3.5.25 \(\int \tan ^3(c+d x) (a+b \tan (c+d x))^2 \, dx\) [425]

Optimal. Leaf size=98 \[ 2 a b x+\frac {\left (a^2-b^2\right ) \log (\cos (c+d x))}{d}-\frac {2 a b \tan (c+d x)}{d}+\frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 d}+\frac {2 a b \tan ^3(c+d x)}{3 d}+\frac {b^2 \tan ^4(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3624, 3609, 3606, 3556} \begin {gather*} \frac {\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 d}+\frac {\left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {2 a b \tan ^3(c+d x)}{3 d}-\frac {2 a b \tan (c+d x)}{d}+2 a b x+\frac {b^2 \tan ^4(c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*(a + b*Tan[c + d*x])^2,x]

[Out]

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

Rule 3556

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

Rule 3606

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 3609

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 3624

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

Rubi steps

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

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Mathematica [A]
time = 0.46, size = 113, normalized size = 1.15 \begin {gather*} \frac {2 a b \text {ArcTan}(\tan (c+d x))}{d}-\frac {2 a b \tan (c+d x)}{d}+\frac {2 a b \tan ^3(c+d x)}{3 d}+\frac {a^2 \left (2 \log (\cos (c+d x))+\tan ^2(c+d x)\right )}{2 d}-\frac {b^2 \left (4 \log (\cos (c+d x))+2 \tan ^2(c+d x)-\tan ^4(c+d x)\right )}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3*(a + b*Tan[c + d*x])^2,x]

[Out]

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

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Maple [A]
time = 0.04, size = 100, normalized size = 1.02

method	result	size
norman	\(2 a b x +\frac {b^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {\left (a^{2}-b^{2}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {2 a b \tan \left (d x +c \right )}{d}+\frac {2 a b \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(98\)
derivativedivides	\(\frac {\frac {b^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {2 a b \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}-2 a b \tan \left (d x +c \right )+\frac {\left (-a^{2}+b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+2 a b \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(100\)
default	\(\frac {\frac {b^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {2 a b \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}-2 a b \tan \left (d x +c \right )+\frac {\left (-a^{2}+b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+2 a b \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(100\)
risch	\(2 a b x -i a^{2} x +i b^{2} x -\frac {2 i a^{2} c}{d}+\frac {2 i b^{2} c}{d}+\frac {-8 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+2 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-4 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-16 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-4 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-\frac {40 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}}{3}+2 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-\frac {16 i a b}{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b^{2}}{d}\)	\(230\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 91, normalized size = 0.93 \begin {gather*} \frac {3 \, b^{2} \tan \left (d x + c\right )^{4} + 8 \, a b \tan \left (d x + c\right )^{3} + 24 \, {\left (d x + c\right )} a b - 24 \, a b \tan \left (d x + c\right ) + 6 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} - 6 \, {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(3*b^2*tan(d*x + c)^4 + 8*a*b*tan(d*x + c)^3 + 24*(d*x + c)*a*b - 24*a*b*tan(d*x + c) + 6*(a^2 - b^2)*tan
(d*x + c)^2 - 6*(a^2 - b^2)*log(tan(d*x + c)^2 + 1))/d

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Fricas [A]
time = 0.79, size = 90, normalized size = 0.92 \begin {gather*} \frac {3 \, b^{2} \tan \left (d x + c\right )^{4} + 8 \, a b \tan \left (d x + c\right )^{3} + 24 \, a b d x - 24 \, a b \tan \left (d x + c\right ) + 6 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} + 6 \, {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(3*b^2*tan(d*x + c)^4 + 8*a*b*tan(d*x + c)^3 + 24*a*b*d*x - 24*a*b*tan(d*x + c) + 6*(a^2 - b^2)*tan(d*x +
c)^2 + 6*(a^2 - b^2)*log(1/(tan(d*x + c)^2 + 1)))/d

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Sympy [A]
time = 0.15, size = 134, normalized size = 1.37 \begin {gather*} \begin {cases} - \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + 2 a b x + \frac {2 a b \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 a b \tan {\left (c + d x \right )}}{d} + \frac {b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right )^{2} \tan ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3*(a+b*tan(d*x+c))**2,x)

[Out]

Piecewise((-a**2*log(tan(c + d*x)**2 + 1)/(2*d) + a**2*tan(c + d*x)**2/(2*d) + 2*a*b*x + 2*a*b*tan(c + d*x)**3
/(3*d) - 2*a*b*tan(c + d*x)/d + b**2*log(tan(c + d*x)**2 + 1)/(2*d) + b**2*tan(c + d*x)**4/(4*d) - b**2*tan(c
+ d*x)**2/(2*d), Ne(d, 0)), (x*(a + b*tan(c))**2*tan(c)**3, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1262 vs. \(2 (92) = 184\).
time = 1.72, size = 1262, normalized size = 12.88 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(24*a*b*d*x*tan(d*x)^4*tan(c)^4 + 6*a^2*log(4*(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(c)^2 + 1))*tan(d*x)^4*tan(c)^4 - 6*b^2*log(4*(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(c)^2 + 1))*tan(d*x
)^4*tan(c)^4 - 96*a*b*d*x*tan(d*x)^3*tan(c)^3 + 6*a^2*tan(d*x)^4*tan(c)^4 - 9*b^2*tan(d*x)^4*tan(c)^4 - 24*a^2
*log(4*(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(c)^2 + 1))*tan(d*x)^3*tan(c)^3 + 24*b^2*log(4*(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(c)^2 + 1))*tan(d*x)^3*tan(c)^3 + 24*a*b*tan(d*x)^4*tan(c)^3 +
24*a*b*tan(d*x)^3*tan(c)^4 + 144*a*b*d*x*tan(d*x)^2*tan(c)^2 + 6*a^2*tan(d*x)^4*tan(c)^2 - 6*b^2*tan(d*x)^4*t
an(c)^2 - 12*a^2*tan(d*x)^3*tan(c)^3 + 24*b^2*tan(d*x)^3*tan(c)^3 + 6*a^2*tan(d*x)^2*tan(c)^4 - 6*b^2*tan(d*x)
^2*tan(c)^4 - 8*a*b*tan(d*x)^4*tan(c) + 36*a^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*t
an(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 - 36*b^2*log(4*(tan(d*x)^4*t
an(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(c)^2 + 1))*tan(
d*x)^2*tan(c)^2 - 96*a*b*tan(d*x)^3*tan(c)^2 - 96*a*b*tan(d*x)^2*tan(c)^3 - 8*a*b*tan(d*x)*tan(c)^4 + 3*b^2*ta
n(d*x)^4 - 96*a*b*d*x*tan(d*x)*tan(c) - 12*a^2*tan(d*x)^3*tan(c) + 24*b^2*tan(d*x)^3*tan(c) + 12*a^2*tan(d*x)^
2*tan(c)^2 - 12*b^2*tan(d*x)^2*tan(c)^2 - 12*a^2*tan(d*x)*tan(c)^3 + 24*b^2*tan(d*x)*tan(c)^3 + 3*b^2*tan(c)^4
+ 8*a*b*tan(d*x)^3 - 24*a^2*log(4*(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(c)^2 + 1))*tan(d*x)*tan(c) + 24*b^2*log(4*(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(c)^2 + 1))*tan(d*x)*tan(c) + 96*a*b*t
an(d*x)^2*tan(c) + 96*a*b*tan(d*x)*tan(c)^2 + 8*a*b*tan(c)^3 + 24*a*b*d*x + 6*a^2*tan(d*x)^2 - 6*b^2*tan(d*x)^
2 - 12*a^2*tan(d*x)*tan(c) + 24*b^2*tan(d*x)*tan(c) + 6*a^2*tan(c)^2 - 6*b^2*tan(c)^2 + 6*a^2*log(4*(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(c)^2 + 1)) -
6*b^2*log(4*(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(c)^2 + 1)) - 24*a*b*tan(d*x) - 24*a*b*tan(c) + 6*a^2 - 9*b^2)/(d*tan(d*x)^4*tan(c)^4 - 4*d*tan(d*x)
^3*tan(c)^3 + 6*d*tan(d*x)^2*tan(c)^2 - 4*d*tan(d*x)*tan(c) + d)

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Mupad [B]
time = 4.07, size = 90, normalized size = 0.92 \begin {gather*} \frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )-\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}-2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+\frac {2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+2\,a\,b\,d\,x}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3*(a + b*tan(c + d*x))^2,x)

[Out]

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

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