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

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

[Out]

1/2*a^4*arctanh(sin(d*x+c))/d-3/4*a^2*b^2*arctanh(sin(d*x+c))/d+1/16*b^4*arctanh(sin(d*x+c))/d+4/3*a^3*b*sec(d
*x+c)^3/d-4/3*a*b^3*sec(d*x+c)^3/d+4/5*a*b^3*sec(d*x+c)^5/d+1/2*a^4*sec(d*x+c)*tan(d*x+c)/d-3/4*a^2*b^2*sec(d*
x+c)*tan(d*x+c)/d+1/16*b^4*sec(d*x+c)*tan(d*x+c)/d+3/2*a^2*b^2*sec(d*x+c)^3*tan(d*x+c)/d-1/8*b^4*sec(d*x+c)^3*
tan(d*x+c)/d+1/6*b^4*sec(d*x+c)^3*tan(d*x+c)^3/d

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Rubi [A]  time = 0.29, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 16, 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} \[ -\frac {3 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {3 a^2 b^2 \tan (c+d x) \sec ^3(c+d x)}{2 d}-\frac {3 a^2 b^2 \tan (c+d x) \sec (c+d x)}{4 d}+\frac {4 a^3 b \sec ^3(c+d x)}{3 d}+\frac {a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^4 \tan (c+d x) \sec (c+d x)}{2 d}+\frac {4 a b^3 \sec ^5(c+d x)}{5 d}-\frac {4 a b^3 \sec ^3(c+d x)}{3 d}+\frac {b^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {b^4 \tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac {b^4 \tan (c+d x) \sec ^3(c+d x)}{8 d}+\frac {b^4 \tan (c+d x) \sec (c+d x)}{16 d} \]

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 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 ^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*}

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Mathematica [B]  time = 6.25, size = 1342, normalized size = 5.20 \[ \text {result too large to display} \]

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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fricas [A]  time = 0.81, size = 187, normalized size = 0.72 \[ \frac {15 \, {\left (8 \, a^{4} - 12 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (8 \, a^{4} - 12 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 384 \, a b^{3} \cos \left (d x + c\right ) + 640 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (3 \, {\left (8 \, a^{4} - 12 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, b^{4} + 2 \, {\left (36 \, a^{2} b^{2} - 7 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

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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giac [B]  time = 0.49, size = 536, normalized size = 2.08 \[ \frac {15 \, {\left (8 \, a^{4} - 12 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (8 \, a^{4} - 12 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (120 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 180 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 15 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 960 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 360 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 900 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 85 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2880 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 1920 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 240 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1080 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 570 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3200 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 1280 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 240 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1080 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 570 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1920 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 360 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 900 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 85 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 960 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 768 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 320 \, a^{3} b - 128 \, a b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \]

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

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maple [A]  time = 65.24, size = 394, normalized size = 1.53 \[ \frac {a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {4 a^{3} b}{3 d \cos \left (d x +c \right )^{3}}+\frac {3 a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{2}}+\frac {3 a^{2} b^{2} \sin \left (d x +c \right )}{4 d}-\frac {3 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}+\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}-\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right ) a \,b^{3}}{15 d}-\frac {8 a \,b^{3} \cos \left (d x +c \right )}{15 d}+\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}+\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{24 d \cos \left (d x +c \right )^{4}}-\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{48 d \cos \left (d x +c \right )^{2}}-\frac {b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{48 d}-\frac {b^{4} \sin \left (d x +c \right )}{16 d}+\frac {b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d} \]

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*sin(d*x+c)^2*cos(d*x+c)*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 = 0.33, size = 251, normalized size = 0.97 \[ -\frac {5 \, b^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{5} + 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, a^{2} b^{2} {\left (\frac {2 \, {\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {640 \, a^{3} b}{\cos \left (d x + c\right )^{3}} + \frac {128 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} a b^{3}}{\cos \left (d x + c\right )^{5}}}{480 \, d} \]

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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mupad [B]  time = 4.27, size = 419, normalized size = 1.62 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^4-\frac {3\,a^2\,b^2}{2}+\frac {b^4}{8}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^4-9\,a^2\,b^2+\frac {19\,b^4}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (2\,a^4-9\,a^2\,b^2+\frac {19\,b^4}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-3\,a^4+\frac {15\,a^2\,b^2}{2}+\frac {17\,b^4}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (-3\,a^4+\frac {15\,a^2\,b^2}{2}+\frac {17\,b^4}{24}\right )+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+\frac {3\,a^2\,b^2}{2}-\frac {b^4}{8}\right )-\frac {16\,a\,b^3}{15}+\frac {8\,a^3\,b}{3}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (a^4+\frac {3\,a^2\,b^2}{2}-\frac {b^4}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {32\,a\,b^3}{5}-8\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (16\,a\,b^3-24\,a^3\,b\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {32\,a\,b^3}{3}-\frac {80\,a^3\,b}{3}\right )+16\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

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)^7,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*(a^4 + b^4/8 - (3*a^2*b^2)/2))/d + (tan(c/2 + (d*x)/2)^5*(2*a^4 + (19*b^4)/4 - 9*a^
2*b^2) + tan(c/2 + (d*x)/2)^7*(2*a^4 + (19*b^4)/4 - 9*a^2*b^2) + tan(c/2 + (d*x)/2)^3*((17*b^4)/24 - 3*a^4 + (
15*a^2*b^2)/2) + tan(c/2 + (d*x)/2)^9*((17*b^4)/24 - 3*a^4 + (15*a^2*b^2)/2) + tan(c/2 + (d*x)/2)*(a^4 - b^4/8
+ (3*a^2*b^2)/2) - (16*a*b^3)/15 + (8*a^3*b)/3 + tan(c/2 + (d*x)/2)^11*(a^4 - b^4/8 + (3*a^2*b^2)/2) + tan(c/
2 + (d*x)/2)^2*((32*a*b^3)/5 - 8*a^3*b) - tan(c/2 + (d*x)/2)^8*(16*a*b^3 - 24*a^3*b) + tan(c/2 + (d*x)/2)^6*((
32*a*b^3)/3 - (80*a^3*b)/3) + 16*a^3*b*tan(c/2 + (d*x)/2)^4 - 8*a^3*b*tan(c/2 + (d*x)/2)^10)/(d*(15*tan(c/2 +
(d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)
^10 + tan(c/2 + (d*x)/2)^12 + 1))

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

[Out]

Timed out

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