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

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

[Out]

a^4*arctanh(sin(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

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Rubi [A]  time = 0.19, antiderivative size = 168, normalized size of antiderivative = 1.00, 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 8

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

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Mathematica [B]  time = 6.23, 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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fricas [A]  time = 0.76, size = 163, normalized size = 0.97 \[ \frac {3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 64 \, a b^{3} \cos \left (d x + c\right ) + 192 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, b^{4} + {\left (24 \, a^{2} b^{2} - 5 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

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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giac [B]  time = 0.48, size = 325, normalized size = 1.93 \[ \frac {3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 33 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 192 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 288 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 256 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, a^{3} b - 64 \, a b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

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

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maple [A]  time = 37.63, size = 297, normalized size = 1.77 \[ \frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a^{3} b}{d \cos \left (d x +c \right )}+\frac {3 a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {3 a^{2} b^{2} \sin \left (d x +c \right )}{d}-\frac {3 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}-\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right ) a \,b^{3}}{3 d}-\frac {8 a \,b^{3} \cos \left (d x +c \right )}{3 d}+\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {3 b^{4} \sin \left (d x +c \right )}{8 d}+\frac {3 b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

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*sin(d*x+c)^2*cos(d*x+c)*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 = 0.33, size = 192, normalized size = 1.14 \[ \frac {3 \, b^{4} {\left (\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {192 \, a^{3} b}{\cos \left (d x + c\right )} - \frac {64 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a b^{3}}{\cos \left (d x + c\right )^{3}}}{48 \, d} \]

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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mupad [B]  time = 4.21, size = 278, normalized size = 1.65 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^4-6\,a^2\,b^2+\frac {3\,b^4}{4}\right )}{d}-\frac {\frac {16\,a\,b^3}{3}-8\,a^3\,b+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,b^4}{4}-6\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {3\,b^4}{4}-6\,a^2\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {11\,b^4}{4}-6\,a^2\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {11\,b^4}{4}-6\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (16\,a\,b^3-24\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {64\,a\,b^3}{3}-24\,a^3\,b\right )+8\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

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)^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))

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

[Out]

Timed out

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