\(\int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\) [84]

Optimal result

Integrand size = 28, antiderivative size = 168 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {a^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {3 a^2 b^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {3 b^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {4 a^3 b \sec (c+d x)}{d}-\frac {4 a b^3 \sec (c+d x)}{d}+\frac {4 a b^3 \sec ^3(c+d x)}{3 d}+\frac {3 a^2 b^2 \sec (c+d x) \tan (c+d x)}{d}-\frac {3 b^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b^4 \sec (c+d x) \tan ^3(c+d x)}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3169, 3855, 2686, 8, 2691} \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {a^4 \text {arctanh}(\sin (c+d x))}{d}+\frac {4 a^3 b \sec (c+d x)}{d}-\frac {3 a^2 b^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {3 a^2 b^2 \tan (c+d x) \sec (c+d x)}{d}+\frac {4 a b^3 \sec ^3(c+d x)}{3 d}-\frac {4 a b^3 \sec (c+d x)}{d}+\frac {3 b^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b^4 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3 b^4 \tan (c+d x) \sec (c+d x)}{8 d} \]

[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 8

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

Rule 2686

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 2691

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 3169

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \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 \text {arctanh}(\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 ) \text {Subst}(\int 1 \, dx,x,\sec (c+d x))}{d}+\frac {\left (4 a b^3\right ) \text {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {a^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {3 a^2 b^2 \text {arctanh}(\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 \text {arctanh}(\sin (c+d x))}{d}-\frac {3 a^2 b^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {3 b^4 \text {arctanh}(\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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(936\) vs. \(2(168)=336\).

Time = 7.52 (sec) , antiderivative size = 936, normalized size of antiderivative = 5.57 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\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 a^2 b^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 a^2 b^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 {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 {\left (72 a^2 b^2+16 a b^3-15 b^4\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 {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 {\left (-72 a^2 b^2+16 a b^3+15 b^4\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 \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} \]

[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)

Maple [A] (verified)

Time = 1.57 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.18

method	result	size
parts	\(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{4} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {4 a^{3} b \sec \left (d x +c \right )}{d}+\frac {4 a \,b^{3} \left (\frac {\sec \left (d x +c \right )^{3}}{3}-\sec \left (d x +c \right )\right )}{d}+\frac {6 a^{2} b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(198\)
derivativedivides	\(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {4 a^{3} b}{\cos \left (d x +c \right )}+6 a^{2} b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a \,b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+b^{4} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(225\)
default	\(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {4 a^{3} b}{\cos \left (d x +c \right )}+6 a^{2} b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a \,b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+b^{4} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(225\)
parallelrisch	\(\frac {-4 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{4}-3 a^{2} b^{2}+\frac {3}{8} b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{4}-3 a^{2} b^{2}+\frac {3}{8} b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+24 b \left (\frac {2 \left (a^{3}-\frac {2}{3} a \,b^{2}\right ) \cos \left (2 d x +2 c \right )}{3}+\frac {\left (\frac {1}{2} a^{3}-\frac {1}{3} a \,b^{2}\right ) \cos \left (4 d x +4 c \right )}{3}+\frac {\left (a^{3}-a \,b^{2}\right ) \cos \left (3 d x +3 c \right )}{3}+\frac {\left (a^{2} b -\frac {5}{24} b^{3}\right ) \sin \left (3 d x +3 c \right )}{4}+\left (a^{3}-\frac {5}{9} a \,b^{2}\right ) \cos \left (d x +c \right )+\frac {\left (a^{2} b +\frac {1}{8} b^{3}\right ) \sin \left (d x +c \right )}{4}+\frac {a^{3}}{2}-\frac {a \,b^{2}}{3}\right )}{d \left (3+\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )\right )}\)	\(274\)
risch	\(\frac {b \,{\mathrm e}^{i \left (d x +c \right )} \left (72 i b \,a^{2}+72 i a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+96 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-96 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-9 i b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+15 i b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+288 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-160 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 i b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-72 i a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+288 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-160 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-72 i a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-15 i b^{3}+96 a^{3}-96 a \,b^{2}\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {\ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right ) a^{4}}{d}-\frac {3 b^{2} \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right ) a^{2}}{d}+\frac {3 b^{4} \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{4}}{d}+\frac {3 b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}-\frac {3 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\)	\(392\)
norman	\(\frac {\frac {24 a^{3} b -16 a \,b^{3}}{3 d}-\frac {4 \left (2 a^{3} b +4 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {2 \left (12 a^{3} b -8 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {4 \left (18 a^{3} b -28 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}+\frac {8 a^{3} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {8 a^{3} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d}+\frac {3 b^{2} \left (8 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {3 b^{2} \left (8 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}-\frac {5 b^{2} \left (24 a^{2}-19 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}-\frac {5 b^{2} \left (24 a^{2}-19 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}+\frac {b^{2} \left (24 a^{2}+37 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {b^{2} \left (24 a^{2}+37 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {b^{2} \left (72 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {b^{2} \left (72 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}-\frac {8 a b \left (9 a^{2}-16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}+\frac {8 a b \left (9 a^{2}-16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {\left (8 a^{4}-24 a^{2} b^{2}+3 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (8 a^{4}-24 a^{2} b^{2}+3 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(547\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.97 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 64 \, a b^{3} \cos \left (d x + c\right ) + 192 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, b^{4} + {\left (24 \, a^{2} b^{2} - 5 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.14 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {3 \, b^{4} {\left (\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {192 \, a^{3} b}{\cos \left (d x + c\right )} - \frac {64 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a b^{3}}{\cos \left (d x + c\right )^{3}}}{48 \, d} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (160) = 320\).

Time = 0.45 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.93 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 33 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 192 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 288 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 256 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, a^{3} b - 64 \, a b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 25.49 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.65 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^4-6\,a^2\,b^2+\frac {3\,b^4}{4}\right )}{d}-\frac {\frac {16\,a\,b^3}{3}-8\,a^3\,b+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,b^4}{4}-6\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {3\,b^4}{4}-6\,a^2\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {11\,b^4}{4}-6\,a^2\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {11\,b^4}{4}-6\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (16\,a\,b^3-24\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {64\,a\,b^3}{3}-24\,a^3\,b\right )+8\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

(atanh(tan(c/2 + (d*x)/2))*(2*a^4 + (3*b^4)/4 - 6*a^2*b^2))/d - ((16*a*b^3)/3 - 8*a^3*b + tan(c/2 + (d*x)/2)*(
(3*b^4)/4 - 6*a^2*b^2) + tan(c/2 + (d*x)/2)^7*((3*b^4)/4 - 6*a^2*b^2) - tan(c/2 + (d*x)/2)^3*((11*b^4)/4 - 6*a
^2*b^2) - tan(c/2 + (d*x)/2)^5*((11*b^4)/4 - 6*a^2*b^2) + tan(c/2 + (d*x)/2)^4*(16*a*b^3 - 24*a^3*b) - tan(c/2
+ (d*x)/2)^2*((64*a*b^3)/3 - 24*a^3*b) + 8*a^3*b*tan(c/2 + (d*x)/2)^6)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2
+ (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))