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

Optimal. Leaf size=124 \[ \frac {\left (2 a^2-b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{d}-\frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {a b \cos ^4(c+d x)}{2 d}+\frac {2 a b \cos ^2(c+d x)}{d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \sec (c+d x)}{d} \]

[Out]

-(a^2-2*b^2)*cos(d*x+c)/d+2*a*b*cos(d*x+c)^2/d+1/3*(2*a^2-b^2)*cos(d*x+c)^3/d-1/2*a*b*cos(d*x+c)^4/d-1/5*a^2*c
os(d*x+c)^5/d-2*a*b*ln(cos(d*x+c))/d+b^2*sec(d*x+c)/d

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Rubi [A]  time = 0.20, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3872, 2837, 12, 948} \[ \frac {\left (2 a^2-b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{d}-\frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {a b \cos ^4(c+d x)}{2 d}+\frac {2 a b \cos ^2(c+d x)}{d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 112, normalized size = 0.90 \[ -\frac {30 \left (5 a^2-14 b^2\right ) \cos (c+d x)-25 a^2 \cos (3 (c+d x))+3 a^2 \cos (5 (c+d x))-180 a b \cos (2 (c+d x))+15 a b \cos (4 (c+d x))+480 a b \log (\cos (c+d x))+20 b^2 \cos (3 (c+d x))-240 b^2 \sec (c+d x)}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/240*(30*(5*a^2 - 14*b^2)*Cos[c + d*x] - 180*a*b*Cos[2*(c + d*x)] - 25*a^2*Cos[3*(c + d*x)] + 20*b^2*Cos[3*(
c + d*x)] + 15*a*b*Cos[4*(c + d*x)] + 3*a^2*Cos[5*(c + d*x)] + 480*a*b*Log[Cos[c + d*x]] - 240*b^2*Sec[c + d*x
])/d

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fricas [A]  time = 0.73, size = 125, normalized size = 1.01 \[ -\frac {48 \, a^{2} \cos \left (d x + c\right )^{6} + 120 \, a b \cos \left (d x + c\right )^{5} - 480 \, a b \cos \left (d x + c\right )^{3} - 80 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 480 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) + 195 \, a b \cos \left (d x + c\right ) + 240 \, {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 240 \, b^{2}}{240 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^5,x, algorithm="fricas")

[Out]

-1/240*(48*a^2*cos(d*x + c)^6 + 120*a*b*cos(d*x + c)^5 - 480*a*b*cos(d*x + c)^3 - 80*(2*a^2 - b^2)*cos(d*x + c
)^4 + 480*a*b*cos(d*x + c)*log(-cos(d*x + c)) + 195*a*b*cos(d*x + c) + 240*(a^2 - 2*b^2)*cos(d*x + c)^2 - 240*
b^2)/(d*cos(d*x + c))

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giac [B]  time = 0.43, size = 418, normalized size = 3.37 \[ \frac {60 \, a b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {60 \, {\left (a b + b^{2} + \frac {a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1} + \frac {32 \, a^{2} + 137 \, a b - 100 \, b^{2} - \frac {160 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {805 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {440 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {320 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1970 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {640 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1970 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {360 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {805 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {60 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {137 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^5,x, algorithm="giac")

[Out]

1/30*(60*a*b*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a*b*log(abs(-(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) - 1)) + 60*(a*b + b^2 + a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((cos(d*x + c) - 1)/(cos(d*x +
c) + 1) + 1) + (32*a^2 + 137*a*b - 100*b^2 - 160*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 805*a*b*(cos(d*x
+ c) - 1)/(cos(d*x + c) + 1) + 440*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 320*a^2*(cos(d*x + c) - 1)^2/(c
os(d*x + c) + 1)^2 + 1970*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 640*b^2*(cos(d*x + c) - 1)^2/(cos(d*
x + c) + 1)^2 - 1970*a*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 360*b^2*(cos(d*x + c) - 1)^3/(cos(d*x + c
) + 1)^3 + 805*a*b*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 60*b^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^
4 - 137*a*b*(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

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maple [A]  time = 0.63, size = 184, normalized size = 1.48 \[ -\frac {8 a^{2} \cos \left (d x +c \right )}{15 d}-\frac {\cos \left (d x +c \right ) a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{5 d}-\frac {4 \cos \left (d x +c \right ) a^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{15 d}-\frac {a b \left (\sin ^{4}\left (d x +c \right )\right )}{2 d}-\frac {a b \left (\sin ^{2}\left (d x +c \right )\right )}{d}-\frac {2 a b \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {b^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {8 b^{2} \cos \left (d x +c \right )}{3 d}+\frac {b^{2} \left (\sin ^{4}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{d}+\frac {4 b^{2} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 105, normalized size = 0.85 \[ -\frac {6 \, a^{2} \cos \left (d x + c\right )^{5} + 15 \, a b \cos \left (d x + c\right )^{4} - 60 \, a b \cos \left (d x + c\right )^{2} - 10 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{3} + 60 \, a b \log \left (\cos \left (d x + c\right )\right ) + 30 \, {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right ) - \frac {30 \, b^{2}}{\cos \left (d x + c\right )}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^5,x, algorithm="maxima")

[Out]

-1/30*(6*a^2*cos(d*x + c)^5 + 15*a*b*cos(d*x + c)^4 - 60*a*b*cos(d*x + c)^2 - 10*(2*a^2 - b^2)*cos(d*x + c)^3
+ 60*a*b*log(cos(d*x + c)) + 30*(a^2 - 2*b^2)*cos(d*x + c) - 30*b^2/cos(d*x + c))/d

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mupad [B]  time = 0.95, size = 104, normalized size = 0.84 \[ -\frac {\cos \left (c+d\,x\right )\,\left (a^2-2\,b^2\right )-{\cos \left (c+d\,x\right )}^3\,\left (\frac {2\,a^2}{3}-\frac {b^2}{3}\right )+\frac {a^2\,{\cos \left (c+d\,x\right )}^5}{5}-\frac {b^2}{\cos \left (c+d\,x\right )}-2\,a\,b\,{\cos \left (c+d\,x\right )}^2+\frac {a\,b\,{\cos \left (c+d\,x\right )}^4}{2}+2\,a\,b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cos(c + d*x)*(a^2 - 2*b^2) - cos(c + d*x)^3*((2*a^2)/3 - b^2/3) + (a^2*cos(c + d*x)^5)/5 - b^2/cos(c + d*x)
- 2*a*b*cos(c + d*x)^2 + (a*b*cos(c + d*x)^4)/2 + 2*a*b*log(cos(c + d*x)))/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((a+b*sec(d*x+c))**2*sin(d*x+c)**5,x)

[Out]

Timed out

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