3.1.8 \(\int (A+C \cos ^2(c+d x)) \sec ^7(c+d x) \, dx\) [8]

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

[Out]

1/16*(5*A+6*C)*arctanh(sin(d*x+c))/d+1/16*(5*A+6*C)*sec(d*x+c)*tan(d*x+c)/d+1/24*(5*A+6*C)*sec(d*x+c)^3*tan(d*
x+c)/d+1/6*A*sec(d*x+c)^5*tan(d*x+c)/d

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Rubi [A]
time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 = {3091, 3853, 3855} \begin {gather*} \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} \end {gather*}

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 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 ^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*}

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Mathematica [A]
time = 0.36, size = 75, normalized size = 0.77 \begin {gather*} \frac {3 (5 A+6 C) \tanh ^{-1}(\sin (c+d x))+\sec (c+d x) \left (3 (5 A+6 C)+2 (5 A+6 C) \sec ^2(c+d x)+8 A \sec ^4(c+d x)\right ) \tan (c+d x)}{48 d} \end {gather*}

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 = 0.30, size = 108, normalized size = 1.10

method	result	size
derivativedivides	\(\frac {A \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(108\)
default	\(\frac {A \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(108\)
risch	\(-\frac {i \left (15 A \,{\mathrm e}^{11 i \left (d x +c \right )}+18 C \,{\mathrm e}^{11 i \left (d x +c \right )}+85 A \,{\mathrm e}^{9 i \left (d x +c \right )}+102 C \,{\mathrm e}^{9 i \left (d x +c \right )}+198 A \,{\mathrm e}^{7 i \left (d x +c \right )}+84 C \,{\mathrm e}^{7 i \left (d x +c \right )}-198 A \,{\mathrm e}^{5 i \left (d x +c \right )}-84 C \,{\mathrm e}^{5 i \left (d x +c \right )}-85 A \,{\mathrm e}^{3 i \left (d x +c \right )}-102 C \,{\mathrm e}^{3 i \left (d x +c \right )}-15 A \,{\mathrm e}^{i \left (d x +c \right )}-18 C \,{\mathrm e}^{i \left (d x +c \right )}\right )}{24 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}-\frac {5 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{16 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {5 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{16 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(242\)
norman	\(\frac {\frac {\left (11 A +10 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (11 A +10 C \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {7 \left (19 A -6 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {7 \left (19 A -6 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (71 A +18 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (71 A +18 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (275 A -6 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (275 A -6 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6}}-\frac {\left (5 A +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d}+\frac {\left (5 A +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d}\)	\(264\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 126, normalized size = 1.29 \begin {gather*} \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 {gather*}

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 = 0.38, size = 114, normalized size = 1.16 \begin {gather*} \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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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)**7,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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Giac [A]
time = 0.44, size = 121, normalized size = 1.23 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.77, size = 102, normalized size = 1.04 \begin {gather*} \frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {5\,A}{16}+\frac {3\,C}{8}\right )}{d}-\frac {\left (\frac {5\,A}{16}+\frac {3\,C}{8}\right )\,{\sin \left (c+d\,x\right )}^5+\left (-\frac {5\,A}{6}-C\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {11\,A}{16}+\frac {5\,C}{8}\right )\,\sin \left (c+d\,x\right )}{d\,\left ({\sin \left (c+d\,x\right )}^6-3\,{\sin \left (c+d\,x\right )}^4+3\,{\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)^7,x)

[Out]

(atanh(sin(c + d*x))*((5*A)/16 + (3*C)/8))/d - (sin(c + d*x)*((11*A)/16 + (5*C)/8) - sin(c + d*x)^3*((5*A)/6 +
C) + sin(c + d*x)^5*((5*A)/16 + (3*C)/8))/(d*(3*sin(c + d*x)^2 - 3*sin(c + d*x)^4 + sin(c + d*x)^6 - 1))

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