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

Optimal. Leaf size=120 \[ \frac {a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d}+\frac {2 a b \sec ^3(c+d x)}{3 d}-\frac {b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {b^2 \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3090, 3768, 3770, 2606, 30, 2611} \[ \frac {a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d}+\frac {2 a b \sec ^3(c+d x)}{3 d}-\frac {b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {b^2 \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*ArcTanh[Sin[c + d*x]])/(2*d) - (b^2*ArcTanh[Sin[c + d*x]])/(8*d) + (2*a*b*Sec[c + d*x]^3)/(3*d) + (a^2*Se
c[c + d*x]*Tan[c + d*x])/(2*d) - (b^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*
d)

Rule 30

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

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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))^2 \, dx &=\int \left (a^2 \sec ^3(c+d x)+2 a b \sec ^3(c+d x) \tan (c+d x)+b^2 \sec ^3(c+d x) \tan ^2(c+d x)\right ) \, dx\\ &=a^2 \int \sec ^3(c+d x) \, dx+(2 a b) \int \sec ^3(c+d x) \tan (c+d x) \, dx+b^2 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{2} a^2 \int \sec (c+d x) \, dx-\frac {1}{4} b^2 \int \sec ^3(c+d x) \, dx+\frac {(2 a b) \operatorname {Subst}\left (\int x^2 \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {2 a b \sec ^3(c+d x)}{3 d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{8} b^2 \int \sec (c+d x) \, dx\\ &=\frac {a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {2 a b \sec ^3(c+d x)}{3 d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 120, normalized size = 1.00 \[ \frac {a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d}+\frac {2 a b \sec ^3(c+d x)}{3 d}-\frac {b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {b^2 \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*ArcTanh[Sin[c + d*x]])/(2*d) - (b^2*ArcTanh[Sin[c + d*x]])/(8*d) + (2*a*b*Sec[c + d*x]^3)/(3*d) + (a^2*Se
c[c + d*x]*Tan[c + d*x])/(2*d) - (b^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*
d)

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fricas [A]  time = 0.59, size = 120, normalized size = 1.00 \[ \frac {3 \, {\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 32 \, a b \cos \left (d x + c\right ) + 6 \, {\left ({\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

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))^2,x, algorithm="fricas")

[Out]

1/48*(3*(4*a^2 - b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*a^2 - b^2)*cos(d*x + c)^4*log(-sin(d*x + c)
+ 1) + 32*a*b*cos(d*x + c) + 6*((4*a^2 - b^2)*cos(d*x + c)^2 + 2*b^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [B]  time = 0.33, size = 249, normalized size = 2.08 \[ \frac {3 \, {\left (4 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 16 \, a b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

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))^2,x, algorithm="giac")

[Out]

1/24*(3*(4*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))
+ 2*(12*a^2*tan(1/2*d*x + 1/2*c)^7 + 3*b^2*tan(1/2*d*x + 1/2*c)^7 - 48*a*b*tan(1/2*d*x + 1/2*c)^6 - 12*a^2*tan
(1/2*d*x + 1/2*c)^5 + 21*b^2*tan(1/2*d*x + 1/2*c)^5 + 48*a*b*tan(1/2*d*x + 1/2*c)^4 - 12*a^2*tan(1/2*d*x + 1/2
*c)^3 + 21*b^2*tan(1/2*d*x + 1/2*c)^3 - 16*a*b*tan(1/2*d*x + 1/2*c)^2 + 12*a^2*tan(1/2*d*x + 1/2*c) + 3*b^2*ta
n(1/2*d*x + 1/2*c) + 16*a*b)/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 1.96, size = 143, normalized size = 1.19 \[ \frac {a^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a b}{3 d \cos \left (d x +c \right )^{3}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {b^{2} \sin \left (d x +c \right )}{8 d}-\frac {b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

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))^2,x)

[Out]

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

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maxima [A]  time = 0.33, size = 129, normalized size = 1.08 \[ \frac {3 \, b^{2} {\left (\frac {2 \, {\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {32 \, a b}{\cos \left (d x + c\right )^{3}}}{48 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

1/48*(3*b^2*(2*(sin(d*x + c)^3 + sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - log(sin(d*x + c) + 1)
+ log(sin(d*x + c) - 1)) - 12*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x
+ c) - 1)) + 32*a*b/cos(d*x + c)^3)/d

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mupad [B]  time = 3.16, size = 216, normalized size = 1.80 \[ \frac {\left (a^2+\frac {b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {7\,b^2}{4}-a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {7\,b^2}{4}-a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\left (a^2+\frac {b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {4\,a\,b}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-\frac {b^2}{4}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((4*a*b)/3 + tan(c/2 + (d*x)/2)*(a^2 + b^2/4) + tan(c/2 + (d*x)/2)^7*(a^2 + b^2/4) - tan(c/2 + (d*x)/2)^3*(a^2
- (7*b^2)/4) - tan(c/2 + (d*x)/2)^5*(a^2 - (7*b^2)/4) - (4*a*b*tan(c/2 + (d*x)/2)^2)/3 + 4*a*b*tan(c/2 + (d*x
)/2)^4 - 4*a*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)) + (atanh(tan(c/2 + (d*x)/2))*(a^2 - b^2/4))/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)**5*(a*cos(d*x+c)+b*sin(d*x+c))**2,x)

[Out]

Timed out

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