3.86 \(\int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=258 \[ -\frac{3 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{3 a^2 b^2 \tan (c+d x) \sec ^3(c+d x)}{2 d}-\frac{3 a^2 b^2 \tan (c+d x) \sec (c+d x)}{4 d}+\frac{4 a^3 b \sec ^3(c+d x)}{3 d}+\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^4 \tan (c+d x) \sec (c+d x)}{2 d}+\frac{4 a b^3 \sec ^5(c+d x)}{5 d}-\frac{4 a b^3 \sec ^3(c+d x)}{3 d}+\frac{b^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{b^4 \tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac{b^4 \tan (c+d x) \sec ^3(c+d x)}{8 d}+\frac{b^4 \tan (c+d x) \sec (c+d x)}{16 d} \]

[Out]

(a^4*ArcTanh[Sin[c + d*x]])/(2*d) - (3*a^2*b^2*ArcTanh[Sin[c + d*x]])/(4*d) + (b^4*ArcTanh[Sin[c + d*x]])/(16*
d) + (4*a^3*b*Sec[c + d*x]^3)/(3*d) - (4*a*b^3*Sec[c + d*x]^3)/(3*d) + (4*a*b^3*Sec[c + d*x]^5)/(5*d) + (a^4*S
ec[c + d*x]*Tan[c + d*x])/(2*d) - (3*a^2*b^2*Sec[c + d*x]*Tan[c + d*x])/(4*d) + (b^4*Sec[c + d*x]*Tan[c + d*x]
)/(16*d) + (3*a^2*b^2*Sec[c + d*x]^3*Tan[c + d*x])/(2*d) - (b^4*Sec[c + d*x]^3*Tan[c + d*x])/(8*d) + (b^4*Sec[
c + d*x]^3*Tan[c + d*x]^3)/(6*d)

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Rubi [A]  time = 0.294117, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 16, 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} \[ -\frac{3 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{3 a^2 b^2 \tan (c+d x) \sec ^3(c+d x)}{2 d}-\frac{3 a^2 b^2 \tan (c+d x) \sec (c+d x)}{4 d}+\frac{4 a^3 b \sec ^3(c+d x)}{3 d}+\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^4 \tan (c+d x) \sec (c+d x)}{2 d}+\frac{4 a b^3 \sec ^5(c+d x)}{5 d}-\frac{4 a b^3 \sec ^3(c+d x)}{3 d}+\frac{b^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{b^4 \tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac{b^4 \tan (c+d x) \sec ^3(c+d x)}{8 d}+\frac{b^4 \tan (c+d x) \sec (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(a^4*ArcTanh[Sin[c + d*x]])/(2*d) - (3*a^2*b^2*ArcTanh[Sin[c + d*x]])/(4*d) + (b^4*ArcTanh[Sin[c + d*x]])/(16*
d) + (4*a^3*b*Sec[c + d*x]^3)/(3*d) - (4*a*b^3*Sec[c + d*x]^3)/(3*d) + (4*a*b^3*Sec[c + d*x]^5)/(5*d) + (a^4*S
ec[c + d*x]*Tan[c + d*x])/(2*d) - (3*a^2*b^2*Sec[c + d*x]*Tan[c + d*x])/(4*d) + (b^4*Sec[c + d*x]*Tan[c + d*x]
)/(16*d) + (3*a^2*b^2*Sec[c + d*x]^3*Tan[c + d*x])/(2*d) - (b^4*Sec[c + d*x]^3*Tan[c + d*x])/(8*d) + (b^4*Sec[
c + d*x]^3*Tan[c + d*x]^3)/(6*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 ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx &=\int \left (a^4 \sec ^3(c+d x)+4 a^3 b \sec ^3(c+d x) \tan (c+d x)+6 a^2 b^2 \sec ^3(c+d x) \tan ^2(c+d x)+4 a b^3 \sec ^3(c+d x) \tan ^3(c+d x)+b^4 \sec ^3(c+d x) \tan ^4(c+d x)\right ) \, dx\\ &=a^4 \int \sec ^3(c+d x) \, dx+\left (4 a^3 b\right ) \int \sec ^3(c+d x) \tan (c+d x) \, dx+\left (6 a^2 b^2\right ) \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx+\left (4 a b^3\right ) \int \sec ^3(c+d x) \tan ^3(c+d x) \, dx+b^4 \int \sec ^3(c+d x) \tan ^4(c+d x) \, dx\\ &=\frac{a^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{3 a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac{b^4 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}+\frac{1}{2} a^4 \int \sec (c+d x) \, dx-\frac{1}{2} \left (3 a^2 b^2\right ) \int \sec ^3(c+d x) \, dx-\frac{1}{2} b^4 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx+\frac{\left (4 a^3 b\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\sec (c+d x)\right )}{d}+\frac{\left (4 a b^3\right ) \operatorname{Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{4 a^3 b \sec ^3(c+d x)}{3 d}+\frac{a^4 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{3 a^2 b^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac{3 a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}-\frac{b^4 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac{b^4 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}-\frac{1}{4} \left (3 a^2 b^2\right ) \int \sec (c+d x) \, dx+\frac{1}{8} b^4 \int \sec ^3(c+d x) \, dx+\frac{\left (4 a b^3\right ) \operatorname{Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{3 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{4 a^3 b \sec ^3(c+d x)}{3 d}-\frac{4 a b^3 \sec ^3(c+d x)}{3 d}+\frac{4 a b^3 \sec ^5(c+d x)}{5 d}+\frac{a^4 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{3 a^2 b^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac{b^4 \sec (c+d x) \tan (c+d x)}{16 d}+\frac{3 a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}-\frac{b^4 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac{b^4 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}+\frac{1}{16} b^4 \int \sec (c+d x) \, dx\\ &=\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{3 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{b^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{4 a^3 b \sec ^3(c+d x)}{3 d}-\frac{4 a b^3 \sec ^3(c+d x)}{3 d}+\frac{4 a b^3 \sec ^5(c+d x)}{5 d}+\frac{a^4 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{3 a^2 b^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac{b^4 \sec (c+d x) \tan (c+d x)}{16 d}+\frac{3 a^2 b^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}-\frac{b^4 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac{b^4 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}\\ \end{align*}

Mathematica [B]  time = 6.25837, size = 1342, normalized size = 5.2 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(a*b*(20*a^2 - 11*b^2)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(30*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((-
8*a^4 + 12*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)/(16*
d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((8*a^4 - 12*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)/(16*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)/(48*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((30*a^
2*b^2 + 8*a*b^3 - 5*b^4)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(80*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4*
(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((120*a^4 + 160*a^3*b - 180*a^2*b^2 - 88*a*b^3 + 15*b^4)*Cos[c + d*x]^4
*(a + b*Tan[c + d*x])^4)/(480*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)/(5*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(
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)/(48*d*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^6*(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)/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((-30*a^2*b^2 +
8*a*b^3 + 5*b^4)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(80*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(a*Cos[c
+ d*x] + b*Sin[c + d*x])^4) + ((-120*a^4 + 160*a^3*b + 180*a^2*b^2 - 88*a*b^3 - 15*b^4)*Cos[c + d*x]^4*(a + b
*Tan[c + d*x])^4)/(480*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*(20*a^3*b*Sin[(c + d*x)/2] - 11*a*b^3*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(30*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*(20*a^3*b*Sin[(c + d*x)/2] -
11*a*b^3*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(30*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*(-20*a^3*b*Sin[(c + d*x)/2] + 11*a*b^3*Sin[(c + d*x)/2])*(a + b*Tan[c
+ d*x])^4)/(30*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*(-20*a^3*b*Sin[(c + d*x)/2] + 11*a*b^3*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(30*d*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)

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Maple [A]  time = 0.146, size = 394, normalized size = 1.5 \begin{align*}{\frac{{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{4\,{a}^{3}b}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{3\,{a}^{2}{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{2}{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{4\,d}}-{\frac{3\,{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{4\,a{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{4\,a{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{4\,a{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\,d\cos \left ( dx+c \right ) }}-{\frac{4\,\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}a{b}^{3}}{15\,d}}-{\frac{8\,a{b}^{3}\cos \left ( dx+c \right ) }{15\,d}}+{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{24\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{48\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{48\,d}}-{\frac{{b}^{4}\sin \left ( dx+c \right ) }{16\,d}}+{\frac{{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

1/2*a^4*sec(d*x+c)*tan(d*x+c)/d+1/2/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+4/3/d*a^3*b/cos(d*x+c)^3+3/2/d*a^2*b^2*sin
(d*x+c)^3/cos(d*x+c)^4+3/4/d*a^2*b^2*sin(d*x+c)^3/cos(d*x+c)^2+3/4*a^2*b^2*sin(d*x+c)/d-3/4/d*a^2*b^2*ln(sec(d
*x+c)+tan(d*x+c))+4/5/d*a*b^3*sin(d*x+c)^4/cos(d*x+c)^5+4/15/d*a*b^3*sin(d*x+c)^4/cos(d*x+c)^3-4/15/d*a*b^3*si
n(d*x+c)^4/cos(d*x+c)-4/15/d*cos(d*x+c)*sin(d*x+c)^2*a*b^3-8/15*a*b^3*cos(d*x+c)/d+1/6/d*b^4*sin(d*x+c)^5/cos(
d*x+c)^6+1/24/d*b^4*sin(d*x+c)^5/cos(d*x+c)^4-1/48/d*b^4*sin(d*x+c)^5/cos(d*x+c)^2-1/48*b^4*sin(d*x+c)^3/d-1/1
6*b^4*sin(d*x+c)/d+1/16/d*b^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.19383, size = 339, normalized size = 1.31 \begin{align*} -\frac{5 \, b^{4}{\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 )} - 180 \, a^{2} b^{2}{\left (\frac{2 \,{\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \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 )} + 120 \, a^{4}{\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 )} - \frac{640 \, a^{3} b}{\cos \left (d x + c\right )^{3}} + \frac{128 \,{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} a b^{3}}{\cos \left (d x + c\right )^{5}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/480*(5*b^4*(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)) - 180*a^2*b^2*(2*(sin(d*x + c)^3 + s
in(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*a^
4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 640*a^3*b/cos(d*x +
c)^3 + 128*(5*cos(d*x + c)^2 - 3)*a*b^3/cos(d*x + c)^5)/d

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Fricas [A]  time = 0.544456, size = 458, normalized size = 1.78 \begin{align*} \frac{15 \,{\left (8 \, a^{4} - 12 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (8 \, a^{4} - 12 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 384 \, a b^{3} \cos \left (d x + c\right ) + 640 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \,{\left (3 \,{\left (8 \, a^{4} - 12 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, b^{4} + 2 \,{\left (36 \, a^{2} b^{2} - 7 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/480*(15*(8*a^4 - 12*a^2*b^2 + b^4)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(8*a^4 - 12*a^2*b^2 + b^4)*cos(
d*x + c)^6*log(-sin(d*x + c) + 1) + 384*a*b^3*cos(d*x + c) + 640*(a^3*b - a*b^3)*cos(d*x + c)^3 + 10*(3*(8*a^4
- 12*a^2*b^2 + b^4)*cos(d*x + c)^4 + 8*b^4 + 2*(36*a^2*b^2 - 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x
+ c)^6)

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

[Out]

Timed out

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Giac [B]  time = 1.28739, size = 724, normalized size = 2.81 \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)^7*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/240*(15*(8*a^4 - 12*a^2*b^2 + b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(8*a^4 - 12*a^2*b^2 + b^4)*log(ab
s(tan(1/2*d*x + 1/2*c) - 1)) + 2*(120*a^4*tan(1/2*d*x + 1/2*c)^11 + 180*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 15*b
^4*tan(1/2*d*x + 1/2*c)^11 - 960*a^3*b*tan(1/2*d*x + 1/2*c)^10 - 360*a^4*tan(1/2*d*x + 1/2*c)^9 + 900*a^2*b^2*
tan(1/2*d*x + 1/2*c)^9 + 85*b^4*tan(1/2*d*x + 1/2*c)^9 + 2880*a^3*b*tan(1/2*d*x + 1/2*c)^8 - 1920*a*b^3*tan(1/
2*d*x + 1/2*c)^8 + 240*a^4*tan(1/2*d*x + 1/2*c)^7 - 1080*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 570*b^4*tan(1/2*d*x
+ 1/2*c)^7 - 3200*a^3*b*tan(1/2*d*x + 1/2*c)^6 + 1280*a*b^3*tan(1/2*d*x + 1/2*c)^6 + 240*a^4*tan(1/2*d*x + 1/2
*c)^5 - 1080*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 570*b^4*tan(1/2*d*x + 1/2*c)^5 + 1920*a^3*b*tan(1/2*d*x + 1/2*c)
^4 - 360*a^4*tan(1/2*d*x + 1/2*c)^3 + 900*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 85*b^4*tan(1/2*d*x + 1/2*c)^3 - 960
*a^3*b*tan(1/2*d*x + 1/2*c)^2 + 768*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 120*a^4*tan(1/2*d*x + 1/2*c) + 180*a^2*b^2*
tan(1/2*d*x + 1/2*c) - 15*b^4*tan(1/2*d*x + 1/2*c) + 320*a^3*b - 128*a*b^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d