3.447 \(\int (a+b \cos (c+d x))^4 \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=188 \[ \frac{\left (60 a^2 b^2+8 a^4+15 b^4\right ) \tan (c+d x)}{15 d}+\frac{a b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 \left (4 a^2+27 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac{a b \left (3 a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{5 d}+\frac{a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

[Out]

(a*b*(3*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + ((8*a^4 + 60*a^2*b^2 + 15*b^4)*Tan[c + d*x])/(15*d) + (a*b
*(3*a^2 + 4*b^2)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^2*(4*a^2 + 27*b^2)*Sec[c + d*x]^2*Tan[c + d*x])/(15*d)
+ (3*a^3*b*Sec[c + d*x]^3*Tan[c + d*x])/(5*d) + (a^2*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

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Rubi [A]  time = 0.36073, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {2792, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{\left (60 a^2 b^2+8 a^4+15 b^4\right ) \tan (c+d x)}{15 d}+\frac{a b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 \left (4 a^2+27 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac{a b \left (3 a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{5 d}+\frac{a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^6,x]

[Out]

(a*b*(3*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + ((8*a^4 + 60*a^2*b^2 + 15*b^4)*Tan[c + d*x])/(15*d) + (a*b
*(3*a^2 + 4*b^2)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^2*(4*a^2 + 27*b^2)*Sec[c + d*x]^2*Tan[c + d*x])/(15*d)
+ (3*a^3*b*Sec[c + d*x]^3*Tan[c + d*x])/(5*d) + (a^2*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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]

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]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x))^4 \sec ^6(c+d x) \, dx &=\frac{a^2 (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+b \cos (c+d x)) \left (12 a^2 b+a \left (4 a^2+15 b^2\right ) \cos (c+d x)+b \left (2 a^2+5 b^2\right ) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{3 a^3 b \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{20} \int \left (-4 a^2 \left (4 a^2+27 b^2\right )-20 a b \left (3 a^2+4 b^2\right ) \cos (c+d x)-4 b^2 \left (2 a^2+5 b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a^2 \left (4 a^2+27 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{3 a^3 b \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{60} \int \left (-60 a b \left (3 a^2+4 b^2\right )-4 \left (8 a^4+60 a^2 b^2+15 b^4\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a^2 \left (4 a^2+27 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{3 a^3 b \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\left (a b \left (3 a^2+4 b^2\right )\right ) \int \sec ^3(c+d x) \, dx-\frac{1}{15} \left (-8 a^4-60 a^2 b^2-15 b^4\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a b \left (3 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a^2 \left (4 a^2+27 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{3 a^3 b \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{2} \left (a b \left (3 a^2+4 b^2\right )\right ) \int \sec (c+d x) \, dx-\frac{\left (8 a^4+60 a^2 b^2+15 b^4\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{a b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\left (8 a^4+60 a^2 b^2+15 b^4\right ) \tan (c+d x)}{15 d}+\frac{a b \left (3 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a^2 \left (4 a^2+27 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{3 a^3 b \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.732868, size = 125, normalized size = 0.66 \[ \frac{15 a b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (20 a^2 \left (a^2+3 b^2\right ) \tan ^2(c+d x)+15 a b \left (3 a^2+4 b^2\right ) \sec (c+d x)+30 \left (6 a^2 b^2+a^4+b^4\right )+30 a^3 b \sec ^3(c+d x)+6 a^4 \tan ^4(c+d x)\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^6,x]

[Out]

(15*a*b*(3*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(30*(a^4 + 6*a^2*b^2 + b^4) + 15*a*b*(3*a^2 + 4*b
^2)*Sec[c + d*x] + 30*a^3*b*Sec[c + d*x]^3 + 20*a^2*(a^2 + 3*b^2)*Tan[c + d*x]^2 + 6*a^4*Tan[c + d*x]^4))/(30*
d)

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Maple [A]  time = 0.068, size = 225, normalized size = 1.2 \begin{align*}{\frac{8\,{a}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{{a}^{3}b \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{3\,{a}^{3}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+4\,{\frac{{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+2\,{\frac{a{b}^{3}\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{d}}+2\,{\frac{a{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{4}\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*sec(d*x+c)^6,x)

[Out]

8/15*a^4*tan(d*x+c)/d+1/5/d*a^4*tan(d*x+c)*sec(d*x+c)^4+4/15/d*a^4*tan(d*x+c)*sec(d*x+c)^2+a^3*b*sec(d*x+c)^3*
tan(d*x+c)/d+3/2*a^3*b*sec(d*x+c)*tan(d*x+c)/d+3/2/d*a^3*b*ln(sec(d*x+c)+tan(d*x+c))+4/d*a^2*b^2*tan(d*x+c)+2/
d*a^2*b^2*tan(d*x+c)*sec(d*x+c)^2+2/d*a*b^3*tan(d*x+c)*sec(d*x+c)+2/d*a*b^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*b^4*
tan(d*x+c)

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Maxima [A]  time = 1.00131, size = 263, normalized size = 1.4 \begin{align*} \frac{4 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{4} + 120 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} b^{2} - 15 \, a^{3} b{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, a b^{3}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 60 \, b^{4} \tan \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/60*(4*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^4 + 120*(tan(d*x + c)^3 + 3*tan(d*x + c))*a
^2*b^2 - 15*a^3*b*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d
*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*a*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1
) + log(sin(d*x + c) - 1)) + 60*b^4*tan(d*x + c))/d

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Fricas [A]  time = 1.9966, size = 443, normalized size = 2.36 \begin{align*} \frac{15 \,{\left (3 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (3 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (30 \, a^{3} b \cos \left (d x + c\right ) + 2 \,{\left (8 \, a^{4} + 60 \, a^{2} b^{2} + 15 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 6 \, a^{4} + 15 \,{\left (3 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (2 \, a^{4} + 15 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/60*(15*(3*a^3*b + 4*a*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*a^3*b + 4*a*b^3)*cos(d*x + c)^5*log(
-sin(d*x + c) + 1) + 2*(30*a^3*b*cos(d*x + c) + 2*(8*a^4 + 60*a^2*b^2 + 15*b^4)*cos(d*x + c)^4 + 6*a^4 + 15*(3
*a^3*b + 4*a*b^3)*cos(d*x + c)^3 + 4*(2*a^4 + 15*a^2*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**4*sec(d*x+c)**6,x)

[Out]

Timed out

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Giac [B]  time = 1.43402, size = 622, normalized size = 3.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/30*(15*(3*a^3*b + 4*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*a^3*b + 4*a*b^3)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(30*a^4*tan(1/2*d*x + 1/2*c)^9 - 75*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 180*a^2*b^2*tan(1/2*d*x +
1/2*c)^9 - 60*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 30*b^4*tan(1/2*d*x + 1/2*c)^9 - 40*a^4*tan(1/2*d*x + 1/2*c)^7 +
30*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 480*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 120*
b^4*tan(1/2*d*x + 1/2*c)^7 + 116*a^4*tan(1/2*d*x + 1/2*c)^5 + 600*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 180*b^4*tan
(1/2*d*x + 1/2*c)^5 - 40*a^4*tan(1/2*d*x + 1/2*c)^3 - 30*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 480*a^2*b^2*tan(1/2*d*
x + 1/2*c)^3 - 120*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 120*b^4*tan(1/2*d*x + 1/2*c)^3 + 30*a^4*tan(1/2*d*x + 1/2*c)
+ 75*a^3*b*tan(1/2*d*x + 1/2*c) + 180*a^2*b^2*tan(1/2*d*x + 1/2*c) + 60*a*b^3*tan(1/2*d*x + 1/2*c) + 30*b^4*t
an(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d