3.99 \(\int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\)

Optimal. Leaf size=205 \[ -\frac {a^5 \sin ^3(c+d x)}{3 d}+\frac {a^5 \sin (c+d x)}{d}-\frac {5 a^4 b \cos ^3(c+d x)}{3 d}+\frac {10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac {10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac {10 a^2 b^3 \cos (c+d x)}{d}-\frac {5 a b^4 \sin ^3(c+d x)}{3 d}-\frac {5 a b^4 \sin (c+d x)}{d}+\frac {5 a b^4 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b^5 \cos ^3(c+d x)}{3 d}+\frac {2 b^5 \cos (c+d x)}{d}+\frac {b^5 \sec (c+d x)}{d} \]

[Out]

5*a*b^4*arctanh(sin(d*x+c))/d-10*a^2*b^3*cos(d*x+c)/d+2*b^5*cos(d*x+c)/d-5/3*a^4*b*cos(d*x+c)^3/d+10/3*a^2*b^3
*cos(d*x+c)^3/d-1/3*b^5*cos(d*x+c)^3/d+b^5*sec(d*x+c)/d+a^5*sin(d*x+c)/d-5*a*b^4*sin(d*x+c)/d-1/3*a^5*sin(d*x+
c)^3/d+10/3*a^3*b^2*sin(d*x+c)^3/d-5/3*a*b^4*sin(d*x+c)^3/d

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Rubi [A]  time = 0.22, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3090, 2633, 2565, 30, 2564, 2592, 302, 206, 2590, 270} \[ \frac {10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac {10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac {10 a^2 b^3 \cos (c+d x)}{d}-\frac {5 a^4 b \cos ^3(c+d x)}{3 d}-\frac {a^5 \sin ^3(c+d x)}{3 d}+\frac {a^5 \sin (c+d x)}{d}-\frac {5 a b^4 \sin ^3(c+d x)}{3 d}-\frac {5 a b^4 \sin (c+d x)}{d}+\frac {5 a b^4 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b^5 \cos ^3(c+d x)}{3 d}+\frac {2 b^5 \cos (c+d x)}{d}+\frac {b^5 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(5*a*b^4*ArcTanh[Sin[c + d*x]])/d - (10*a^2*b^3*Cos[c + d*x])/d + (2*b^5*Cos[c + d*x])/d - (5*a^4*b*Cos[c + d*
x]^3)/(3*d) + (10*a^2*b^3*Cos[c + d*x]^3)/(3*d) - (b^5*Cos[c + d*x]^3)/(3*d) + (b^5*Sec[c + d*x])/d + (a^5*Sin
[c + d*x])/d - (5*a*b^4*Sin[c + d*x])/d - (a^5*Sin[c + d*x]^3)/(3*d) + (10*a^3*b^2*Sin[c + d*x]^3)/(3*d) - (5*
a*b^4*Sin[c + d*x]^3)/(3*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=\int \left (a^5 \cos ^3(c+d x)+5 a^4 b \cos ^2(c+d x) \sin (c+d x)+10 a^3 b^2 \cos (c+d x) \sin ^2(c+d x)+10 a^2 b^3 \sin ^3(c+d x)+5 a b^4 \sin ^3(c+d x) \tan (c+d x)+b^5 \sin ^3(c+d x) \tan ^2(c+d x)\right ) \, dx\\ &=a^5 \int \cos ^3(c+d x) \, dx+\left (5 a^4 b\right ) \int \cos ^2(c+d x) \sin (c+d x) \, dx+\left (10 a^3 b^2\right ) \int \cos (c+d x) \sin ^2(c+d x) \, dx+\left (10 a^2 b^3\right ) \int \sin ^3(c+d x) \, dx+\left (5 a b^4\right ) \int \sin ^3(c+d x) \tan (c+d x) \, dx+b^5 \int \sin ^3(c+d x) \tan ^2(c+d x) \, dx\\ &=-\frac {a^5 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac {\left (5 a^4 b\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (10 a^3 b^2\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (10 a^2 b^3\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (5 a b^4\right ) \operatorname {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {b^5 \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {10 a^2 b^3 \cos (c+d x)}{d}-\frac {5 a^4 b \cos ^3(c+d x)}{3 d}+\frac {10 a^2 b^3 \cos ^3(c+d x)}{3 d}+\frac {a^5 \sin (c+d x)}{d}-\frac {a^5 \sin ^3(c+d x)}{3 d}+\frac {10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac {\left (5 a b^4\right ) \operatorname {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac {b^5 \operatorname {Subst}\left (\int \left (-2+\frac {1}{x^2}+x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {10 a^2 b^3 \cos (c+d x)}{d}+\frac {2 b^5 \cos (c+d x)}{d}-\frac {5 a^4 b \cos ^3(c+d x)}{3 d}+\frac {10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac {b^5 \cos ^3(c+d x)}{3 d}+\frac {b^5 \sec (c+d x)}{d}+\frac {a^5 \sin (c+d x)}{d}-\frac {5 a b^4 \sin (c+d x)}{d}-\frac {a^5 \sin ^3(c+d x)}{3 d}+\frac {10 a^3 b^2 \sin ^3(c+d x)}{3 d}-\frac {5 a b^4 \sin ^3(c+d x)}{3 d}+\frac {\left (5 a b^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {5 a b^4 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {10 a^2 b^3 \cos (c+d x)}{d}+\frac {2 b^5 \cos (c+d x)}{d}-\frac {5 a^4 b \cos ^3(c+d x)}{3 d}+\frac {10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac {b^5 \cos ^3(c+d x)}{3 d}+\frac {b^5 \sec (c+d x)}{d}+\frac {a^5 \sin (c+d x)}{d}-\frac {5 a b^4 \sin (c+d x)}{d}-\frac {a^5 \sin ^3(c+d x)}{3 d}+\frac {10 a^3 b^2 \sin ^3(c+d x)}{3 d}-\frac {5 a b^4 \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [B]  time = 6.30, size = 632, normalized size = 3.08 \[ -\frac {b \left (5 a^4+30 a^2 b^2-7 b^4\right ) \cos ^6(c+d x) (a+b \tan (c+d x))^5}{4 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {a \left (a^4-10 a^2 b^2+5 b^4\right ) \sin (3 (c+d x)) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{12 d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b \left (5 a^4-10 a^2 b^2+b^4\right ) \cos (3 (c+d x)) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{12 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {a \left (3 a^4+10 a^2 b^2-25 b^4\right ) \sin (c+d x) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{4 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {b^5 \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {b^5 \cos ^5(c+d x) (a+b \tan (c+d x))^5}{d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b^5 \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {5 a b^4 \cos ^5(c+d x) (a+b \tan (c+d x))^5 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {5 a b^4 \cos ^5(c+d x) (a+b \tan (c+d x))^5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+b \sin (c+d x))^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(b^5*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) - (b*(5*a^4 + 30*a^2*b^2 -
7*b^4)*Cos[c + d*x]^6*(a + b*Tan[c + d*x])^5)/(4*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) - (b*(5*a^4 - 10*a^2*
b^2 + b^4)*Cos[c + d*x]^5*Cos[3*(c + d*x)]*(a + b*Tan[c + d*x])^5)/(12*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5)
- (5*a*b^4*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^5)/(d*(a*Cos[c + d*x]
+ b*Sin[c + d*x])^5) + (5*a*b^4*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^5
)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (b^5*Cos[c + d*x]^5*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^5)/(d*(C
os[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) - (b^5*Cos[c + d*x]^5*Sin[(c + d*x)/2
]*(a + b*Tan[c + d*x])^5)/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (a*(
3*a^4 + 10*a^2*b^2 - 25*b^4)*Cos[c + d*x]^5*Sin[c + d*x]*(a + b*Tan[c + d*x])^5)/(4*d*(a*Cos[c + d*x] + b*Sin[
c + d*x])^5) + (a*(a^4 - 10*a^2*b^2 + 5*b^4)*Cos[c + d*x]^5*Sin[3*(c + d*x)]*(a + b*Tan[c + d*x])^5)/(12*d*(a*
Cos[c + d*x] + b*Sin[c + d*x])^5)

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fricas [A]  time = 0.61, size = 177, normalized size = 0.86 \[ \frac {15 \, a b^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a b^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, b^{5} - 2 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - 12 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left ({\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{5} + 5 \, a^{3} b^{2} - 10 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas")

[Out]

1/6*(15*a*b^4*cos(d*x + c)*log(sin(d*x + c) + 1) - 15*a*b^4*cos(d*x + c)*log(-sin(d*x + c) + 1) + 6*b^5 - 2*(5
*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^4 - 12*(5*a^2*b^3 - b^5)*cos(d*x + c)^2 + 2*((a^5 - 10*a^3*b^2 + 5*a*b
^4)*cos(d*x + c)^3 + 2*(a^5 + 5*a^3*b^2 - 10*a*b^4)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))

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giac [A]  time = 0.56, size = 283, normalized size = 1.38 \[ \frac {15 \, a b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, b^{5}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 50 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{4} b - 20 \, a^{2} b^{3} + 5 \, b^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")

[Out]

1/3*(15*a*b^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a*b^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*b^5/(tan(1/
2*d*x + 1/2*c)^2 - 1) + 2*(3*a^5*tan(1/2*d*x + 1/2*c)^5 - 15*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 15*a^4*b*tan(1/2*d
*x + 1/2*c)^4 + 3*b^5*tan(1/2*d*x + 1/2*c)^4 + 2*a^5*tan(1/2*d*x + 1/2*c)^3 + 40*a^3*b^2*tan(1/2*d*x + 1/2*c)^
3 - 50*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 60*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 + 12*b^5*tan(1/2*d*x + 1/2*c)^2 + 3*a^
5*tan(1/2*d*x + 1/2*c) - 15*a*b^4*tan(1/2*d*x + 1/2*c) - 5*a^4*b - 20*a^2*b^3 + 5*b^5)/(tan(1/2*d*x + 1/2*c)^2
+ 1)^3)/d

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maple [A]  time = 0.24, size = 251, normalized size = 1.22 \[ \frac {\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{5}}{3 d}+\frac {2 a^{5} \sin \left (d x +c \right )}{3 d}-\frac {5 a^{4} b \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {10 a^{3} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {10 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b^{3}}{3 d}-\frac {20 a^{2} b^{3} \cos \left (d x +c \right )}{3 d}-\frac {5 a \,b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {5 a \,b^{4} \sin \left (d x +c \right )}{d}+\frac {5 a \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {8 b^{5} \cos \left (d x +c \right )}{3 d}+\frac {b^{5} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{d}+\frac {4 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) b^{5}}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x)

[Out]

1/3/d*cos(d*x+c)^2*sin(d*x+c)*a^5+2/3*a^5*sin(d*x+c)/d-5/3*a^4*b*cos(d*x+c)^3/d+10/3*a^3*b^2*sin(d*x+c)^3/d-10
/3/d*cos(d*x+c)*sin(d*x+c)^2*a^2*b^3-20/3*a^2*b^3*cos(d*x+c)/d-5/3*a*b^4*sin(d*x+c)^3/d-5*a*b^4*sin(d*x+c)/d+5
/d*a*b^4*ln(sec(d*x+c)+tan(d*x+c))+1/d*b^5*sin(d*x+c)^6/cos(d*x+c)+8/3*b^5*cos(d*x+c)/d+1/d*b^5*cos(d*x+c)*sin
(d*x+c)^4+4/3/d*cos(d*x+c)*sin(d*x+c)^2*b^5

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maxima [A]  time = 0.33, size = 162, normalized size = 0.79 \[ -\frac {10 \, a^{4} b \cos \left (d x + c\right )^{3} - 20 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{5} - 20 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{2} b^{3} + 5 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a b^{4} + 2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} b^{5}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/6*(10*a^4*b*cos(d*x + c)^3 - 20*a^3*b^2*sin(d*x + c)^3 + 2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^5 - 20*(cos(
d*x + c)^3 - 3*cos(d*x + c))*a^2*b^3 + 5*(2*sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)
+ 6*sin(d*x + c))*a*b^4 + 2*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*b^5)/d

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mupad [B]  time = 3.98, size = 277, normalized size = 1.35 \[ \frac {10\,a\,b^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (10\,a\,b^4-2\,a^5\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (10\,a^4\,b-40\,a^2\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {2\,a^5}{3}-\frac {80\,a^3\,b^2}{3}+\frac {70\,a\,b^4}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {2\,a^5}{3}-\frac {80\,a^3\,b^2}{3}+\frac {70\,a\,b^4}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {10\,a^4\,b}{3}-\frac {80\,a^2\,b^3}{3}+\frac {32\,b^5}{3}\right )+\frac {10\,a^4\,b}{3}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (10\,a\,b^4-2\,a^5\right )-\frac {16\,b^5}{3}+\frac {40\,a^2\,b^3}{3}-10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^5/cos(c + d*x)^2,x)

[Out]

(10*a*b^4*atanh(tan(c/2 + (d*x)/2)))/d - (tan(c/2 + (d*x)/2)*(10*a*b^4 - 2*a^5) + tan(c/2 + (d*x)/2)^4*(10*a^4
*b - 40*a^2*b^3) + tan(c/2 + (d*x)/2)^3*((70*a*b^4)/3 + (2*a^5)/3 - (80*a^3*b^2)/3) - tan(c/2 + (d*x)/2)^5*((7
0*a*b^4)/3 + (2*a^5)/3 - (80*a^3*b^2)/3) - tan(c/2 + (d*x)/2)^2*((10*a^4*b)/3 + (32*b^5)/3 - (80*a^2*b^3)/3) +
(10*a^4*b)/3 - tan(c/2 + (d*x)/2)^7*(10*a*b^4 - 2*a^5) - (16*b^5)/3 + (40*a^2*b^3)/3 - 10*a^4*b*tan(c/2 + (d*
x)/2)^6)/(d*(2*tan(c/2 + (d*x)/2)^2 - 2*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)**2*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)

[Out]

Timed out

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