3.7 \(\int (A+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=70 \[ \frac{(3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(3 A+4 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{A \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.0468188, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3012, 3768, 3770} \[ \frac{(3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(3 A+4 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{A \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A + 4*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((3*A + 4*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (A*Sec[c + d*x]^3*
Tan[c + d*x])/(4*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 ^5(c+d x) \, dx &=\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} (3 A+4 C) \int \sec ^3(c+d x) \, dx\\ &=\frac{(3 A+4 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{8} (3 A+4 C) \int \sec (c+d x) \, dx\\ &=\frac{(3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(3 A+4 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.114558, size = 54, normalized size = 0.77 \[ \frac{(3 A+4 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x) \left (2 A \sec ^2(c+d x)+3 A+4 C\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.079, size = 98, normalized size = 1.4 \begin{align*}{\frac{A \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{C\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{2\,d}}+{\frac{C\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+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

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

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Maxima [A]  time = 1.03508, size = 131, normalized size = 1.87 \begin{align*} \frac{{\left (3 \, A + 4 \, C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, A + 4 \, C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left ({\left (3 \, A + 4 \, C\right )} \sin \left (d x + c\right )^{3} -{\left (5 \, A + 4 \, C\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, 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)^5,x, algorithm="maxima")

[Out]

1/16*((3*A + 4*C)*log(sin(d*x + c) + 1) - (3*A + 4*C)*log(sin(d*x + c) - 1) - 2*((3*A + 4*C)*sin(d*x + c)^3 -
(5*A + 4*C)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

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Fricas [A]  time = 1.66661, size = 243, normalized size = 3.47 \begin{align*} \frac{{\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left ({\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, A\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \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)^5,x, algorithm="fricas")

[Out]

1/16*((3*A + 4*C)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*A + 4*C)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2
*((3*A + 4*C)*cos(d*x + c)^2 + 2*A)*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+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [A]  time = 1.23869, size = 132, normalized size = 1.89 \begin{align*} \frac{{\left (3 \, A + 4 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (3 \, A + 4 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, A \sin \left (d x + c\right )^{3} + 4 \, C \sin \left (d x + c\right )^{3} - 5 \, A \sin \left (d x + c\right ) - 4 \, C \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, 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)^5,x, algorithm="giac")

[Out]

1/16*((3*A + 4*C)*log(abs(sin(d*x + c) + 1)) - (3*A + 4*C)*log(abs(sin(d*x + c) - 1)) - 2*(3*A*sin(d*x + c)^3
+ 4*C*sin(d*x + c)^3 - 5*A*sin(d*x + c) - 4*C*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d