3.426 \(\int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=110 \[ \frac{\left (3 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (3 a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{2 a b \tan ^3(c+d x)}{3 d}+\frac{2 a b \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0950722, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2789, 3767, 3012, 3768, 3770} \[ \frac{\left (3 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (3 a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{2 a b \tan ^3(c+d x)}{3 d}+\frac{2 a b \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^5,x]

[Out]

((3*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (2*a*b*Tan[c + d*x])/d + ((3*a^2 + 4*b^2)*Sec[c + d*x]*Tan[c +
d*x])/(8*d) + (a^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (2*a*b*Tan[c + d*x]^3)/(3*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 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 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 (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx &=(2 a b) \int \sec ^4(c+d x) \, dx+\int \left (a^2+b^2 \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \left (3 a^2+4 b^2\right ) \int \sec ^3(c+d x) \, dx-\frac{(2 a b) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac{2 a b \tan (c+d x)}{d}+\frac{\left (3 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{2 a b \tan ^3(c+d x)}{3 d}+\frac{1}{8} \left (3 a^2+4 b^2\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (3 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{2 a b \tan (c+d x)}{d}+\frac{\left (3 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{2 a b \tan ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.275158, size = 82, normalized size = 0.75 \[ \frac{3 \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 \left (3 a^2+4 b^2\right ) \sec (c+d x)+6 a^2 \sec ^3(c+d x)+16 a b \left (\tan ^2(c+d x)+3\right )\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^5,x]

[Out]

(3*(3*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*(3*a^2 + 4*b^2)*Sec[c + d*x] + 6*a^2*Sec[c + d*x]^3
+ 16*a*b*(3 + Tan[c + d*x]^2)))/(24*d)

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Maple [A]  time = 0.062, size = 142, normalized size = 1.3 \begin{align*}{\frac{{a}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{4\,ab\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,ab\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{b}^{2}\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{2}\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((a+b*cos(d*x+c))^2*sec(d*x+c)^5,x)

[Out]

1/4*a^2*sec(d*x+c)^3*tan(d*x+c)/d+3/8/d*a^2*sec(d*x+c)*tan(d*x+c)+3/8/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+4/3*a*b*
tan(d*x+c)/d+2/3/d*a*b*tan(d*x+c)*sec(d*x+c)^2+1/2/d*b^2*tan(d*x+c)*sec(d*x+c)+1/2/d*b^2*ln(sec(d*x+c)+tan(d*x
+c))

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Maxima [A]  time = 0.984156, size = 194, normalized size = 1.76 \begin{align*} \frac{32 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b - 3 \, a^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, b^{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 )}}{48 \, 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)^5,x, algorithm="maxima")

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*a*b - 3*a^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4
- 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*b^2*(2*sin(d*x + c)/(sin(d*x
+ c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 2.10362, size = 332, normalized size = 3.02 \begin{align*} \frac{3 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (32 \, a b \cos \left (d x + c\right )^{3} + 16 \, a b \cos \left (d x + c\right ) + 3 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, a^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \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)^5,x, algorithm="fricas")

[Out]

1/48*(3*(3*a^2 + 4*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*a^2 + 4*b^2)*cos(d*x + c)^4*log(-sin(d*x +
c) + 1) + 2*(32*a*b*cos(d*x + c)^3 + 16*a*b*cos(d*x + c) + 3*(3*a^2 + 4*b^2)*cos(d*x + c)^2 + 6*a^2)*sin(d*x
+ c))/(d*cos(d*x + c)^4)

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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)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.43151, size = 348, normalized size = 3.16 \begin{align*} \frac{3 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 80 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 80 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 48 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, b^{2} \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 )}^{4}}}{24 \, 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)^5,x, algorithm="giac")

[Out]

1/24*(3*(3*a^2 + 4*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*a^2 + 4*b^2)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) + 2*(15*a^2*tan(1/2*d*x + 1/2*c)^7 - 48*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*b^2*tan(1/2*d*x + 1/2*c)^7 + 9*a^2
*tan(1/2*d*x + 1/2*c)^5 + 80*a*b*tan(1/2*d*x + 1/2*c)^5 - 12*b^2*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*tan(1/2*d*x +
1/2*c)^3 - 80*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*b^2*tan(1/2*d*x + 1/2*c)^3 + 15*a^2*tan(1/2*d*x + 1/2*c) + 48*a*
b*tan(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d