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

Optimal. Leaf size=97 \[ \frac{b \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac{a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}-b x \left (3 a^2-b^2\right )+\frac{(a+b \tan (c+d x))^3}{3 d}+\frac{a (a+b \tan (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0818862, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3528, 3525, 3475} \[ \frac{b \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac{a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}-b x \left (3 a^2-b^2\right )+\frac{(a+b \tan (c+d x))^3}{3 d}+\frac{a (a+b \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3475

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

Rubi steps

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

Mathematica [C]  time = 0.560727, size = 100, normalized size = 1.03 \[ \frac{-6 b \left (b^2-3 a^2\right ) \tan (c+d x)+9 a b^2 \tan ^2(c+d x)+3 \left ((a-i b)^3 \log (\tan (c+d x)+i)+(a+i b)^3 \log (-\tan (c+d x)+i)\right )+2 b^3 \tan ^3(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 133, normalized size = 1.4 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}{b}^{3}}{3\,d}}+{\frac{3\,a{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+3\,{\frac{\tan \left ( dx+c \right ){a}^{2}b}{d}}-{\frac{{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d}}-{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) a{b}^{2}}{2\,d}}-3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}b}{d}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{3}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.72276, size = 128, normalized size = 1.32 \begin{align*} \frac{2 \, b^{3} \tan \left (d x + c\right )^{3} + 9 \, a b^{2} \tan \left (d x + c\right )^{2} - 6 \,{\left (3 \, a^{2} b - b^{3}\right )}{\left (d x + c\right )} + 3 \,{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 \,{\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^3*tan(d*x + c)^3 + 9*a*b^2*tan(d*x + c)^2 - 6*(3*a^2*b - b^3)*(d*x + c) + 3*(a^3 - 3*a*b^2)*log(tan(d
*x + c)^2 + 1) + 6*(3*a^2*b - b^3)*tan(d*x + c))/d

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Fricas [A]  time = 1.75936, size = 216, normalized size = 2.23 \begin{align*} \frac{2 \, b^{3} \tan \left (d x + c\right )^{3} + 9 \, a b^{2} \tan \left (d x + c\right )^{2} - 6 \,{\left (3 \, a^{2} b - b^{3}\right )} d x - 3 \,{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 6 \,{\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^3*tan(d*x + c)^3 + 9*a*b^2*tan(d*x + c)^2 - 6*(3*a^2*b - b^3)*d*x - 3*(a^3 - 3*a*b^2)*log(1/(tan(d*x
+ c)^2 + 1)) + 6*(3*a^2*b - b^3)*tan(d*x + c))/d

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Sympy [A]  time = 0.461567, size = 128, normalized size = 1.32 \begin{align*} \begin{cases} \frac{a^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 3 a^{2} b x + \frac{3 a^{2} b \tan{\left (c + d x \right )}}{d} - \frac{3 a b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{3 a b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + b^{3} x + \frac{b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac{b^{3} \tan{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tan{\left (c \right )}\right )^{3} \tan{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 3.07469, size = 1341, normalized size = 13.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(18*a^2*b*d*x*tan(d*x)^3*tan(c)^3 - 6*b^3*d*x*tan(d*x)^3*tan(c)^3 + 3*a^3*log(4*(tan(c)^2 + 1)/(tan(d*x)^
4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^3*tan(c
)^3 - 9*a*b^2*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)
^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^3*tan(c)^3 - 54*a^2*b*d*x*tan(d*x)^2*tan(c)^2 + 18*b^3*d*x*tan(d*x)^2*ta
n(c)^2 - 9*a*b^2*tan(d*x)^3*tan(c)^3 - 9*a^3*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) +
tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^2*tan(c)^2 + 27*a*b^2*log(4*(tan(c)^2 + 1
)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(
d*x)^2*tan(c)^2 + 18*a^2*b*tan(d*x)^3*tan(c)^2 - 6*b^3*tan(d*x)^3*tan(c)^2 + 18*a^2*b*tan(d*x)^2*tan(c)^3 - 6*
b^3*tan(d*x)^2*tan(c)^3 + 54*a^2*b*d*x*tan(d*x)*tan(c) - 18*b^3*d*x*tan(d*x)*tan(c) - 9*a*b^2*tan(d*x)^3*tan(c
) + 9*a*b^2*tan(d*x)^2*tan(c)^2 - 9*a*b^2*tan(d*x)*tan(c)^3 + 2*b^3*tan(d*x)^3 + 9*a^3*log(4*(tan(c)^2 + 1)/(t
an(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)
*tan(c) - 27*a*b^2*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan
(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)*tan(c) - 36*a^2*b*tan(d*x)^2*tan(c) + 18*b^3*tan(d*x)^2*tan(c) - 36
*a^2*b*tan(d*x)*tan(c)^2 + 18*b^3*tan(d*x)*tan(c)^2 + 2*b^3*tan(c)^3 - 18*a^2*b*d*x + 6*b^3*d*x + 9*a*b^2*tan(
d*x)^2 - 9*a*b^2*tan(d*x)*tan(c) + 9*a*b^2*tan(c)^2 - 3*a^3*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(
d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)) + 9*a*b^2*log(4*(tan(c)^2 + 1)/(tan
(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)) + 18*a^2*b
*tan(d*x) - 6*b^3*tan(d*x) + 18*a^2*b*tan(c) - 6*b^3*tan(c) + 9*a*b^2)/(d*tan(d*x)^3*tan(c)^3 - 3*d*tan(d*x)^2
*tan(c)^2 + 3*d*tan(d*x)*tan(c) - d)