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

Optimal. Leaf size=59 \[ \frac{\left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 \tan (c+d x) \sec (c+d x)}{2 d}+\frac{2 a b \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0784044, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2789, 3767, 8, 3012, 3770} \[ \frac{\left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 \tan (c+d x) \sec (c+d x)}{2 d}+\frac{2 a b \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

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]

Rule 8

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

Rule 3012

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*Cos[e
+ f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

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

Mathematica [A]  time = 0.0132483, size = 67, normalized size = 1.14 \[ \frac{a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 \tan (c+d x) \sec (c+d x)}{2 d}+\frac{2 a b \tan (c+d x)}{d}+\frac{b^2 \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.059, size = 78, normalized size = 1.3 \begin{align*}{\frac{{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{ab\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.958304, size = 117, normalized size = 1.98 \begin{align*} -\frac{a^{2}{\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 )} - 2 \, b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 8 \, a b \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 2*b^2*(log(s
in(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 8*a*b*tan(d*x + c))/d

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Fricas [A]  time = 1.99757, size = 236, normalized size = 4. \begin{align*} \frac{{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (4 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.4239, size = 171, normalized size = 2.9 \begin{align*} \frac{{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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