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

Optimal. Leaf size=120 \[ \frac{a^2 \sec ^6(c+d x)}{6 d}+\frac{2 a^2 \sec ^5(c+d x)}{5 d}-\frac{a^2 \sec ^4(c+d x)}{4 d}-\frac{4 a^2 \sec ^3(c+d x)}{3 d}-\frac{a^2 \sec ^2(c+d x)}{2 d}+\frac{2 a^2 \sec (c+d x)}{d}-\frac{a^2 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.072974, antiderivative size = 120, normalized size of antiderivative = 1., 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 = {3879, 88} \[ \frac{a^2 \sec ^6(c+d x)}{6 d}+\frac{2 a^2 \sec ^5(c+d x)}{5 d}-\frac{a^2 \sec ^4(c+d x)}{4 d}-\frac{4 a^2 \sec ^3(c+d x)}{3 d}-\frac{a^2 \sec ^2(c+d x)}{2 d}+\frac{2 a^2 \sec (c+d x)}{d}-\frac{a^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3879

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^2 \tan ^5(c+d x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(a-a x)^2 (a+a x)^4}{x^7} \, dx,x,\cos (c+d x)\right )}{a^4 d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{a^6}{x^7}+\frac{2 a^6}{x^6}-\frac{a^6}{x^5}-\frac{4 a^6}{x^4}-\frac{a^6}{x^3}+\frac{2 a^6}{x^2}+\frac{a^6}{x}\right ) \, dx,x,\cos (c+d x)\right )}{a^4 d}\\ &=-\frac{a^2 \log (\cos (c+d x))}{d}+\frac{2 a^2 \sec (c+d x)}{d}-\frac{a^2 \sec ^2(c+d x)}{2 d}-\frac{4 a^2 \sec ^3(c+d x)}{3 d}-\frac{a^2 \sec ^4(c+d x)}{4 d}+\frac{2 a^2 \sec ^5(c+d x)}{5 d}+\frac{a^2 \sec ^6(c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.409315, size = 125, normalized size = 1.04 \[ \frac{a^2 \sec ^6(c+d x) (312 \cos (c+d x)-5 (-28 \cos (3 (c+d x))+6 \cos (4 (c+d x))-12 \cos (5 (c+d x))+18 \cos (4 (c+d x)) \log (\cos (c+d x))+3 \cos (6 (c+d x)) \log (\cos (c+d x))+30 \log (\cos (c+d x))+9 \cos (2 (c+d x)) (5 \log (\cos (c+d x))+4)+14))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(312*Cos[c + d*x] - 5*(14 - 28*Cos[3*(c + d*x)] + 6*Cos[4*(c + d*x)] - 12*Cos[5*(c + d*x)] + 30*Log[Cos[c
+ d*x]] + 18*Cos[4*(c + d*x)]*Log[Cos[c + d*x]] + 3*Cos[6*(c + d*x)]*Log[Cos[c + d*x]] + 9*Cos[2*(c + d*x)]*(
4 + 5*Log[Cos[c + d*x]])))*Sec[c + d*x]^6)/(480*d)

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Maple [A]  time = 0.05, size = 203, normalized size = 1.7 \begin{align*}{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{2\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-{\frac{2\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{15\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{2\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{5\,d\cos \left ( dx+c \right ) }}+{\frac{16\,{a}^{2}\cos \left ( dx+c \right ) }{15\,d}}+{\frac{2\,{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{8\,{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.22873, size = 131, normalized size = 1.09 \begin{align*} -\frac{60 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac{120 \, a^{2} \cos \left (d x + c\right )^{5} - 30 \, a^{2} \cos \left (d x + c\right )^{4} - 80 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} \cos \left (d x + c\right )^{2} + 24 \, a^{2} \cos \left (d x + c\right ) + 10 \, a^{2}}{\cos \left (d x + c\right )^{6}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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^2*cos(d*x + c)^5 - 30*a^2*cos(d*x + c)^4 - 80*a^2*cos(d*x + c)^3 - 15
*a^2*cos(d*x + c)^2 + 24*a^2*cos(d*x + c) + 10*a^2)/cos(d*x + c)^6)/d

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Fricas [A]  time = 0.951059, size = 266, normalized size = 2.22 \begin{align*} -\frac{60 \, a^{2} \cos \left (d x + c\right )^{6} \log \left (-\cos \left (d x + c\right )\right ) - 120 \, a^{2} \cos \left (d x + c\right )^{5} + 30 \, a^{2} \cos \left (d x + c\right )^{4} + 80 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} \cos \left (d x + c\right )^{2} - 24 \, a^{2} \cos \left (d x + c\right ) - 10 \, a^{2}}{60 \, d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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^2*cos(d*x + c)^5 + 30*a^2*cos(d*x + c)^4 + 80*a^2*cos(
d*x + c)^3 + 15*a^2*cos(d*x + c)^2 - 24*a^2*cos(d*x + c) - 10*a^2)/(d*cos(d*x + c)^6)

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Sympy [A]  time = 11.612, size = 189, normalized size = 1.58 \begin{align*} \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 )} \sec ^{2}{\left (c + d x \right )}}{6 d} + \frac{2 a^{2} \tan ^{4}{\left (c + d x \right )} \sec{\left (c + d x \right )}}{5 d} + \frac{a^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac{a^{2} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{6 d} - \frac{8 a^{2} \tan ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}}{15 d} - \frac{a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac{a^{2} \sec ^{2}{\left (c + d x \right )}}{6 d} + \frac{16 a^{2} \sec{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a \sec{\left (c \right )} + a\right )^{2} \tan ^{5}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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*sec(c + d*x)**2/(6*d) + 2*a**2*tan(c + d
*x)**4*sec(c + d*x)/(5*d) + a**2*tan(c + d*x)**4/(4*d) - a**2*tan(c + d*x)**2*sec(c + d*x)**2/(6*d) - 8*a**2*t
an(c + d*x)**2*sec(c + d*x)/(15*d) - a**2*tan(c + d*x)**2/(2*d) + a**2*sec(c + d*x)**2/(6*d) + 16*a**2*sec(c +
d*x)/(15*d), Ne(d, 0)), (x*(a*sec(c) + a)**2*tan(c)**5, True))

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Giac [B]  time = 3.86297, size = 327, normalized size = 2.72 \begin{align*} \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{275 \, a^{2} + \frac{1770 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{4845 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{4780 \, a^{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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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)) + (275*a^2 + 1770*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 4845*a^2*(cos(d*x + c) - 1)^2/
(cos(d*x + c) + 1)^2 + 4780*a^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