3.82 \(\int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=151 \[ \frac {a^4 \sin (c+d x)}{d}-\frac {4 a^3 b \cos (c+d x)}{d}-\frac {6 a^2 b^2 \sin (c+d x)}{d}+\frac {6 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {4 a b^3 \cos (c+d x)}{d}+\frac {4 a b^3 \sec (c+d x)}{d}+\frac {3 b^4 \sin (c+d x)}{2 d}+\frac {b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}-\frac {3 b^4 \tanh ^{-1}(\sin (c+d x))}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3090, 2637, 2638, 2592, 321, 206, 2590, 14, 288} \[ -\frac {6 a^2 b^2 \sin (c+d x)}{d}+\frac {6 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {4 a^3 b \cos (c+d x)}{d}+\frac {a^4 \sin (c+d x)}{d}+\frac {4 a b^3 \cos (c+d x)}{d}+\frac {4 a b^3 \sec (c+d x)}{d}+\frac {3 b^4 \sin (c+d x)}{2 d}+\frac {b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}-\frac {3 b^4 \tanh ^{-1}(\sin (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(6*a^2*b^2*ArcTanh[Sin[c + d*x]])/d - (3*b^4*ArcTanh[Sin[c + d*x]])/(2*d) - (4*a^3*b*Cos[c + d*x])/d + (4*a*b^
3*Cos[c + d*x])/d + (4*a*b^3*Sec[c + d*x])/d + (a^4*Sin[c + d*x])/d - (6*a^2*b^2*Sin[c + d*x])/d + (3*b^4*Sin[
c + d*x])/(2*d) + (b^4*Sin[c + d*x]*Tan[c + d*x]^2)/(2*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 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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

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

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

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Mathematica [A]  time = 2.56, size = 268, normalized size = 1.77 \[ \frac {4 a^4 \sin (c+d x)-24 a^2 b^2 \sin (c+d x)-16 a b \left (a^2-b^2\right ) \cos (c+d x)-24 a^2 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+24 a^2 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+32 a b^3 \sin ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)+16 a b^3+4 b^4 \sin (c+d x)+\frac {b^4}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b^4}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+6 b^4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 b^4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(16*a*b^3 - 16*a*b*(a^2 - b^2)*Cos[c + d*x] - 24*a^2*b^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*b^4*Log[
Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 24*a^2*b^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 6*b^4*Log[Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]] + b^4/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + 32*a*b^3*Sec[c + d*x]*Sin[(c + d
*x)/2]^2 - b^4/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 4*a^4*Sin[c + d*x] - 24*a^2*b^2*Sin[c + d*x] + 4*b^4*
Sin[c + d*x])/(4*d)

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fricas [A]  time = 1.05, size = 153, normalized size = 1.01 \[ \frac {16 \, a b^{3} \cos \left (d x + c\right ) - 16 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (b^{4} + 2 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/4*(16*a*b^3*cos(d*x + c) - 16*(a^3*b - a*b^3)*cos(d*x + c)^3 + 3*(4*a^2*b^2 - b^4)*cos(d*x + c)^2*log(sin(d*
x + c) + 1) - 3*(4*a^2*b^2 - b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(b^4 + 2*(a^4 - 6*a^2*b^2 + b^4)*c
os(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^2)

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giac [A]  time = 0.40, size = 206, normalized size = 1.36 \[ \frac {3 \, {\left (4 \, a^{2} b^{2} - b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, a^{2} b^{2} - b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {4 \, {\left (a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{3} b + 4 \, a b^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {2 \, {\left (b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, a b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/2*(3*(4*a^2*b^2 - b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*a^2*b^2 - b^4)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 4*(a^4*tan(1/2*d*x + 1/2*c) - 6*a^2*b^2*tan(1/2*d*x + 1/2*c) + b^4*tan(1/2*d*x + 1/2*c) - 4*a^3*b + 4
*a*b^3)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(b^4*tan(1/2*d*x + 1/2*c)^3 - 8*a*b^3*tan(1/2*d*x + 1/2*c)^2 + b^4*ta
n(1/2*d*x + 1/2*c) + 8*a*b^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d

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maple [A]  time = 29.43, size = 211, normalized size = 1.40 \[ \frac {a^{4} \sin \left (d x +c \right )}{d}-\frac {4 a^{3} b \cos \left (d x +c \right )}{d}+\frac {6 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {6 a^{2} b^{2} \sin \left (d x +c \right )}{d}+\frac {4 a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right ) a \,b^{3}}{d}+\frac {8 a \,b^{3} \cos \left (d x +c \right )}{d}+\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d}+\frac {3 b^{4} \sin \left (d x +c \right )}{2 d}-\frac {3 b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

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

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maxima [A]  time = 0.35, size = 142, normalized size = 0.94 \[ -\frac {b^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )} - 16 \, a b^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 12 \, a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} + 16 \, a^{3} b \cos \left (d x + c\right ) - 4 \, a^{4} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/4*(b^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + 3*log(sin(d*x + c) + 1) - 3*log(sin(d*x + c) - 1) - 4*sin(d*x
+ c)) - 16*a*b^3*(1/cos(d*x + c) + cos(d*x + c)) - 12*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)
- 2*sin(d*x + c)) + 16*a^3*b*cos(d*x + c) - 4*a^4*sin(d*x + c))/d

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mupad [B]  time = 2.96, size = 221, normalized size = 1.46 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a^4-24\,a^2\,b^2+2\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^4-12\,a^2\,b^2+3\,b^4\right )-16\,a\,b^3+8\,a^3\,b-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^4-12\,a^2\,b^2+3\,b^4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (16\,a\,b^3-16\,a^3\,b\right )+8\,a^3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,b^4-12\,a^2\,b^2\right )}{d} \]

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)^3,x)

[Out]

(tan(c/2 + (d*x)/2)^3*(4*a^4 + 2*b^4 - 24*a^2*b^2) - tan(c/2 + (d*x)/2)^5*(2*a^4 + 3*b^4 - 12*a^2*b^2) - 16*a*
b^3 + 8*a^3*b - tan(c/2 + (d*x)/2)*(2*a^4 + 3*b^4 - 12*a^2*b^2) + tan(c/2 + (d*x)/2)^2*(16*a*b^3 - 16*a^3*b) +
8*a^3*b*tan(c/2 + (d*x)/2)^4)/(d*(tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2)^6 - 1)) -
(atanh(tan(c/2 + (d*x)/2))*(3*b^4 - 12*a^2*b^2))/d

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

[Out]

Timed out

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