3.55 \(\int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=168 \[ \frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {2 a b \sec ^5(c+d x)}{5 d}-\frac {b^2 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {b^2 \tan (c+d x) \sec ^5(c+d x)}{6 d}-\frac {b^2 \tan (c+d x) \sec ^3(c+d x)}{24 d}-\frac {b^2 \tan (c+d x) \sec (c+d x)}{16 d} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 11, 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} \[ \frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {2 a b \sec ^5(c+d x)}{5 d}-\frac {b^2 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {b^2 \tan (c+d x) \sec ^5(c+d x)}{6 d}-\frac {b^2 \tan (c+d x) \sec ^3(c+d x)}{24 d}-\frac {b^2 \tan (c+d x) \sec (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(3*a^2*ArcTanh[Sin[c + d*x]])/(8*d) - (b^2*ArcTanh[Sin[c + d*x]])/(16*d) + (2*a*b*Sec[c + d*x]^5)/(5*d) + (3*a
^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (b^2*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a^2*Sec[c + d*x]^3*Tan[c + d*x
])/(4*d) - (b^2*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + (b^2*Sec[c + d*x]^5*Tan[c + d*x])/(6*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 ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx &=\int \left (a^2 \sec ^5(c+d x)+2 a b \sec ^5(c+d x) \tan (c+d x)+b^2 \sec ^5(c+d x) \tan ^2(c+d x)\right ) \, dx\\ &=a^2 \int \sec ^5(c+d x) \, dx+(2 a b) \int \sec ^5(c+d x) \tan (c+d x) \, dx+b^2 \int \sec ^5(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b^2 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{4} \left (3 a^2\right ) \int \sec ^3(c+d x) \, dx-\frac {1}{6} b^2 \int \sec ^5(c+d x) \, dx+\frac {(2 a b) \operatorname {Subst}\left (\int x^4 \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {2 a b \sec ^5(c+d x)}{5 d}+\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {b^2 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {b^2 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{8} \left (3 a^2\right ) \int \sec (c+d x) \, dx-\frac {1}{8} b^2 \int \sec ^3(c+d x) \, dx\\ &=\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {2 a b \sec ^5(c+d x)}{5 d}+\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{8 d}-\frac {b^2 \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {b^2 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {b^2 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{16} b^2 \int \sec (c+d x) \, dx\\ &=\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {b^2 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {2 a b \sec ^5(c+d x)}{5 d}+\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{8 d}-\frac {b^2 \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {b^2 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {b^2 \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 104, normalized size = 0.62 \[ \frac {15 \left (6 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))+10 \left (6 a^2-b^2\right ) \tan (c+d x) \sec ^3(c+d x)+15 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)+8 b \sec ^5(c+d x) (12 a+5 b \tan (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(15*(6*a^2 - b^2)*ArcTanh[Sin[c + d*x]] + 15*(6*a^2 - b^2)*Sec[c + d*x]*Tan[c + d*x] + 10*(6*a^2 - b^2)*Sec[c
+ d*x]^3*Tan[c + d*x] + 8*b*Sec[c + d*x]^5*(12*a + 5*b*Tan[c + d*x]))/(240*d)

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fricas [A]  time = 0.53, size = 142, normalized size = 0.85 \[ \frac {15 \, {\left (6 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (6 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 192 \, a b \cos \left (d x + c\right ) + 10 \, {\left (3 \, {\left (6 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (6 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/480*(15*(6*a^2 - b^2)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(6*a^2 - b^2)*cos(d*x + c)^6*log(-sin(d*x +
c) + 1) + 192*a*b*cos(d*x + c) + 10*(3*(6*a^2 - b^2)*cos(d*x + c)^4 + 2*(6*a^2 - b^2)*cos(d*x + c)^2 + 8*b^2)*
sin(d*x + c))/(d*cos(d*x + c)^6)

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giac [B]  time = 0.53, size = 343, normalized size = 2.04 \[ \frac {15 \, {\left (6 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (6 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 15 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 480 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 210 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 235 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 480 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 390 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 960 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 390 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 960 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 210 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 235 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 96 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, a b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/240*(15*(6*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(6*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1
)) + 2*(150*a^2*tan(1/2*d*x + 1/2*c)^11 + 15*b^2*tan(1/2*d*x + 1/2*c)^11 - 480*a*b*tan(1/2*d*x + 1/2*c)^10 - 2
10*a^2*tan(1/2*d*x + 1/2*c)^9 + 235*b^2*tan(1/2*d*x + 1/2*c)^9 + 480*a*b*tan(1/2*d*x + 1/2*c)^8 + 60*a^2*tan(1
/2*d*x + 1/2*c)^7 + 390*b^2*tan(1/2*d*x + 1/2*c)^7 - 960*a*b*tan(1/2*d*x + 1/2*c)^6 + 60*a^2*tan(1/2*d*x + 1/2
*c)^5 + 390*b^2*tan(1/2*d*x + 1/2*c)^5 + 960*a*b*tan(1/2*d*x + 1/2*c)^4 - 210*a^2*tan(1/2*d*x + 1/2*c)^3 + 235
*b^2*tan(1/2*d*x + 1/2*c)^3 - 96*a*b*tan(1/2*d*x + 1/2*c)^2 + 150*a^2*tan(1/2*d*x + 1/2*c) + 15*b^2*tan(1/2*d*
x + 1/2*c) + 96*a*b)/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d

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maple [A]  time = 10.99, size = 189, normalized size = 1.12 \[ \frac {a^{2} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 a b}{5 d \cos \left (d x +c \right )^{5}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{4}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{2}}+\frac {b^{2} \sin \left (d x +c \right )}{16 d}-\frac {b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^2,x)

[Out]

1/4*a^2*sec(d*x+c)^3*tan(d*x+c)/d+3/8*a^2*sec(d*x+c)*tan(d*x+c)/d+3/8/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+2/5/d*a*
b/cos(d*x+c)^5+1/6/d*b^2*sin(d*x+c)^3/cos(d*x+c)^6+1/8/d*b^2*sin(d*x+c)^3/cos(d*x+c)^4+1/16/d*b^2*sin(d*x+c)^3
/cos(d*x+c)^2+1/16*b^2*sin(d*x+c)/d-1/16/d*b^2*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.34, size = 180, normalized size = 1.07 \[ \frac {5 \, b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {192 \, a b}{\cos \left (d x + c\right )^{5}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/480*(5*b^2*(2*(3*sin(d*x + c)^5 - 8*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*
sin(d*x + c)^2 - 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 30*a^2*(2*(3*sin(d*x + c)^3 - 5*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)) + 192*
a*b/cos(d*x + c)^5)/d

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mupad [B]  time = 3.26, size = 328, normalized size = 1.95 \[ \frac {\left (\frac {5\,a^2}{4}+\frac {b^2}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {47\,b^2}{24}-\frac {7\,a^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (\frac {a^2}{2}+\frac {13\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {a^2}{2}+\frac {13\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {47\,b^2}{24}-\frac {7\,a^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}+\left (\frac {5\,a^2}{4}+\frac {b^2}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {4\,a\,b}{5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^2}{4}-\frac {b^2}{8}\right )}{d} \]

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)^7,x)

[Out]

((4*a*b)/5 + tan(c/2 + (d*x)/2)^5*(a^2/2 + (13*b^2)/4) + tan(c/2 + (d*x)/2)^7*(a^2/2 + (13*b^2)/4) + tan(c/2 +
(d*x)/2)^11*((5*a^2)/4 + b^2/8) - tan(c/2 + (d*x)/2)^3*((7*a^2)/4 - (47*b^2)/24) - tan(c/2 + (d*x)/2)^9*((7*a
^2)/4 - (47*b^2)/24) + tan(c/2 + (d*x)/2)*((5*a^2)/4 + b^2/8) - (4*a*b*tan(c/2 + (d*x)/2)^2)/5 + 8*a*b*tan(c/2
+ (d*x)/2)^4 - 8*a*b*tan(c/2 + (d*x)/2)^6 + 4*a*b*tan(c/2 + (d*x)/2)^8 - 4*a*b*tan(c/2 + (d*x)/2)^10)/(d*(15*
tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2
+ (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) + (atanh(tan(c/2 + (d*x)/2))*((3*a^2)/4 - b^2/8))/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)**7*(a*cos(d*x+c)+b*sin(d*x+c))**2,x)

[Out]

Timed out

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