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

Optimal. Leaf size=84 \[ \frac{\tan ^4(c+d x) (5 a+4 b \sec (c+d x))}{20 d}-\frac{\tan ^2(c+d x) (15 a+8 b \sec (c+d x))}{30 d}-\frac{a \log (\cos (c+d x))}{d}+\frac{8 b \sec (c+d x)}{15 d} \]

[Out]

-((a*Log[Cos[c + d*x]])/d) + (8*b*Sec[c + d*x])/(15*d) - ((15*a + 8*b*Sec[c + d*x])*Tan[c + d*x]^2)/(30*d) + (
(5*a + 4*b*Sec[c + d*x])*Tan[c + d*x]^4)/(20*d)

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Rubi [A]  time = 0.0935648, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3881, 3884, 3475, 2606, 8} \[ \frac{\tan ^4(c+d x) (5 a+4 b \sec (c+d x))}{20 d}-\frac{\tan ^2(c+d x) (15 a+8 b \sec (c+d x))}{30 d}-\frac{a \log (\cos (c+d x))}{d}+\frac{8 b \sec (c+d x)}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Log[Cos[c + d*x]])/d) + (8*b*Sec[c + d*x])/(15*d) - ((15*a + 8*b*Sec[c + d*x])*Tan[c + d*x]^2)/(30*d) + (
(5*a + 4*b*Sec[c + d*x])*Tan[c + d*x]^4)/(20*d)

Rule 3881

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(e*(e*Cot[
c + d*x])^(m - 1)*(a*m + b*(m - 1)*Csc[c + d*x]))/(d*m*(m - 1)), x] - Dist[e^2/m, Int[(e*Cot[c + d*x])^(m - 2)
*(a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]

Rule 3884

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(e*
Cot[c + d*x])^m, x], x] + Dist[b, Int[(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 8

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

Rubi steps

\begin{align*} \int (a+b \sec (c+d x)) \tan ^5(c+d x) \, dx &=\frac{(5 a+4 b \sec (c+d x)) \tan ^4(c+d x)}{20 d}-\frac{1}{5} \int (5 a+4 b \sec (c+d x)) \tan ^3(c+d x) \, dx\\ &=-\frac{(15 a+8 b \sec (c+d x)) \tan ^2(c+d x)}{30 d}+\frac{(5 a+4 b \sec (c+d x)) \tan ^4(c+d x)}{20 d}+\frac{1}{15} \int (15 a+8 b \sec (c+d x)) \tan (c+d x) \, dx\\ &=-\frac{(15 a+8 b \sec (c+d x)) \tan ^2(c+d x)}{30 d}+\frac{(5 a+4 b \sec (c+d x)) \tan ^4(c+d x)}{20 d}+a \int \tan (c+d x) \, dx+\frac{1}{15} (8 b) \int \sec (c+d x) \tan (c+d x) \, dx\\ &=-\frac{a \log (\cos (c+d x))}{d}-\frac{(15 a+8 b \sec (c+d x)) \tan ^2(c+d x)}{30 d}+\frac{(5 a+4 b \sec (c+d x)) \tan ^4(c+d x)}{20 d}+\frac{(8 b) \operatorname{Subst}(\int 1 \, dx,x,\sec (c+d x))}{15 d}\\ &=-\frac{a \log (\cos (c+d x))}{d}+\frac{8 b \sec (c+d x)}{15 d}-\frac{(15 a+8 b \sec (c+d x)) \tan ^2(c+d x)}{30 d}+\frac{(5 a+4 b \sec (c+d x)) \tan ^4(c+d x)}{20 d}\\ \end{align*}

Mathematica [A]  time = 0.189151, size = 82, normalized size = 0.98 \[ -\frac{a \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d}+\frac{b \sec ^5(c+d x)}{5 d}-\frac{2 b \sec ^3(c+d x)}{3 d}+\frac{b \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.04, size = 161, normalized size = 1.9 \begin{align*}{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{2}a}{2\,d}}-{\frac{a\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{15\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{5\,d\cos \left ( dx+c \right ) }}+{\frac{8\,b\cos \left ( dx+c \right ) }{15\,d}}+{\frac{b\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,b\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.988523, size = 97, normalized size = 1.15 \begin{align*} -\frac{60 \, a \log \left (\cos \left (d x + c\right )\right ) - \frac{60 \, b \cos \left (d x + c\right )^{4} - 60 \, a \cos \left (d x + c\right )^{3} - 40 \, b \cos \left (d x + c\right )^{2} + 15 \, a \cos \left (d x + c\right ) + 12 \, b}{\cos \left (d x + c\right )^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*a*log(cos(d*x + c)) - (60*b*cos(d*x + c)^4 - 60*a*cos(d*x + c)^3 - 40*b*cos(d*x + c)^2 + 15*a*cos(d*
x + c) + 12*b)/cos(d*x + c)^5)/d

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Fricas [A]  time = 1.34255, size = 216, normalized size = 2.57 \begin{align*} -\frac{60 \, a \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) - 60 \, b \cos \left (d x + c\right )^{4} + 60 \, a \cos \left (d x + c\right )^{3} + 40 \, b \cos \left (d x + c\right )^{2} - 15 \, a \cos \left (d x + c\right ) - 12 \, b}{60 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*a*cos(d*x + c)^5*log(-cos(d*x + c)) - 60*b*cos(d*x + c)^4 + 60*a*cos(d*x + c)^3 + 40*b*cos(d*x + c)^
2 - 15*a*cos(d*x + c) - 12*b)/(d*cos(d*x + c)^5)

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Sympy [A]  time = 5.30308, size = 112, normalized size = 1.33 \begin{align*} \begin{cases} \frac{a \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{a \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac{a \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac{b \tan ^{4}{\left (c + d x \right )} \sec{\left (c + d x \right )}}{5 d} - \frac{4 b \tan ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}}{15 d} + \frac{8 b \sec{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a + b \sec{\left (c \right )}\right ) \tan ^{5}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 3.54973, size = 335, normalized size = 3.99 \begin{align*} \frac{60 \, a \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{137 \, a + 64 \, b + \frac{805 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{320 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{1970 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{640 \, b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{1970 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{805 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{137 \, a{\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}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1)) + (137*a + 64*b + 805*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 320*b*(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) + 1970*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 640*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)
^2 + 1970*a*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 805*a*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 137*
a*(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