3.8 \(\int (A+C \cos ^2(c+d x)) \sec ^7(c+d x) \, dx\)

Optimal. Leaf size=98 \[ \frac{(5 A+6 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{(5 A+6 C) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{(5 A+6 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{A \tan (c+d x) \sec ^5(c+d x)}{6 d} \]

[Out]

((5*A + 6*C)*ArcTanh[Sin[c + d*x]])/(16*d) + ((5*A + 6*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + ((5*A + 6*C)*Sec
[c + d*x]^3*Tan[c + d*x])/(24*d) + (A*Sec[c + d*x]^5*Tan[c + d*x])/(6*d)

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Rubi [A]  time = 0.0601627, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3012, 3768, 3770} \[ \frac{(5 A+6 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{(5 A+6 C) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{(5 A+6 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{A \tan (c+d x) \sec ^5(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

((5*A + 6*C)*ArcTanh[Sin[c + d*x]])/(16*d) + ((5*A + 6*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + ((5*A + 6*C)*Sec
[c + d*x]^3*Tan[c + d*x])/(24*d) + (A*Sec[c + d*x]^5*Tan[c + d*x])/(6*d)

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 \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\frac{A \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{6} (5 A+6 C) \int \sec ^5(c+d x) \, dx\\ &=\frac{(5 A+6 C) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{A \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{8} (5 A+6 C) \int \sec ^3(c+d x) \, dx\\ &=\frac{(5 A+6 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{(5 A+6 C) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{A \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{16} (5 A+6 C) \int \sec (c+d x) \, dx\\ &=\frac{(5 A+6 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{(5 A+6 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{(5 A+6 C) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{A \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.300422, size = 75, normalized size = 0.77 \[ \frac{3 (5 A+6 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x) \left (2 (5 A+6 C) \sec ^2(c+d x)+8 A \sec ^4(c+d x)+3 (5 A+6 C)\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

(3*(5*A + 6*C)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(3*(5*A + 6*C) + 2*(5*A + 6*C)*Sec[c + d*x]^2 + 8*A*Sec[c
+ d*x]^4)*Tan[c + d*x])/(48*d)

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Maple [A]  time = 0.067, size = 138, normalized size = 1.4 \begin{align*}{\frac{A \left ( \sec \left ( dx+c \right ) \right ) ^{5}\tan \left ( dx+c \right ) }{6\,d}}+{\frac{5\,A \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{24\,d}}+{\frac{5\,A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{5\,A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,C\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{8\,d}}+{\frac{3\,C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x)

[Out]

1/6*A*sec(d*x+c)^5*tan(d*x+c)/d+5/24*A*sec(d*x+c)^3*tan(d*x+c)/d+5/16*A*sec(d*x+c)*tan(d*x+c)/d+5/16/d*A*ln(se
c(d*x+c)+tan(d*x+c))+1/4/d*C*tan(d*x+c)*sec(d*x+c)^3+3/8/d*C*tan(d*x+c)*sec(d*x+c)+3/8/d*C*ln(sec(d*x+c)+tan(d
*x+c))

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Maxima [A]  time = 1.06679, size = 170, normalized size = 1.73 \begin{align*} \frac{3 \,{\left (5 \, A + 6 \, C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (5 \, A + 6 \, C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (3 \,{\left (5 \, A + 6 \, C\right )} \sin \left (d x + c\right )^{5} - 8 \,{\left (5 \, A + 6 \, C\right )} \sin \left (d x + c\right )^{3} + 3 \,{\left (11 \, A + 10 \, C\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="maxima")

[Out]

1/96*(3*(5*A + 6*C)*log(sin(d*x + c) + 1) - 3*(5*A + 6*C)*log(sin(d*x + c) - 1) - 2*(3*(5*A + 6*C)*sin(d*x + c
)^5 - 8*(5*A + 6*C)*sin(d*x + c)^3 + 3*(11*A + 10*C)*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(
d*x + c)^2 - 1))/d

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Fricas [A]  time = 1.73048, size = 293, normalized size = 2.99 \begin{align*} \frac{3 \,{\left (5 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (5 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \,{\left (5 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (5 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, A\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="fricas")

[Out]

1/96*(3*(5*A + 6*C)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(5*A + 6*C)*cos(d*x + c)^6*log(-sin(d*x + c) + 1)
+ 2*(3*(5*A + 6*C)*cos(d*x + c)^4 + 2*(5*A + 6*C)*cos(d*x + c)^2 + 8*A)*sin(d*x + c))/(d*cos(d*x + c)^6)

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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+C*cos(d*x+c)**2)*sec(d*x+c)**7,x)

[Out]

Timed out

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Giac [A]  time = 1.20344, size = 163, normalized size = 1.66 \begin{align*} \frac{3 \,{\left (5 \, A + 6 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \,{\left (5 \, A + 6 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (15 \, A \sin \left (d x + c\right )^{5} + 18 \, C \sin \left (d x + c\right )^{5} - 40 \, A \sin \left (d x + c\right )^{3} - 48 \, C \sin \left (d x + c\right )^{3} + 33 \, A \sin \left (d x + c\right ) + 30 \, C \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="giac")

[Out]

1/96*(3*(5*A + 6*C)*log(abs(sin(d*x + c) + 1)) - 3*(5*A + 6*C)*log(abs(sin(d*x + c) - 1)) - 2*(15*A*sin(d*x +
c)^5 + 18*C*sin(d*x + c)^5 - 40*A*sin(d*x + c)^3 - 48*C*sin(d*x + c)^3 + 33*A*sin(d*x + c) + 30*C*sin(d*x + c)
)/(sin(d*x + c)^2 - 1)^3)/d