3.2.68 \(\int (a+b \sec (c+d x)) \sin ^6(c+d x) \, dx\) [168]

Optimal. Leaf size=127 \[ \frac {5 a x}{16}+\frac {b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {b \sin ^5(c+d x)}{5 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d} \]

[Out]

5/16*a*x+b*arctanh(sin(d*x+c))/d-b*sin(d*x+c)/d-5/16*a*cos(d*x+c)*sin(d*x+c)/d-1/3*b*sin(d*x+c)^3/d-5/24*a*cos
(d*x+c)*sin(d*x+c)^3/d-1/5*b*sin(d*x+c)^5/d-1/6*a*cos(d*x+c)*sin(d*x+c)^5/d

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Rubi [A]
time = 0.10, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3957, 2917, 2672, 308, 212, 2715, 8} \begin {gather*} -\frac {a \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {5 a \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {5 a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a x}{16}-\frac {b \sin ^5(c+d x)}{5 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin (c+d x)}{d}+\frac {b \tanh ^{-1}(\sin (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])*Sin[c + d*x]^6,x]

[Out]

(5*a*x)/16 + (b*ArcTanh[Sin[c + d*x]])/d - (b*Sin[c + d*x])/d - (5*a*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (b*Si
n[c + d*x]^3)/(3*d) - (5*a*Cos[c + d*x]*Sin[c + d*x]^3)/(24*d) - (b*Sin[c + d*x]^5)/(5*d) - (a*Cos[c + d*x]*Si
n[c + d*x]^5)/(6*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 212

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

Rule 308

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 2672

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 2715

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

Rule 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x)) \sin ^6(c+d x) \, dx &=-\int (-b-a \cos (c+d x)) \sin ^5(c+d x) \tan (c+d x) \, dx\\ &=a \int \sin ^6(c+d x) \, dx+b \int \sin ^5(c+d x) \tan (c+d x) \, dx\\ &=-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {1}{6} (5 a) \int \sin ^4(c+d x) \, dx+\frac {b \text {Subst}\left (\int \frac {x^6}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {1}{8} (5 a) \int \sin ^2(c+d x) \, dx+\frac {b \text {Subst}\left (\int \left (-1-x^2-x^4+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {b \sin (c+d x)}{d}-\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {b \sin ^5(c+d x)}{5 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {1}{16} (5 a) \int 1 \, dx+\frac {b \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {5 a x}{16}+\frac {b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {b \sin ^5(c+d x)}{5 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 118, normalized size = 0.93 \begin {gather*} \frac {5 a (c+d x)}{16 d}+\frac {b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin ^5(c+d x)}{5 d}-\frac {15 a \sin (2 (c+d x))}{64 d}+\frac {3 a \sin (4 (c+d x))}{64 d}-\frac {a \sin (6 (c+d x))}{192 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])*Sin[c + d*x]^6,x]

[Out]

(5*a*(c + d*x))/(16*d) + (b*ArcTanh[Sin[c + d*x]])/d - (b*Sin[c + d*x])/d - (b*Sin[c + d*x]^3)/(3*d) - (b*Sin[
c + d*x]^5)/(5*d) - (15*a*Sin[2*(c + d*x)])/(64*d) + (3*a*Sin[4*(c + d*x)])/(64*d) - (a*Sin[6*(c + d*x)])/(192
*d)

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Maple [A]
time = 0.10, size = 96, normalized size = 0.76

method	result	size
derivativedivides	\(\frac {b \left (-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(96\)
default	\(\frac {b \left (-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(96\)
risch	\(\frac {5 a x}{16}+\frac {11 i b \,{\mathrm e}^{i \left (d x +c \right )}}{16 d}-\frac {11 i b \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a \sin \left (6 d x +6 c \right )}{192 d}-\frac {b \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 a \sin \left (4 d x +4 c \right )}{64 d}+\frac {7 b \sin \left (3 d x +3 c \right )}{48 d}-\frac {15 a \sin \left (2 d x +2 c \right )}{64 d}\)	\(150\)
norman	\(\frac {\frac {5 a x}{16}+\frac {15 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {75 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {25 a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {75 a x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {15 a x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {5 a x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {\left (5 a -16 b \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {\left (5 a +16 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (85 a -304 b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {\left (85 a +304 b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (165 a -688 b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}-\frac {\left (165 a +688 b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(284\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))*sin(d*x+c)^6,x,method=_RETURNVERBOSE)

[Out]

1/d*(b*(-1/5*sin(d*x+c)^5-1/3*sin(d*x+c)^3-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+a*(-1/6*(sin(d*x+c)^5+5/4*sin
(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+5/16*d*x+5/16*c))

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Maxima [A]
time = 0.27, size = 106, normalized size = 0.83 \begin {gather*} \frac {5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a - 32 \, {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} b}{960 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*sin(d*x+c)^6,x, algorithm="maxima")

[Out]

1/960*(5*(4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a - 32*(6*sin(d*x +
c)^5 + 10*sin(d*x + c)^3 - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1) + 30*sin(d*x + c))*b)/d

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Fricas [A]
time = 4.98, size = 102, normalized size = 0.80 \begin {gather*} \frac {75 \, a d x + 120 \, b \log \left (\sin \left (d x + c\right ) + 1\right ) - 120 \, b \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (40 \, a \cos \left (d x + c\right )^{5} + 48 \, b \cos \left (d x + c\right )^{4} - 130 \, a \cos \left (d x + c\right )^{3} - 176 \, b \cos \left (d x + c\right )^{2} + 165 \, a \cos \left (d x + c\right ) + 368 \, b\right )} \sin \left (d x + c\right )}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*sin(d*x+c)^6,x, algorithm="fricas")

[Out]

1/240*(75*a*d*x + 120*b*log(sin(d*x + c) + 1) - 120*b*log(-sin(d*x + c) + 1) - (40*a*cos(d*x + c)^5 + 48*b*cos
(d*x + c)^4 - 130*a*cos(d*x + c)^3 - 176*b*cos(d*x + c)^2 + 165*a*cos(d*x + c) + 368*b)*sin(d*x + c))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sec {\left (c + d x \right )}\right ) \sin ^{6}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*sin(d*x+c)**6,x)

[Out]

Integral((a + b*sec(c + d*x))*sin(c + d*x)**6, x)

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Giac [A]
time = 0.45, size = 228, normalized size = 1.80 \begin {gather*} \frac {75 \, {\left (d x + c\right )} a + 240 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 240 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (75 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 240 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 425 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1520 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 990 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4128 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 990 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4128 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 425 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1520 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 75 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 240 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*sin(d*x+c)^6,x, algorithm="giac")

[Out]

1/240*(75*(d*x + c)*a + 240*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 240*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) +
2*(75*a*tan(1/2*d*x + 1/2*c)^11 - 240*b*tan(1/2*d*x + 1/2*c)^11 + 425*a*tan(1/2*d*x + 1/2*c)^9 - 1520*b*tan(1/
2*d*x + 1/2*c)^9 + 990*a*tan(1/2*d*x + 1/2*c)^7 - 4128*b*tan(1/2*d*x + 1/2*c)^7 - 990*a*tan(1/2*d*x + 1/2*c)^5
- 4128*b*tan(1/2*d*x + 1/2*c)^5 - 425*a*tan(1/2*d*x + 1/2*c)^3 - 1520*b*tan(1/2*d*x + 1/2*c)^3 - 75*a*tan(1/2
*d*x + 1/2*c) - 240*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d

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Mupad [B]
time = 2.17, size = 332, normalized size = 2.61 \begin {gather*} \frac {5\,a\,\mathrm {atan}\left (\frac {125\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,\left (\frac {125\,a^3}{64}+20\,a\,b^2\right )}+\frac {20\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {125\,a^3}{64}+20\,a\,b^2}\right )}{8\,d}+\frac {2\,b\,\mathrm {atanh}\left (\frac {64\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {25\,a^2\,b}{4}+64\,b^3}+\frac {25\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {25\,a^2\,b}{4}+64\,b^3\right )}\right )}{d}-\frac {\left (2\,b-\frac {5\,a}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {38\,b}{3}-\frac {85\,a}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {172\,b}{5}-\frac {33\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {33\,a}{4}+\frac {172\,b}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {85\,a}{24}+\frac {38\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,a}{8}+2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^6*(a + b/cos(c + d*x)),x)

[Out]

(5*a*atan((125*a^3*tan(c/2 + (d*x)/2))/(64*(20*a*b^2 + (125*a^3)/64)) + (20*a*b^2*tan(c/2 + (d*x)/2))/(20*a*b^
2 + (125*a^3)/64)))/(8*d) + (2*b*atanh((64*b^3*tan(c/2 + (d*x)/2))/((25*a^2*b)/4 + 64*b^3) + (25*a^2*b*tan(c/2
+ (d*x)/2))/(4*((25*a^2*b)/4 + 64*b^3))))/d - (tan(c/2 + (d*x)/2)*((5*a)/8 + 2*b) - tan(c/2 + (d*x)/2)^11*((5
*a)/8 - 2*b) + tan(c/2 + (d*x)/2)^3*((85*a)/24 + (38*b)/3) - tan(c/2 + (d*x)/2)^9*((85*a)/24 - (38*b)/3) + tan
(c/2 + (d*x)/2)^5*((33*a)/4 + (172*b)/5) - tan(c/2 + (d*x)/2)^7*((33*a)/4 - (172*b)/5))/(d*(6*tan(c/2 + (d*x)/
2)^2 + 15*tan(c/2 + (d*x)/2)^4 + 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))

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