3.185 \(\int (a+b \sec (c+d x))^3 \sin ^5(c+d x) \, dx\)

Optimal. Leaf size=170 \[ \frac{a \left (2 a^2-3 b^2\right ) \cos ^3(c+d x)}{3 d}+\frac{b \left (6 a^2-b^2\right ) \cos ^2(c+d x)}{2 d}-\frac{a \left (a^2-6 b^2\right ) \cos (c+d x)}{d}-\frac{b \left (3 a^2-2 b^2\right ) \log (\cos (c+d x))}{d}-\frac{3 a^2 b \cos ^4(c+d x)}{4 d}-\frac{a^3 \cos ^5(c+d x)}{5 d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{b^3 \sec ^2(c+d x)}{2 d} \]

[Out]

-((a*(a^2 - 6*b^2)*Cos[c + d*x])/d) + (b*(6*a^2 - b^2)*Cos[c + d*x]^2)/(2*d) + (a*(2*a^2 - 3*b^2)*Cos[c + d*x]
^3)/(3*d) - (3*a^2*b*Cos[c + d*x]^4)/(4*d) - (a^3*Cos[c + d*x]^5)/(5*d) - (b*(3*a^2 - 2*b^2)*Log[Cos[c + d*x]]
)/d + (3*a*b^2*Sec[c + d*x])/d + (b^3*Sec[c + d*x]^2)/(2*d)

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Rubi [A]  time = 0.255129, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3872, 2837, 12, 948} \[ \frac{a \left (2 a^2-3 b^2\right ) \cos ^3(c+d x)}{3 d}+\frac{b \left (6 a^2-b^2\right ) \cos ^2(c+d x)}{2 d}-\frac{a \left (a^2-6 b^2\right ) \cos (c+d x)}{d}-\frac{b \left (3 a^2-2 b^2\right ) \log (\cos (c+d x))}{d}-\frac{3 a^2 b \cos ^4(c+d x)}{4 d}-\frac{a^3 \cos ^5(c+d x)}{5 d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{b^3 \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^3*Sin[c + d*x]^5,x]

[Out]

-((a*(a^2 - 6*b^2)*Cos[c + d*x])/d) + (b*(6*a^2 - b^2)*Cos[c + d*x]^2)/(2*d) + (a*(2*a^2 - 3*b^2)*Cos[c + d*x]
^3)/(3*d) - (3*a^2*b*Cos[c + d*x]^4)/(4*d) - (a^3*Cos[c + d*x]^5)/(5*d) - (b*(3*a^2 - 2*b^2)*Log[Cos[c + d*x]]
)/d + (3*a*b^2*Sec[c + d*x])/d + (b^3*Sec[c + d*x]^2)/(2*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 948

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

Rubi steps

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

Mathematica [A]  time = 0.63709, size = 154, normalized size = 0.91 \[ \frac{-60 a \left (5 a^2-42 b^2\right ) \cos (c+d x)+60 \left (9 a^2 b-2 b^3\right ) \cos (2 (c+d x))-45 a^2 b \cos (4 (c+d x))-1440 a^2 b \log (\cos (c+d x))+50 a^3 \cos (3 (c+d x))-6 a^3 \cos (5 (c+d x))-120 a b^2 \cos (3 (c+d x))+1440 a b^2 \sec (c+d x)+240 b^3 \sec ^2(c+d x)+960 b^3 \log (\cos (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^3*Sin[c + d*x]^5,x]

[Out]

(-60*a*(5*a^2 - 42*b^2)*Cos[c + d*x] + 60*(9*a^2*b - 2*b^3)*Cos[2*(c + d*x)] + 50*a^3*Cos[3*(c + d*x)] - 120*a
*b^2*Cos[3*(c + d*x)] - 45*a^2*b*Cos[4*(c + d*x)] - 6*a^3*Cos[5*(c + d*x)] - 1440*a^2*b*Log[Cos[c + d*x]] + 96
0*b^3*Log[Cos[c + d*x]] + 1440*a*b^2*Sec[c + d*x] + 240*b^3*Sec[c + d*x]^2)/(480*d)

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Maple [A]  time = 0.049, size = 266, normalized size = 1.6 \begin{align*} -{\frac{8\,{a}^{3}\cos \left ( dx+c \right ) }{15\,d}}-{\frac{{a}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{4\,{a}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{3\,{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{3\,{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-3\,{\frac{{a}^{2}b\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d\cos \left ( dx+c \right ) }}+8\,{\frac{\cos \left ( dx+c \right ) a{b}^{2}}{d}}+3\,{\frac{a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}\cos \left ( dx+c \right ) }{d}}+4\,{\frac{\cos \left ( dx+c \right ) a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{2\,d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}+2\,{\frac{{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^3*sin(d*x+c)^5,x)

[Out]

-8/15*a^3*cos(d*x+c)/d-1/5/d*a^3*cos(d*x+c)*sin(d*x+c)^4-4/15/d*a^3*cos(d*x+c)*sin(d*x+c)^2-3/4/d*a^2*b*sin(d*
x+c)^4-3/2/d*a^2*b*sin(d*x+c)^2-3*a^2*b*ln(cos(d*x+c))/d+3/d*a*b^2*sin(d*x+c)^6/cos(d*x+c)+8/d*cos(d*x+c)*a*b^
2+3/d*a*b^2*sin(d*x+c)^4*cos(d*x+c)+4/d*a*b^2*cos(d*x+c)*sin(d*x+c)^2+1/2/d*b^3*sin(d*x+c)^6/cos(d*x+c)^2+1/2/
d*b^3*sin(d*x+c)^4+1/d*b^3*sin(d*x+c)^2+2/d*b^3*ln(cos(d*x+c))

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Maxima [A]  time = 0.999803, size = 192, normalized size = 1.13 \begin{align*} -\frac{12 \, a^{3} \cos \left (d x + c\right )^{5} + 45 \, a^{2} b \cos \left (d x + c\right )^{4} - 20 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 30 \,{\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 60 \,{\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right ) + 60 \,{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \log \left (\cos \left (d x + c\right )\right ) - \frac{30 \,{\left (6 \, a b^{2} \cos \left (d x + c\right ) + b^{3}\right )}}{\cos \left (d x + c\right )^{2}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^5,x, algorithm="maxima")

[Out]

-1/60*(12*a^3*cos(d*x + c)^5 + 45*a^2*b*cos(d*x + c)^4 - 20*(2*a^3 - 3*a*b^2)*cos(d*x + c)^3 - 30*(6*a^2*b - b
^3)*cos(d*x + c)^2 + 60*(a^3 - 6*a*b^2)*cos(d*x + c) + 60*(3*a^2*b - 2*b^3)*log(cos(d*x + c)) - 30*(6*a*b^2*co
s(d*x + c) + b^3)/cos(d*x + c)^2)/d

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Fricas [A]  time = 1.92754, size = 437, normalized size = 2.57 \begin{align*} -\frac{96 \, a^{3} \cos \left (d x + c\right )^{7} + 360 \, a^{2} b \cos \left (d x + c\right )^{6} - 160 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 240 \,{\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} - 1440 \, a b^{2} \cos \left (d x + c\right ) + 480 \,{\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 480 \,{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) - 240 \, b^{3} + 15 \,{\left (39 \, a^{2} b - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{2}}{480 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^5,x, algorithm="fricas")

[Out]

-1/480*(96*a^3*cos(d*x + c)^7 + 360*a^2*b*cos(d*x + c)^6 - 160*(2*a^3 - 3*a*b^2)*cos(d*x + c)^5 - 240*(6*a^2*b
- b^3)*cos(d*x + c)^4 - 1440*a*b^2*cos(d*x + c) + 480*(a^3 - 6*a*b^2)*cos(d*x + c)^3 + 480*(3*a^2*b - 2*b^3)*
cos(d*x + c)^2*log(-cos(d*x + c)) - 240*b^3 + 15*(39*a^2*b - 8*b^3)*cos(d*x + c)^2)/(d*cos(d*x + c)^2)

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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((a+b*sec(d*x+c))**3*sin(d*x+c)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.34467, size = 938, normalized size = 5.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*sin(d*x+c)^5,x, algorithm="giac")

[Out]

1/60*(60*(3*a^2*b - 2*b^3)*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*(3*a^2*b - 2*b^3)*log(abs
(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)) + 30*(9*a^2*b + 12*a*b^2 - 6*b^3 + 18*a^2*b*(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) + 12*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 16*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) +
1) + 9*a^2*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 6*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/((c
os(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2 + (64*a^3 + 411*a^2*b - 600*a*b^2 - 274*b^3 - 320*a^3*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1) - 2415*a^2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2640*a*b^2*(cos(d*x + c) - 1)/(c
os(d*x + c) + 1) + 1490*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 640*a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c)
+ 1)^2 + 5910*a^2*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 3840*a*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c)
+ 1)^2 - 3100*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 5910*a^2*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) +
1)^3 + 2160*a*b^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 3100*b^3*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1
)^3 + 2415*a^2*b*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 360*a*b^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)
^4 - 1490*b^3*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 411*a^2*b*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5
+ 274*b^3*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^5)/d