3.15 \(\int (A+C \cos ^2(c+d x)) \sec ^8(c+d x) \, dx\)

Optimal. Leaf size=87 \[ \frac {(6 A+7 C) \tan ^5(c+d x)}{35 d}+\frac {2 (6 A+7 C) \tan ^3(c+d x)}{21 d}+\frac {(6 A+7 C) \tan (c+d x)}{7 d}+\frac {A \tan (c+d x) \sec ^6(c+d x)}{7 d} \]

[Out]

1/7*(6*A+7*C)*tan(d*x+c)/d+1/7*A*sec(d*x+c)^6*tan(d*x+c)/d+2/21*(6*A+7*C)*tan(d*x+c)^3/d+1/35*(6*A+7*C)*tan(d*
x+c)^5/d

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Rubi [A]  time = 0.05, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3012, 3767} \[ \frac {(6 A+7 C) \tan ^5(c+d x)}{35 d}+\frac {2 (6 A+7 C) \tan ^3(c+d x)}{21 d}+\frac {(6 A+7 C) \tan (c+d x)}{7 d}+\frac {A \tan (c+d x) \sec ^6(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^8,x]

[Out]

((6*A + 7*C)*Tan[c + d*x])/(7*d) + (A*Sec[c + d*x]^6*Tan[c + d*x])/(7*d) + (2*(6*A + 7*C)*Tan[c + d*x]^3)/(21*
d) + ((6*A + 7*C)*Tan[c + d*x]^5)/(35*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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \left (A+C \cos ^2(c+d x)\right ) \sec ^8(c+d x) \, dx &=\frac {A \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {1}{7} (6 A+7 C) \int \sec ^6(c+d x) \, dx\\ &=\frac {A \sec ^6(c+d x) \tan (c+d x)}{7 d}-\frac {(6 A+7 C) \operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{7 d}\\ &=\frac {(6 A+7 C) \tan (c+d x)}{7 d}+\frac {A \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {2 (6 A+7 C) \tan ^3(c+d x)}{21 d}+\frac {(6 A+7 C) \tan ^5(c+d x)}{35 d}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 81, normalized size = 0.93 \[ \frac {A \left (\frac {1}{7} \tan ^7(c+d x)+\frac {3}{5} \tan ^5(c+d x)+\tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac {C \left (\frac {1}{5} \tan ^5(c+d x)+\frac {2}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^8,x]

[Out]

(C*(Tan[c + d*x] + (2*Tan[c + d*x]^3)/3 + Tan[c + d*x]^5/5))/d + (A*(Tan[c + d*x] + Tan[c + d*x]^3 + (3*Tan[c
+ d*x]^5)/5 + Tan[c + d*x]^7/7))/d

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fricas [A]  time = 0.41, size = 74, normalized size = 0.85 \[ \frac {{\left (8 \, {\left (6 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (6 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 15 \, A\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x, algorithm="fricas")

[Out]

1/105*(8*(6*A + 7*C)*cos(d*x + c)^6 + 4*(6*A + 7*C)*cos(d*x + c)^4 + 3*(6*A + 7*C)*cos(d*x + c)^2 + 15*A)*sin(
d*x + c)/(d*cos(d*x + c)^7)

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giac [A]  time = 0.21, size = 79, normalized size = 0.91 \[ \frac {15 \, A \tan \left (d x + c\right )^{7} + 63 \, A \tan \left (d x + c\right )^{5} + 21 \, C \tan \left (d x + c\right )^{5} + 105 \, A \tan \left (d x + c\right )^{3} + 70 \, C \tan \left (d x + c\right )^{3} + 105 \, A \tan \left (d x + c\right ) + 105 \, C \tan \left (d x + c\right )}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x, algorithm="giac")

[Out]

1/105*(15*A*tan(d*x + c)^7 + 63*A*tan(d*x + c)^5 + 21*C*tan(d*x + c)^5 + 105*A*tan(d*x + c)^3 + 70*C*tan(d*x +
c)^3 + 105*A*tan(d*x + c) + 105*C*tan(d*x + c))/d

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maple [A]  time = 0.16, size = 78, normalized size = 0.90 \[ \frac {-A \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )-C \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x)

[Out]

1/d*(-A*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c)^4-8/35*sec(d*x+c)^2)*tan(d*x+c)-C*(-8/15-1/5*sec(d*x+c)^4-4/1
5*sec(d*x+c)^2)*tan(d*x+c))

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maxima [A]  time = 0.31, size = 60, normalized size = 0.69 \[ \frac {15 \, A \tan \left (d x + c\right )^{7} + 21 \, {\left (3 \, A + C\right )} \tan \left (d x + c\right )^{5} + 35 \, {\left (3 \, A + 2 \, C\right )} \tan \left (d x + c\right )^{3} + 105 \, {\left (A + C\right )} \tan \left (d x + c\right )}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x, algorithm="maxima")

[Out]

1/105*(15*A*tan(d*x + c)^7 + 21*(3*A + C)*tan(d*x + c)^5 + 35*(3*A + 2*C)*tan(d*x + c)^3 + 105*(A + C)*tan(d*x
+ c))/d

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mupad [B]  time = 0.66, size = 56, normalized size = 0.64 \[ \frac {\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^7}{7}+\left (\frac {3\,A}{5}+\frac {C}{5}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^5+\left (A+\frac {2\,C}{3}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (A+C\right )\,\mathrm {tan}\left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)/cos(c + d*x)^8,x)

[Out]

((A*tan(c + d*x)^7)/7 + tan(c + d*x)^3*(A + (2*C)/3) + tan(c + d*x)*(A + C) + tan(c + d*x)^5*((3*A)/5 + C/5))/
d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**8,x)

[Out]

Timed out

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