3.1.7 \(\int (A+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\) [7]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 70, normalized size of antiderivative = 1.00, 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 = {3091, 3853, 3855} \begin {gather*} \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} \end {gather*}

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 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 3853

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 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 ^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*}

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Mathematica [A]
time = 0.15, size = 54, normalized size = 0.77 \begin {gather*} \frac {(3 A+4 C) \tanh ^{-1}(\sin (c+d x))+\sec (c+d x) \left (3 A+4 C+2 A \sec ^2(c+d x)\right ) \tan (c+d x)}{8 d} \end {gather*}

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 = 0.25, size = 85, normalized size = 1.21

method	result	size
derivativedivides	\(\frac {A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(85\)
default	\(\frac {A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(85\)
risch	\(-\frac {i \left (3 A \,{\mathrm e}^{7 i \left (d x +c \right )}+4 C \,{\mathrm e}^{7 i \left (d x +c \right )}+11 A \,{\mathrm e}^{5 i \left (d x +c \right )}+4 C \,{\mathrm e}^{5 i \left (d x +c \right )}-11 A \,{\mathrm e}^{3 i \left (d x +c \right )}-4 C \,{\mathrm e}^{3 i \left (d x +c \right )}-3 A \,{\mathrm e}^{i \left (d x +c \right )}-4 C \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\)	\(194\)
norman	\(\frac {\frac {\left (5 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (5 A +4 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (7 A -4 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (7 A -4 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (13 A +4 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (13 A +4 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (3 A +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 A +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(218\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 97, normalized size = 1.39 \begin {gather*} \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 {gather*}

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 = 0.36, size = 95, normalized size = 1.36 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{5}{\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)**5,x)

[Out]

Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)**5, x)

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Giac [A]
time = 0.43, size = 98, normalized size = 1.40 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.74, size = 77, normalized size = 1.10 \begin {gather*} \frac {\sin \left (c+d\,x\right )\,\left (\frac {5\,A}{8}+\frac {C}{2}\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {3\,A}{8}+\frac {C}{2}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {3\,A}{8}+\frac {C}{2}\right )}{d} \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)^5,x)

[Out]

(sin(c + d*x)*((5*A)/8 + C/2) - sin(c + d*x)^3*((3*A)/8 + C/2))/(d*(sin(c + d*x)^4 - 2*sin(c + d*x)^2 + 1)) +
(atanh(sin(c + d*x))*((3*A)/8 + C/2))/d

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