3.1.6 \(\int (A+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\) [6]

Optimal. Leaf size=40 \[ \frac {(A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

1/2*(A+2*C)*arctanh(sin(d*x+c))/d+1/2*A*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]
time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3091, 3855} \begin {gather*} \frac {(A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A \tan (c+d x) \sec (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3091

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 3855

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 ^3(c+d x) \, dx &=\frac {A \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} (A+2 C) \int \sec (c+d x) \, dx\\ &=\frac {(A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 48, normalized size = 1.20 \begin {gather*} \frac {A \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.17, size = 55, normalized size = 1.38

method	result	size
derivativedivides	\(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(55\)
default	\(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(55\)
risch	\(-\frac {i A \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\)	\(118\)
norman	\(\frac {\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {A \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {\left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+C*ln(sec(d*x+c)+tan(d*x+c)))

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Maxima [A]
time = 0.29, size = 58, normalized size = 1.45 \begin {gather*} \frac {{\left (A + 2 \, C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A + 2 \, C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, A \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 72, normalized size = 1.80 \begin {gather*} \frac {{\left (A + 2 \, C\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A + 2 \, C\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, A \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.45, size = 60, normalized size = 1.50 \begin {gather*} \frac {{\left (A + 2 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (A + 2 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, A \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.10, size = 41, normalized size = 1.02 \begin {gather*} \frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {A}{2}+C\right )}{d}-\frac {A\,\sin \left (c+d\,x\right )}{2\,d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)/cos(c + d*x)^3,x)

[Out]

(atanh(sin(c + d*x))*(A/2 + C))/d - (A*sin(c + d*x))/(2*d*(sin(c + d*x)^2 - 1))

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