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

Optimal. Leaf size=115 \[ \frac {a^2 \sec ^4(c+d x)}{4 d}-\frac {a^2 \sec ^2(c+d x)}{d}-\frac {a^2 \log (\cos (c+d x))}{d}+\frac {2 a b \sec ^5(c+d x)}{5 d}-\frac {4 a b \sec ^3(c+d x)}{3 d}+\frac {2 a b \sec (c+d x)}{d}+\frac {b^2 \tan ^6(c+d x)}{6 d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 131, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3885, 948} \[ \frac {\left (a^2-2 b^2\right ) \sec ^4(c+d x)}{4 d}-\frac {\left (2 a^2-b^2\right ) \sec ^2(c+d x)}{2 d}-\frac {a^2 \log (\cos (c+d x))}{d}+\frac {2 a b \sec ^5(c+d x)}{5 d}-\frac {4 a b \sec ^3(c+d x)}{3 d}+\frac {2 a b \sec (c+d x)}{d}+\frac {b^2 \sec ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 105, normalized size = 0.91 \[ \frac {15 \left (a^2-2 b^2\right ) \sec ^4(c+d x)+30 \left (b^2-2 a^2\right ) \sec ^2(c+d x)-60 a^2 \log (\cos (c+d x))+24 a b \sec ^5(c+d x)-80 a b \sec ^3(c+d x)+120 a b \sec (c+d x)+10 b^2 \sec ^6(c+d x)}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-60*a^2*Log[Cos[c + d*x]] + 120*a*b*Sec[c + d*x] + 30*(-2*a^2 + b^2)*Sec[c + d*x]^2 - 80*a*b*Sec[c + d*x]^3 +
15*(a^2 - 2*b^2)*Sec[c + d*x]^4 + 24*a*b*Sec[c + d*x]^5 + 10*b^2*Sec[c + d*x]^6)/(60*d)

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fricas [A]  time = 0.56, size = 115, normalized size = 1.00 \[ -\frac {60 \, a^{2} \cos \left (d x + c\right )^{6} \log \left (-\cos \left (d x + c\right )\right ) - 120 \, a b \cos \left (d x + c\right )^{5} + 80 \, a b \cos \left (d x + c\right )^{3} + 30 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} - 24 \, a b \cos \left (d x + c\right ) - 15 \, {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 10 \, b^{2}}{60 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*a^2*cos(d*x + c)^6*log(-cos(d*x + c)) - 120*a*b*cos(d*x + c)^5 + 80*a*b*cos(d*x + c)^3 + 30*(2*a^2 -
b^2)*cos(d*x + c)^4 - 24*a*b*cos(d*x + c) - 15*(a^2 - 2*b^2)*cos(d*x + c)^2 - 10*b^2)/(d*cos(d*x + c)^6)

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giac [B]  time = 5.39, size = 341, normalized size = 2.97 \[ \frac {60 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {147 \, a^{2} + 128 \, a b + \frac {1002 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {768 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2925 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1920 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4140 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1280 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {640 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2925 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {1002 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {147 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{6}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) - 1)) + (147*a^2 + 128*a*b + 1002*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 768*a*b*(cos(d*x + c
) - 1)/(cos(d*x + c) + 1) + 2925*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1920*a*b*(cos(d*x + c) - 1)^2
/(cos(d*x + c) + 1)^2 + 4140*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 1280*a*b*(cos(d*x + c) - 1)^3/(co
s(d*x + c) + 1)^3 - 640*b^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 2925*a^2*(cos(d*x + c) - 1)^4/(cos(d*x
+ c) + 1)^4 + 1002*a^2*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 147*a^2*(cos(d*x + c) - 1)^6/(cos(d*x + c)
+ 1)^6)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^6)/d

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maple [A]  time = 0.66, size = 197, normalized size = 1.71 \[ \frac {a^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {2 a b \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}-\frac {2 a b \left (\sin ^{6}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}+\frac {2 a b \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )}+\frac {16 a b \cos \left (d x +c \right )}{15 d}+\frac {2 a b \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{5 d}+\frac {8 a b \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{15 d}+\frac {b^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*tan(d*x+c)^4/d-1/2*a^2*tan(d*x+c)^2/d-a^2*ln(cos(d*x+c))/d+2/5/d*a*b*sin(d*x+c)^6/cos(d*x+c)^5-2/15/d*
a*b*sin(d*x+c)^6/cos(d*x+c)^3+2/5/d*a*b*sin(d*x+c)^6/cos(d*x+c)+16/15*a*b*cos(d*x+c)/d+2/5/d*a*b*cos(d*x+c)*si
n(d*x+c)^4+8/15/d*a*b*cos(d*x+c)*sin(d*x+c)^2+1/6/d*b^2*sin(d*x+c)^6/cos(d*x+c)^6

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maxima [A]  time = 0.33, size = 108, normalized size = 0.94 \[ -\frac {60 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {120 \, a b \cos \left (d x + c\right )^{5} - 80 \, a b \cos \left (d x + c\right )^{3} - 30 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 24 \, a b \cos \left (d x + c\right ) + 15 \, {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 10 \, b^{2}}{\cos \left (d x + c\right )^{6}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*a^2*log(cos(d*x + c)) - (120*a*b*cos(d*x + c)^5 - 80*a*b*cos(d*x + c)^3 - 30*(2*a^2 - b^2)*cos(d*x +
c)^4 + 24*a*b*cos(d*x + c) + 15*(a^2 - 2*b^2)*cos(d*x + c)^2 + 10*b^2)/cos(d*x + c)^6)/d

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mupad [B]  time = 4.97, size = 215, normalized size = 1.87 \[ \frac {2\,a^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}+\frac {\frac {32\,a\,b}{15}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,a^2+32\,b\,a\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+\frac {64\,b\,a}{5}\right )+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (20\,a^2+\frac {64\,a\,b}{3}-\frac {32\,b^2}{3}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*atanh(tan(c/2 + (d*x)/2)^2))/d + ((32*a*b)/15 + tan(c/2 + (d*x)/2)^4*(32*a*b + 12*a^2) - tan(c/2 + (d*x
)/2)^2*((64*a*b)/5 + 2*a^2) + 12*a^2*tan(c/2 + (d*x)/2)^8 - 2*a^2*tan(c/2 + (d*x)/2)^10 - tan(c/2 + (d*x)/2)^6
*((64*a*b)/3 + 20*a^2 - (32*b^2)/3))/(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))

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sympy [A]  time = 5.44, size = 189, normalized size = 1.64 \[ \begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {2 a b \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{5 d} - \frac {8 a b \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{15 d} + \frac {16 a b \sec {\left (c + d x \right )}}{15 d} + \frac {b^{2} \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{6 d} - \frac {b^{2} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{6 d} + \frac {b^{2} \sec ^{2}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a + b \sec {\relax (c )}\right )^{2} \tan ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**2*tan(d*x+c)**5,x)

[Out]

Piecewise((a**2*log(tan(c + d*x)**2 + 1)/(2*d) + a**2*tan(c + d*x)**4/(4*d) - a**2*tan(c + d*x)**2/(2*d) + 2*a
*b*tan(c + d*x)**4*sec(c + d*x)/(5*d) - 8*a*b*tan(c + d*x)**2*sec(c + d*x)/(15*d) + 16*a*b*sec(c + d*x)/(15*d)
+ b**2*tan(c + d*x)**4*sec(c + d*x)**2/(6*d) - b**2*tan(c + d*x)**2*sec(c + d*x)**2/(6*d) + b**2*sec(c + d*x)
**2/(6*d), Ne(d, 0)), (x*(a + b*sec(c))**2*tan(c)**5, True))

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