3.258 \(\int (a+b \sec (c+d x)) \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=55 \[ \frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}+\frac {a \log (\cos (c+d x))}{d}-\frac {2 b \sec (c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ \frac {\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}+\frac {a \log (\cos (c+d x))}{d}-\frac {2 b \sec (c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3475

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 55, normalized size = 1.00 \[ \frac {a \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d}+\frac {b \sec ^3(c+d x)}{3 d}-\frac {b \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 57, normalized size = 1.04 \[ \frac {6 \, a \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) - 6 \, b \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + 2 \, b}{6 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*a*cos(d*x + c)^3*log(-cos(d*x + c)) - 6*b*cos(d*x + c)^2 + 3*a*cos(d*x + c) + 2*b)/(d*cos(d*x + c)^3)

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giac [B]  time = 0.95, size = 179, normalized size = 3.25 \[ -\frac {6 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 6 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {11 \, a + 8 \, b + \frac {45 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {24 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {45 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {11 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 6*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) - 1)) + (11*a + 8*b + 45*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 24*b*(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) + 45*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 11*a*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)/((cos
(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^3)/d

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maple [B]  time = 0.64, size = 104, normalized size = 1.89 \[ \frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {b \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}-\frac {b \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )}-\frac {b \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{3 d}-\frac {2 b \cos \left (d x +c \right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 50, normalized size = 0.91 \[ \frac {6 \, a \log \left (\cos \left (d x + c\right )\right ) - \frac {6 \, b \cos \left (d x + c\right )^{2} - 3 \, a \cos \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*a*log(cos(d*x + c)) - (6*b*cos(d*x + c)^2 - 3*a*cos(d*x + c) - 2*b)/cos(d*x + c)^3)/d

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mupad [B]  time = 2.23, size = 102, normalized size = 1.85 \[ \frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-2\,a-4\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {4\,b}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,a\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3*(a + b/cos(c + d*x)),x)

[Out]

((4*b)/3 - tan(c/2 + (d*x)/2)^2*(2*a + 4*b) + 2*a*tan(c/2 + (d*x)/2)^4)/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2
+ (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1)) - (2*a*atanh(tan(c/2 + (d*x)/2)^2))/d

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sympy [A]  time = 0.92, size = 76, normalized size = 1.38 \[ \begin {cases} - \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {b \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{3 d} - \frac {2 b \sec {\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a + b \sec {\relax (c )}\right ) \tan ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*tan(d*x+c)**3,x)

[Out]

Piecewise((-a*log(tan(c + d*x)**2 + 1)/(2*d) + a*tan(c + d*x)**2/(2*d) + b*tan(c + d*x)**2*sec(c + d*x)/(3*d)
- 2*b*sec(c + d*x)/(3*d), Ne(d, 0)), (x*(a + b*sec(c))*tan(c)**3, True))

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