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

Optimal. Leaf size=150 \[ \frac{2 a^2 b^2 \sin ^3(c+d x)}{d}-\frac{4 a^3 b \cos ^3(c+d x)}{3 d}-\frac{a^4 \sin ^3(c+d x)}{3 d}+\frac{a^4 \sin (c+d x)}{d}+\frac{4 a b^3 \cos ^3(c+d x)}{3 d}-\frac{4 a b^3 \cos (c+d x)}{d}-\frac{b^4 \sin ^3(c+d x)}{3 d}-\frac{b^4 \sin (c+d x)}{d}+\frac{b^4 \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.15296, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3090, 2633, 2565, 30, 2564, 2592, 302, 206} \[ \frac{2 a^2 b^2 \sin ^3(c+d x)}{d}-\frac{4 a^3 b \cos ^3(c+d x)}{3 d}-\frac{a^4 \sin ^3(c+d x)}{3 d}+\frac{a^4 \sin (c+d x)}{d}+\frac{4 a b^3 \cos ^3(c+d x)}{3 d}-\frac{4 a b^3 \cos (c+d x)}{d}-\frac{b^4 \sin ^3(c+d x)}{3 d}-\frac{b^4 \sin (c+d x)}{d}+\frac{b^4 \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^4*ArcTanh[Sin[c + d*x]])/d - (4*a*b^3*Cos[c + d*x])/d - (4*a^3*b*Cos[c + d*x]^3)/(3*d) + (4*a*b^3*Cos[c + d
*x]^3)/(3*d) + (a^4*Sin[c + d*x])/d - (b^4*Sin[c + d*x])/d - (a^4*Sin[c + d*x]^3)/(3*d) + (2*a^2*b^2*Sin[c + d
*x]^3)/d - (b^4*Sin[c + d*x]^3)/(3*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 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 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 30

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

Rubi steps

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

Mathematica [A]  time = 0.994209, size = 181, normalized size = 1.21 \[ \frac{18 a^2 b^2 \sin (c+d x)-6 a^2 b^2 \sin (3 (c+d x))-12 a b \left (a^2+3 b^2\right ) \cos (c+d x)+\left (4 a b^3-4 a^3 b\right ) \cos (3 (c+d x))+9 a^4 \sin (c+d x)+a^4 \sin (3 (c+d x))-15 b^4 \sin (c+d x)+b^4 \sin (3 (c+d x))-12 b^4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 b^4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*a*b*(a^2 + 3*b^2)*Cos[c + d*x] + (-4*a^3*b + 4*a*b^3)*Cos[3*(c + d*x)] - 12*b^4*Log[Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]] + 12*b^4*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 9*a^4*Sin[c + d*x] + 18*a^2*b^2*Sin[c + d*
x] - 15*b^4*Sin[c + d*x] + a^4*Sin[3*(c + d*x)] - 6*a^2*b^2*Sin[3*(c + d*x)] + b^4*Sin[3*(c + d*x)])/(12*d)

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Maple [A]  time = 0.12, size = 163, normalized size = 1.1 \begin{align*}{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{4}}{3\,d}}+{\frac{2\,{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}-{\frac{4\,{a}^{3}b \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+2\,{\frac{{a}^{2}{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}-{\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 ) ^{3}}{3\,d}}-{\frac{{b}^{4}\sin \left ( dx+c \right ) }{d}}+{\frac{{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.17797, size = 170, normalized size = 1.13 \begin{align*} -\frac{8 \, a^{3} b \cos \left (d x + c\right )^{3} - 12 \, a^{2} 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^{4} - 8 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a b^{3} +{\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 )} b^{4}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(8*a^3*b*cos(d*x + c)^3 - 12*a^2*b^2*sin(d*x + c)^3 + 2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^4 - 8*(cos(d*
x + c)^3 - 3*cos(d*x + c))*a*b^3 + (2*sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1) + 6*s
in(d*x + c))*b^4)/d

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Fricas [A]  time = 0.523691, size = 289, normalized size = 1.93 \begin{align*} -\frac{24 \, a b^{3} \cos \left (d x + c\right ) - 3 \, b^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, b^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} - 2 \,{\left (2 \, a^{4} + 6 \, a^{2} b^{2} - 4 \, b^{4} +{\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 )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.22915, size = 293, normalized size = 1.95 \begin{align*} \frac{3 \, b^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, b^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 10 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a^{3} b - 8 \, a b^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*b^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*b^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(3*a^4*tan(1/2*d*
x + 1/2*c)^5 - 3*b^4*tan(1/2*d*x + 1/2*c)^5 - 12*a^3*b*tan(1/2*d*x + 1/2*c)^4 + 2*a^4*tan(1/2*d*x + 1/2*c)^3 +
24*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 10*b^4*tan(1/2*d*x + 1/2*c)^3 - 24*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^4*t
an(1/2*d*x + 1/2*c) - 3*b^4*tan(1/2*d*x + 1/2*c) - 4*a^3*b - 8*a*b^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d