3.4 \(\int \sec ^3(c+d x) (A+C \sec ^2(c+d x)) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0467736, 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 = {4046, 3768, 3770} \[ \frac{(4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2),x]

[Out]

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

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[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 \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} (4 A+3 C) \int \sec ^3(c+d x) \, dx\\ &=\frac{(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{8} (4 A+3 C) \int \sec (c+d x) \, dx\\ &=\frac{(4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{C \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.11571, size = 54, normalized size = 0.77 \[ \frac{(4 A+3 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x) \left (4 A+2 C \sec ^2(c+d x)+3 C\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2),x]

[Out]

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

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Maple [A]  time = 0.028, size = 98, normalized size = 1.4 \begin{align*}{\frac{A\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{2\,d}}+{\frac{A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{C \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,C\sec \left ( dx+c \right ) \tan \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(sec(d*x+c)^3*(A+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.929826, size = 131, normalized size = 1.87 \begin{align*} \frac{{\left (4 \, A + 3 \, C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (4 \, A + 3 \, C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left ({\left (4 \, A + 3 \, C\right )} \sin \left (d x + c\right )^{3} -{\left (4 \, A + 5 \, 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(sec(d*x+c)^3*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/16*((4*A + 3*C)*log(sin(d*x + c) + 1) - (4*A + 3*C)*log(sin(d*x + c) - 1) - 2*((4*A + 3*C)*sin(d*x + c)^3 -
(4*A + 5*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 = 0.504777, size = 243, normalized size = 3.47 \begin{align*} \frac{{\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left ({\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, C\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(sec(d*x+c)^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3, x)

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Giac [A]  time = 1.25076, size = 132, normalized size = 1.89 \begin{align*} \frac{{\left (4 \, A + 3 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (4 \, A + 3 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (4 \, A \sin \left (d x + c\right )^{3} + 3 \, C \sin \left (d x + c\right )^{3} - 4 \, A \sin \left (d x + c\right ) - 5 \, 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(sec(d*x+c)^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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