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

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

[Out]

3/8*a^4*arctanh(sin(d*x+c))/d-3/8*a^2*b^2*arctanh(sin(d*x+c))/d+3/128*b^4*arctanh(sin(d*x+c))/d+4/5*a^3*b*sec(
d*x+c)^5/d-4/5*a*b^3*sec(d*x+c)^5/d+4/7*a*b^3*sec(d*x+c)^7/d+3/8*a^4*sec(d*x+c)*tan(d*x+c)/d-3/8*a^2*b^2*sec(d
*x+c)*tan(d*x+c)/d+3/128*b^4*sec(d*x+c)*tan(d*x+c)/d+1/4*a^4*sec(d*x+c)^3*tan(d*x+c)/d-1/4*a^2*b^2*sec(d*x+c)^
3*tan(d*x+c)/d+1/64*b^4*sec(d*x+c)^3*tan(d*x+c)/d+a^2*b^2*sec(d*x+c)^5*tan(d*x+c)/d-1/16*b^4*sec(d*x+c)^5*tan(
d*x+c)/d+1/8*b^4*sec(d*x+c)^5*tan(d*x+c)^3/d

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Rubi [A]  time = 0.34, antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]]

Rule 30

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

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 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 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]

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*}

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Mathematica [B]  time = 6.39, 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)

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fricas [A]  time = 0.56, size = 214, normalized size = 0.65 \[ \frac {105 \, {\left (16 \, a^{4} - 16 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{8} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (16 \, a^{4} - 16 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{8} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 5120 \, a b^{3} \cos \left (d x + c\right ) + 7168 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 70 \, {\left (3 \, {\left (16 \, a^{4} - 16 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} + 2 \, {\left (16 \, a^{4} - 16 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 16 \, b^{4} + 8 \, {\left (16 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8960 \, d \cos \left (d x + c\right )^{8}} \]

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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giac [B]  time = 0.51, size = 706, normalized size = 2.14 \[ \text {result too large to display} \]

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

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maple [A]  time = 68.46, size = 491, normalized size = 1.49 \[ \frac {a^{4} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {4 a^{3} b}{5 d \cos \left (d x +c \right )^{5}}+\frac {a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{6}}+\frac {3 a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {3 a^{2} b^{2} \sin \left (d x +c \right )}{8 d}-\frac {3 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{7 d \cos \left (d x +c \right )^{7}}+\frac {12 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{35 d \cos \left (d x +c \right )^{5}}+\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{35 d \cos \left (d x +c \right )^{3}}-\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{35 d \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right ) a \,b^{3}}{35 d}-\frac {8 a \,b^{3} \cos \left (d x +c \right )}{35 d}+\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{8}}+\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{6}}+\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{64 d \cos \left (d x +c \right )^{4}}-\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{128 d \cos \left (d x +c \right )^{2}}-\frac {b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{128 d}-\frac {3 b^{4} \sin \left (d x +c \right )}{128 d}+\frac {3 b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128 d} \]

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*sin(d*x+c)^2*cos(d*x+c)*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 = 0.34, size = 322, normalized size = 0.98 \[ -\frac {35 \, b^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{7} - 11 \, \sin \left (d x + c\right )^{5} - 11 \, \sin \left (d x + c\right )^{3} + 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{8} - 4 \, \sin \left (d x + c\right )^{6} + 6 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 560 \, a^{2} b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 560 \, a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {7168 \, a^{3} b}{\cos \left (d x + c\right )^{5}} + \frac {1024 \, {\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} a b^{3}}{\cos \left (d x + c\right )^{7}}}{8960 \, d} \]

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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mupad [B]  time = 4.40, size = 566, normalized size = 1.72 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}\,\left (\frac {5\,a^4}{4}+\frac {3\,a^2\,b^2}{4}-\frac {3\,b^4}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-\frac {17\,a^4}{4}+\frac {41\,a^2\,b^2}{4}+\frac {23\,b^4}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (-\frac {17\,a^4}{4}+\frac {41\,a^2\,b^2}{4}+\frac {23\,b^4}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {21\,a^4}{4}-\frac {13\,a^2\,b^2}{4}+\frac {333\,b^4}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {21\,a^4}{4}-\frac {13\,a^2\,b^2}{4}+\frac {333\,b^4}{64}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {9\,a^4}{4}+\frac {31\,a^2\,b^2}{4}-\frac {671\,b^4}{64}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {9\,a^4}{4}+\frac {31\,a^2\,b^2}{4}-\frac {671\,b^4}{64}\right )-\frac {16\,a\,b^3}{35}+\frac {8\,a^3\,b}{5}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,a^4}{4}+\frac {3\,a^2\,b^2}{4}-\frac {3\,b^4}{64}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (16\,a\,b^3-24\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (16\,a\,b^3-56\,a^3\,b\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {104\,a^3\,b}{5}+\frac {16\,a\,b^3}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {128\,a\,b^3}{35}-\frac {24\,a^3\,b}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {128\,a\,b^3}{5}-\frac {248\,a^3\,b}{5}\right )-40\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-8\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+70\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^4}{4}-\frac {3\,a^2\,b^2}{4}+\frac {3\,b^4}{64}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^15*((5*a^4)/4 - (3*b^4)/64 + (3*a^2*b^2)/4) + tan(c/2 + (d*x)/2)^3*((23*b^4)/64 - (17*a^4)
/4 + (41*a^2*b^2)/4) + tan(c/2 + (d*x)/2)^13*((23*b^4)/64 - (17*a^4)/4 + (41*a^2*b^2)/4) + tan(c/2 + (d*x)/2)^
5*((21*a^4)/4 + (333*b^4)/64 - (13*a^2*b^2)/4) + tan(c/2 + (d*x)/2)^11*((21*a^4)/4 + (333*b^4)/64 - (13*a^2*b^
2)/4) - tan(c/2 + (d*x)/2)^7*((9*a^4)/4 - (671*b^4)/64 + (31*a^2*b^2)/4) - tan(c/2 + (d*x)/2)^9*((9*a^4)/4 - (
671*b^4)/64 + (31*a^2*b^2)/4) - (16*a*b^3)/35 + (8*a^3*b)/5 + tan(c/2 + (d*x)/2)*((5*a^4)/4 - (3*b^4)/64 + (3*
a^2*b^2)/4) - tan(c/2 + (d*x)/2)^12*(16*a*b^3 - 24*a^3*b) - tan(c/2 + (d*x)/2)^8*(16*a*b^3 - 56*a^3*b) + tan(c
/2 + (d*x)/2)^4*((16*a*b^3)/5 + (104*a^3*b)/5) + tan(c/2 + (d*x)/2)^2*((128*a*b^3)/35 - (24*a^3*b)/5) + tan(c/
2 + (d*x)/2)^6*((128*a*b^3)/5 - (248*a^3*b)/5) - 40*a^3*b*tan(c/2 + (d*x)/2)^10 - 8*a^3*b*tan(c/2 + (d*x)/2)^1
4)/(d*(28*tan(c/2 + (d*x)/2)^4 - 8*tan(c/2 + (d*x)/2)^2 - 56*tan(c/2 + (d*x)/2)^6 + 70*tan(c/2 + (d*x)/2)^8 -
56*tan(c/2 + (d*x)/2)^10 + 28*tan(c/2 + (d*x)/2)^12 - 8*tan(c/2 + (d*x)/2)^14 + tan(c/2 + (d*x)/2)^16 + 1)) +
(atanh(tan(c/2 + (d*x)/2))*((3*a^4)/4 + (3*b^4)/64 - (3*a^2*b^2)/4))/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(sec(d*x+c)**9*(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

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