3.84 \(\int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=168 \[ -\frac{3 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{3 a^2 b^2 \tan (c+d x) \sec (c+d x)}{d}+\frac{4 a^3 b \sec (c+d x)}{d}+\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{4 a b^3 \sec ^3(c+d x)}{3 d}-\frac{4 a b^3 \sec (c+d x)}{d}+\frac{3 b^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^4 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac{3 b^4 \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

(a^4*ArcTanh[Sin[c + d*x]])/d - (3*a^2*b^2*ArcTanh[Sin[c + d*x]])/d + (3*b^4*ArcTanh[Sin[c + d*x]])/(8*d) + (4
*a^3*b*Sec[c + d*x])/d - (4*a*b^3*Sec[c + d*x])/d + (4*a*b^3*Sec[c + d*x]^3)/(3*d) + (3*a^2*b^2*Sec[c + d*x]*T
an[c + d*x])/d - (3*b^4*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b^4*Sec[c + d*x]*Tan[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.194526, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3090, 3770, 2606, 8, 2611} \[ -\frac{3 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{3 a^2 b^2 \tan (c+d x) \sec (c+d x)}{d}+\frac{4 a^3 b \sec (c+d x)}{d}+\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{4 a b^3 \sec ^3(c+d x)}{3 d}-\frac{4 a b^3 \sec (c+d x)}{d}+\frac{3 b^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^4 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac{3 b^4 \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(a^4*ArcTanh[Sin[c + d*x]])/d - (3*a^2*b^2*ArcTanh[Sin[c + d*x]])/d + (3*b^4*ArcTanh[Sin[c + d*x]])/(8*d) + (4
*a^3*b*Sec[c + d*x])/d - (4*a*b^3*Sec[c + d*x])/d + (4*a*b^3*Sec[c + d*x]^3)/(3*d) + (3*a^2*b^2*Sec[c + d*x]*T
an[c + d*x])/d - (3*b^4*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b^4*Sec[c + d*x]*Tan[c + d*x]^3)/(4*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 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 8

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

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]

Rubi steps

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

Mathematica [B]  time = 6.23842, size = 936, normalized size = 5.57 \[ \frac{2 a b \left (6 a^2-5 b^2\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^4}{3 d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac{\left (-8 a^4+24 b^2 a^2-3 b^4\right ) \cos ^4(c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4}{8 d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac{\left (8 a^4-24 b^2 a^2+3 b^4\right ) \cos ^4(c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4}{8 d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac{2 a b^3 \cos ^4(c+d x) \sin \left (\frac{1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4}{3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 (a \cos (c+d x)+b \sin (c+d x))^4}+\frac{2 \cos ^4(c+d x) \left (6 a^3 b \sin \left (\frac{1}{2} (c+d x)\right )-5 a b^3 \sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4}{3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^4}-\frac{2 \cos ^4(c+d x) \left (6 a^3 b \sin \left (\frac{1}{2} (c+d x)\right )-5 a b^3 \sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4}{3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^4}+\frac{\left (-15 b^4+16 a b^3+72 a^2 b^2\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^4}{48 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^4}+\frac{\left (15 b^4+16 a b^3-72 a^2 b^2\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^4}{48 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^4}-\frac{2 a b^3 \cos ^4(c+d x) \sin \left (\frac{1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4}{3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 (a \cos (c+d x)+b \sin (c+d x))^4}+\frac{b^4 \cos ^4(c+d x) (a+b \tan (c+d x))^4}{16 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4 (a \cos (c+d x)+b \sin (c+d x))^4}-\frac{b^4 \cos ^4(c+d x) (a+b \tan (c+d x))^4}{16 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^4 (a \cos (c+d x)+b \sin (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(2*a*b*(6*a^2 - 5*b^2)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(3*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((-8
*a^4 + 24*a^2*b^2 - 3*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)/(8*
d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((8*a^4 - 24*a^2*b^2 + 3*b^4)*Cos[c + d*x]^4*Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4)/(8*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)/(16*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((72*a
^2*b^2 + 16*a*b^3 - 15*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])
^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (2*a*b^3*Cos[c + d*x]^4*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(3*
d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (b^4*Cos[c + d*x]^4*(a + b*Ta
n[c + d*x])^4)/(16*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (2*a*b^3*C
os[c + d*x]^4*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(a*Cos[c +
d*x] + b*Sin[c + d*x])^4) + ((-72*a^2*b^2 + 16*a*b^3 + 15*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])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (2*Cos[c + d*x]^4*(6*a^3*b*Sin[(
c + d*x)/2] - 5*a*b^3*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*
Cos[c + d*x] + b*Sin[c + d*x])^4) - (2*Cos[c + d*x]^4*(6*a^3*b*Sin[(c + d*x)/2] - 5*a*b^3*Sin[(c + d*x)/2])*(a
+ b*Tan[c + d*x])^4)/(3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)

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Maple [A]  time = 0.147, size = 297, normalized size = 1.8 \begin{align*}{\frac{{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{{a}^{3}b}{d\cos \left ( dx+c \right ) }}+3\,{\frac{{a}^{2}{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{d}}-3\,{\frac{{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{4\,a{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{4\,a{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d\cos \left ( dx+c \right ) }}-{\frac{4\,\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}a{b}^{3}}{3\,d}}-{\frac{8\,a{b}^{3}\cos \left ( dx+c \right ) }{3\,d}}+{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{3\,{b}^{4}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

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

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Maxima [A]  time = 1.20938, size = 259, normalized size = 1.54 \begin{align*} \frac{3 \, b^{4}{\left (\frac{2 \,{\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \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 )} - 72 \, a^{2} b^{2}{\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 )} + 24 \, a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac{192 \, a^{3} b}{\cos \left (d x + c\right )} - \frac{64 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a b^{3}}{\cos \left (d x + c\right )^{3}}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/48*(3*b^4*(2*(5*sin(d*x + c)^3 - 3*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)) - 72*a^2*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) -
log(sin(d*x + c) - 1)) + 24*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 192*a^3*b/cos(d*x + c) - 64*
(3*cos(d*x + c)^2 - 1)*a*b^3/cos(d*x + c)^3)/d

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Fricas [A]  time = 0.512719, size = 392, normalized size = 2.33 \begin{align*} \frac{3 \,{\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 64 \, a b^{3} \cos \left (d x + c\right ) + 192 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (2 \, b^{4} +{\left (24 \, a^{2} b^{2} - 5 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/48*(3*(8*a^4 - 24*a^2*b^2 + 3*b^4)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(8*a^4 - 24*a^2*b^2 + 3*b^4)*cos
(d*x + c)^4*log(-sin(d*x + c) + 1) + 64*a*b^3*cos(d*x + c) + 192*(a^3*b - a*b^3)*cos(d*x + c)^3 + 6*(2*b^4 + (
24*a^2*b^2 - 5*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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

[Out]

Timed out

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Giac [B]  time = 1.25489, size = 439, normalized size = 2.61 \begin{align*} \frac{3 \,{\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 9 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 96 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 33 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 288 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 192 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 33 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 288 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 256 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 96 \, a^{3} b - 64 \, a b^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/24*(3*(8*a^4 - 24*a^2*b^2 + 3*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(8*a^4 - 24*a^2*b^2 + 3*b^4)*log(a
bs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(72*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 9*b^4*tan(1/2*d*x + 1/2*c)^7 - 96*a^3*b
*tan(1/2*d*x + 1/2*c)^6 - 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 33*b^4*tan(1/2*d*x + 1/2*c)^5 + 288*a^3*b*tan(1/
2*d*x + 1/2*c)^4 - 192*a*b^3*tan(1/2*d*x + 1/2*c)^4 - 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 33*b^4*tan(1/2*d*x +
1/2*c)^3 - 288*a^3*b*tan(1/2*d*x + 1/2*c)^2 + 256*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 72*a^2*b^2*tan(1/2*d*x + 1/2
*c) - 9*b^4*tan(1/2*d*x + 1/2*c) + 96*a^3*b - 64*a*b^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d