3.88 \(\int \sec ^9(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=330 \[ -\frac{3 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 b^2 \tan (c+d x) \sec ^5(c+d x)}{d}-\frac{a^2 b^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac{3 a^2 b^2 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{4 a^3 b \sec ^5(c+d x)}{5 d}+\frac{3 a^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^4 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 a^4 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{4 a b^3 \sec ^7(c+d x)}{7 d}-\frac{4 a b^3 \sec ^5(c+d x)}{5 d}+\frac{3 b^4 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac{b^4 \tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac{b^4 \tan (c+d x) \sec ^5(c+d x)}{16 d}+\frac{b^4 \tan (c+d x) \sec ^3(c+d x)}{64 d}+\frac{3 b^4 \tan (c+d x) \sec (c+d x)}{128 d} \]

[Out]

(3*a^4*ArcTanh[Sin[c + d*x]])/(8*d) - (3*a^2*b^2*ArcTanh[Sin[c + d*x]])/(8*d) + (3*b^4*ArcTanh[Sin[c + d*x]])/
(128*d) + (4*a^3*b*Sec[c + d*x]^5)/(5*d) - (4*a*b^3*Sec[c + d*x]^5)/(5*d) + (4*a*b^3*Sec[c + d*x]^7)/(7*d) + (
3*a^4*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (3*a^2*b^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (3*b^4*Sec[c + d*x]*Tan
[c + d*x])/(128*d) + (a^4*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) - (a^2*b^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (
b^4*Sec[c + d*x]^3*Tan[c + d*x])/(64*d) + (a^2*b^2*Sec[c + d*x]^5*Tan[c + d*x])/d - (b^4*Sec[c + d*x]^5*Tan[c
+ d*x])/(16*d) + (b^4*Sec[c + d*x]^5*Tan[c + d*x]^3)/(8*d)

________________________________________________________________________________________

Rubi [A]  time = 0.34335, antiderivative size = 330, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3090, 3768, 3770, 2606, 30, 2611, 14} \[ -\frac{3 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 b^2 \tan (c+d x) \sec ^5(c+d x)}{d}-\frac{a^2 b^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac{3 a^2 b^2 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{4 a^3 b \sec ^5(c+d x)}{5 d}+\frac{3 a^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^4 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 a^4 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{4 a b^3 \sec ^7(c+d x)}{7 d}-\frac{4 a b^3 \sec ^5(c+d x)}{5 d}+\frac{3 b^4 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac{b^4 \tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac{b^4 \tan (c+d x) \sec ^5(c+d x)}{16 d}+\frac{b^4 \tan (c+d x) \sec ^3(c+d x)}{64 d}+\frac{3 b^4 \tan (c+d x) \sec (c+d x)}{128 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(3*a^4*ArcTanh[Sin[c + d*x]])/(8*d) - (3*a^2*b^2*ArcTanh[Sin[c + d*x]])/(8*d) + (3*b^4*ArcTanh[Sin[c + d*x]])/
(128*d) + (4*a^3*b*Sec[c + d*x]^5)/(5*d) - (4*a*b^3*Sec[c + d*x]^5)/(5*d) + (4*a*b^3*Sec[c + d*x]^7)/(7*d) + (
3*a^4*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (3*a^2*b^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (3*b^4*Sec[c + d*x]*Tan
[c + d*x])/(128*d) + (a^4*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) - (a^2*b^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (
b^4*Sec[c + d*x]^3*Tan[c + d*x])/(64*d) + (a^2*b^2*Sec[c + d*x]^5*Tan[c + d*x])/d - (b^4*Sec[c + d*x]^5*Tan[c
+ d*x])/(16*d) + (b^4*Sec[c + d*x]^5*Tan[c + d*x]^3)/(8*d)

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

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 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \sec ^9(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx &=\int \left (a^4 \sec ^5(c+d x)+4 a^3 b \sec ^5(c+d x) \tan (c+d x)+6 a^2 b^2 \sec ^5(c+d x) \tan ^2(c+d x)+4 a b^3 \sec ^5(c+d x) \tan ^3(c+d x)+b^4 \sec ^5(c+d x) \tan ^4(c+d x)\right ) \, dx\\ &=a^4 \int \sec ^5(c+d x) \, dx+\left (4 a^3 b\right ) \int \sec ^5(c+d x) \tan (c+d x) \, dx+\left (6 a^2 b^2\right ) \int \sec ^5(c+d x) \tan ^2(c+d x) \, dx+\left (4 a b^3\right ) \int \sec ^5(c+d x) \tan ^3(c+d x) \, dx+b^4 \int \sec ^5(c+d x) \tan ^4(c+d x) \, dx\\ &=\frac{a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{a^2 b^2 \sec ^5(c+d x) \tan (c+d x)}{d}+\frac{b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}+\frac{1}{4} \left (3 a^4\right ) \int \sec ^3(c+d x) \, dx-\left (a^2 b^2\right ) \int \sec ^5(c+d x) \, dx-\frac{1}{8} \left (3 b^4\right ) \int \sec ^5(c+d x) \tan ^2(c+d x) \, dx+\frac{\left (4 a^3 b\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,\sec (c+d x)\right )}{d}+\frac{\left (4 a b^3\right ) \operatorname{Subst}\left (\int x^4 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{4 a^3 b \sec ^5(c+d x)}{5 d}+\frac{3 a^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{a^2 b^2 \sec ^5(c+d x) \tan (c+d x)}{d}-\frac{b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}+\frac{1}{8} \left (3 a^4\right ) \int \sec (c+d x) \, dx-\frac{1}{4} \left (3 a^2 b^2\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{16} b^4 \int \sec ^5(c+d x) \, dx+\frac{\left (4 a b^3\right ) \operatorname{Subst}\left (\int \left (-x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{3 a^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{4 a^3 b \sec ^5(c+d x)}{5 d}-\frac{4 a b^3 \sec ^5(c+d x)}{5 d}+\frac{4 a b^3 \sec ^7(c+d x)}{7 d}+\frac{3 a^4 \sec (c+d x) \tan (c+d x)}{8 d}-\frac{3 a^2 b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{b^4 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac{a^2 b^2 \sec ^5(c+d x) \tan (c+d x)}{d}-\frac{b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}-\frac{1}{8} \left (3 a^2 b^2\right ) \int \sec (c+d x) \, dx+\frac{1}{64} \left (3 b^4\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{3 a^4 \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{3 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{4 a^3 b \sec ^5(c+d x)}{5 d}-\frac{4 a b^3 \sec ^5(c+d x)}{5 d}+\frac{4 a b^3 \sec ^7(c+d x)}{7 d}+\frac{3 a^4 \sec (c+d x) \tan (c+d x)}{8 d}-\frac{3 a^2 b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{3 b^4 \sec (c+d x) \tan (c+d x)}{128 d}+\frac{a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{b^4 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac{a^2 b^2 \sec ^5(c+d x) \tan (c+d x)}{d}-\frac{b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}+\frac{1}{128} \left (3 b^4\right ) \int \sec (c+d x) \, dx\\ &=\frac{3 a^4 \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{3 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 b^4 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac{4 a^3 b \sec ^5(c+d x)}{5 d}-\frac{4 a b^3 \sec ^5(c+d x)}{5 d}+\frac{4 a b^3 \sec ^7(c+d x)}{7 d}+\frac{3 a^4 \sec (c+d x) \tan (c+d x)}{8 d}-\frac{3 a^2 b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{3 b^4 \sec (c+d x) \tan (c+d x)}{128 d}+\frac{a^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{b^4 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac{a^2 b^2 \sec ^5(c+d x) \tan (c+d x)}{d}-\frac{b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac{b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}\\ \end{align*}

Mathematica [B]  time = 6.38568, size = 1732, normalized size = 5.25 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(a*b*(42*a^2 - 17*b^2)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(140*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (3
*(16*a^4 - 16*a^2*b^2 + b^4)*Cos[c + d*x]^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4)/(
128*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (3*(16*a^4 - 16*a^2*b^2 + b^4)*Cos[c + d*x]^4*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4)/(128*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (b^4*Cos[c + d*x]^
4*(a + b*Tan[c + d*x])^4)/(128*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^8*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)
+ ((56*a^2*b^2 + 16*a*b^3 - 7*b^4)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(448*d*(Cos[(c + d*x)/2] - Sin[(c +
d*x)/2])^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((560*a^4 + 896*a^3*b - 256*a*b^3 - 35*b^4)*Cos[c + d*x]^4*(
a + b*Tan[c + d*x])^4)/(8960*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) +
((1680*a^4 + 1344*a^3*b - 1680*a^2*b^2 - 544*a*b^3 + 105*b^4)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(8960*d*(
Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (a*b^3*Cos[c + d*x]^4*Sin[(c + d
*x)/2]*(a + b*Tan[c + d*x])^4)/(14*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^7*(a*Cos[c + d*x] + b*Sin[c + d*x])
^4) - (b^4*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(128*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8*(a*Cos[c + d*
x] + b*Sin[c + d*x])^4) - (a*b^3*Cos[c + d*x]^4*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(14*d*(Cos[(c + d*x)/
2] + Sin[(c + d*x)/2])^7*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((-56*a^2*b^2 + 16*a*b^3 + 7*b^4)*Cos[c + d*x]
^4*(a + b*Tan[c + d*x])^4)/(448*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)
+ ((-560*a^4 + 896*a^3*b - 256*a*b^3 + 35*b^4)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(8960*d*(Cos[(c + d*x)/
2] + Sin[(c + d*x)/2])^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((-1680*a^4 + 1344*a^3*b + 1680*a^2*b^2 - 544*
a*b^3 - 105*b^4)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(8960*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a*Cos
[c + d*x] + b*Sin[c + d*x])^4) + (Cos[c + d*x]^4*(42*a^3*b*Sin[(c + d*x)/2] - 17*a*b^3*Sin[(c + d*x)/2])*(a +
b*Tan[c + d*x])^4)/(140*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (Cos[
c + d*x]^4*(42*a^3*b*Sin[(c + d*x)/2] - 17*a*b^3*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(140*d*(Cos[(c + d*
x)/2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (Cos[c + d*x]^4*(7*a^3*b*Sin[(c + d*x)/2] - 2
*a*b^3*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(35*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(a*Cos[c + d*x]
+ b*Sin[c + d*x])^4) + (Cos[c + d*x]^4*(-7*a^3*b*Sin[(c + d*x)/2] + 2*a*b^3*Sin[(c + d*x)/2])*(a + b*Tan[c +
d*x])^4)/(35*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (Cos[c + d*x]^4*
(-42*a^3*b*Sin[(c + d*x)/2] + 17*a*b^3*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(140*d*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (Cos[c + d*x]^4*(-42*a^3*b*Sin[(c + d*x)/2] + 17*a*b^
3*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(140*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*S
in[c + d*x])^4)

________________________________________________________________________________________

Maple [A]  time = 0.159, size = 491, normalized size = 1.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

1/4*a^4*sec(d*x+c)^3*tan(d*x+c)/d+3/8*a^4*sec(d*x+c)*tan(d*x+c)/d+3/8/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+4/5/d*a^
3*b/cos(d*x+c)^5+1/d*a^2*b^2*sin(d*x+c)^3/cos(d*x+c)^6+3/4/d*a^2*b^2*sin(d*x+c)^3/cos(d*x+c)^4+3/8/d*a^2*b^2*s
in(d*x+c)^3/cos(d*x+c)^2+3/8*a^2*b^2*sin(d*x+c)/d-3/8/d*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+4/7/d*a*b^3*sin(d*x+
c)^4/cos(d*x+c)^7+12/35/d*a*b^3*sin(d*x+c)^4/cos(d*x+c)^5+4/35/d*a*b^3*sin(d*x+c)^4/cos(d*x+c)^3-4/35/d*a*b^3*
sin(d*x+c)^4/cos(d*x+c)-4/35/d*cos(d*x+c)*sin(d*x+c)^2*a*b^3-8/35*a*b^3*cos(d*x+c)/d+1/8/d*b^4*sin(d*x+c)^5/co
s(d*x+c)^8+1/16/d*b^4*sin(d*x+c)^5/cos(d*x+c)^6+1/64/d*b^4*sin(d*x+c)^5/cos(d*x+c)^4-1/128/d*b^4*sin(d*x+c)^5/
cos(d*x+c)^2-1/128*b^4*sin(d*x+c)^3/d-3/128*b^4*sin(d*x+c)/d+3/128/d*b^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.17099, size = 435, normalized size = 1.32 \begin{align*} -\frac{35 \, b^{4}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{7} - 11 \, \sin \left (d x + c\right )^{5} - 11 \, \sin \left (d x + c\right )^{3} + 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{8} - 4 \, \sin \left (d x + c\right )^{6} + 6 \, \sin \left (d x + c\right )^{4} - 4 \, \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 )} - 560 \, a^{2} b^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 8 \, \sin \left (d x + c\right )^{3} - 3 \, \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} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 560 \, a^{4}{\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 )} - \frac{7168 \, a^{3} b}{\cos \left (d x + c\right )^{5}} + \frac{1024 \,{\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} a b^{3}}{\cos \left (d x + c\right )^{7}}}{8960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/8960*(35*b^4*(2*(3*sin(d*x + c)^7 - 11*sin(d*x + c)^5 - 11*sin(d*x + c)^3 + 3*sin(d*x + c))/(sin(d*x + c)^8
- 4*sin(d*x + c)^6 + 6*sin(d*x + c)^4 - 4*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c)
- 1)) - 560*a^2*b^2*(2*(3*sin(d*x + c)^5 - 8*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)
^4 + 3*sin(d*x + c)^2 - 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) + 560*a^4*(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)
) - 7168*a^3*b/cos(d*x + c)^5 + 1024*(7*cos(d*x + c)^2 - 5)*a*b^3/cos(d*x + c)^7)/d

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Fricas [A]  time = 0.584715, size = 533, normalized size = 1.62 \begin{align*} \frac{105 \,{\left (16 \, a^{4} - 16 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{8} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left (16 \, a^{4} - 16 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{8} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 5120 \, a b^{3} \cos \left (d x + c\right ) + 7168 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 70 \,{\left (3 \,{\left (16 \, a^{4} - 16 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} + 2 \,{\left (16 \, a^{4} - 16 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 16 \, b^{4} + 8 \,{\left (16 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8960 \, d \cos \left (d x + c\right )^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/8960*(105*(16*a^4 - 16*a^2*b^2 + b^4)*cos(d*x + c)^8*log(sin(d*x + c) + 1) - 105*(16*a^4 - 16*a^2*b^2 + b^4)
*cos(d*x + c)^8*log(-sin(d*x + c) + 1) + 5120*a*b^3*cos(d*x + c) + 7168*(a^3*b - a*b^3)*cos(d*x + c)^3 + 70*(3
*(16*a^4 - 16*a^2*b^2 + b^4)*cos(d*x + c)^6 + 2*(16*a^4 - 16*a^2*b^2 + b^4)*cos(d*x + c)^4 + 16*b^4 + 8*(16*a^
2*b^2 - 3*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^8)

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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(sec(d*x+c)**9*(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

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Giac [B]  time = 1.28803, size = 953, normalized size = 2.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/4480*(105*(16*a^4 - 16*a^2*b^2 + b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(16*a^4 - 16*a^2*b^2 + b^4)*l
og(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(2800*a^4*tan(1/2*d*x + 1/2*c)^15 + 1680*a^2*b^2*tan(1/2*d*x + 1/2*c)^15
- 105*b^4*tan(1/2*d*x + 1/2*c)^15 - 17920*a^3*b*tan(1/2*d*x + 1/2*c)^14 - 9520*a^4*tan(1/2*d*x + 1/2*c)^13 +
22960*a^2*b^2*tan(1/2*d*x + 1/2*c)^13 + 805*b^4*tan(1/2*d*x + 1/2*c)^13 + 53760*a^3*b*tan(1/2*d*x + 1/2*c)^12
- 35840*a*b^3*tan(1/2*d*x + 1/2*c)^12 + 11760*a^4*tan(1/2*d*x + 1/2*c)^11 - 7280*a^2*b^2*tan(1/2*d*x + 1/2*c)^
11 + 11655*b^4*tan(1/2*d*x + 1/2*c)^11 - 89600*a^3*b*tan(1/2*d*x + 1/2*c)^10 - 5040*a^4*tan(1/2*d*x + 1/2*c)^9
- 17360*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 23485*b^4*tan(1/2*d*x + 1/2*c)^9 + 125440*a^3*b*tan(1/2*d*x + 1/2*c)
^8 - 35840*a*b^3*tan(1/2*d*x + 1/2*c)^8 - 5040*a^4*tan(1/2*d*x + 1/2*c)^7 - 17360*a^2*b^2*tan(1/2*d*x + 1/2*c)
^7 + 23485*b^4*tan(1/2*d*x + 1/2*c)^7 - 111104*a^3*b*tan(1/2*d*x + 1/2*c)^6 + 57344*a*b^3*tan(1/2*d*x + 1/2*c)
^6 + 11760*a^4*tan(1/2*d*x + 1/2*c)^5 - 7280*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 11655*b^4*tan(1/2*d*x + 1/2*c)^5
+ 46592*a^3*b*tan(1/2*d*x + 1/2*c)^4 + 7168*a*b^3*tan(1/2*d*x + 1/2*c)^4 - 9520*a^4*tan(1/2*d*x + 1/2*c)^3 +
22960*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 805*b^4*tan(1/2*d*x + 1/2*c)^3 - 10752*a^3*b*tan(1/2*d*x + 1/2*c)^2 + 8
192*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 2800*a^4*tan(1/2*d*x + 1/2*c) + 1680*a^2*b^2*tan(1/2*d*x + 1/2*c) - 105*b^4
*tan(1/2*d*x + 1/2*c) + 3584*a^3*b - 1024*a*b^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^8)/d