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - B \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(B*cos(d*x + c)^2*log(sin(d*x + c) + 1) - B*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*B*sin(d*x + c))/(d*c
os(d*x + c)^2)

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giac [A]  time = 0.59, size = 52, normalized size = 1.44 \[ \frac {B \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - B \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, B \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(B*log(abs(sin(d*x + c) + 1)) - B*log(abs(sin(d*x + c) - 1)) - 2*B*sin(d*x + c)/(sin(d*x + c)^2 - 1))/d

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maple [A]  time = 0.09, size = 40, normalized size = 1.11 \[ \frac {B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*sec(d*x+c)*tan(d*x+c)/d+1/2/d*B*ln(sec(d*x+c)+tan(d*x+c))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 0.86, size = 73, normalized size = 2.03 \[ \frac {B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*tan(c/2 + (d*x)/2) + B*tan(c/2 + (d*x)/2)^3)/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) + (B*a
tanh(tan(c/2 + (d*x)/2)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*Integral(sec(c + d*x)**3, x)

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