3.5.24 \(\int \tan ^4(c+d x) (a+b \tan (c+d x))^2 \, dx\) [424]

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

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, 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 ^3(c+d x)}{3 d}-\frac {\left (a^2-b^2\right ) \tan (c+d x)}{d}+x \left (a^2-b^2\right )+\frac {a b \tan ^4(c+d x)}{2 d}-\frac {a b \tan ^2(c+d x)}{d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \tan ^5(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4*(a + b*Tan[c + d*x])^2,x]

[Out]

(a^2 - b^2)*x - (2*a*b*Log[Cos[c + d*x]])/d - ((a^2 - b^2)*Tan[c + d*x])/d - (a*b*Tan[c + d*x]^2)/d + ((a^2 -
b^2)*Tan[c + d*x]^3)/(3*d) + (a*b*Tan[c + d*x]^4)/(2*d) + (b^2*Tan[c + d*x]^5)/(5*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 ^4(c+d x) (a+b \tan (c+d x))^2 \, dx &=\frac {b^2 \tan ^5(c+d x)}{5 d}+\int \tan ^4(c+d x) \left (a^2-b^2+2 a b \tan (c+d x)\right ) \, dx\\ &=\frac {a b \tan ^4(c+d x)}{2 d}+\frac {b^2 \tan ^5(c+d x)}{5 d}+\int \tan ^3(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 ^3(c+d x)}{3 d}+\frac {a b \tan ^4(c+d x)}{2 d}+\frac {b^2 \tan ^5(c+d x)}{5 d}+\int \tan ^2(c+d x) \left (-a^2+b^2-2 a b \tan (c+d x)\right ) \, dx\\ &=-\frac {a b \tan ^2(c+d x)}{d}+\frac {\left (a^2-b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {a b \tan ^4(c+d x)}{2 d}+\frac {b^2 \tan ^5(c+d x)}{5 d}+\int \tan (c+d x) \left (2 a b-\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=\left (a^2-b^2\right ) x-\frac {\left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac {a b \tan ^2(c+d x)}{d}+\frac {\left (a^2-b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {a b \tan ^4(c+d x)}{2 d}+\frac {b^2 \tan ^5(c+d x)}{5 d}+(2 a b) \int \tan (c+d x) \, dx\\ &=\left (a^2-b^2\right ) x-\frac {2 a b \log (\cos (c+d x))}{d}-\frac {\left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac {a b \tan ^2(c+d x)}{d}+\frac {\left (a^2-b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {a b \tan ^4(c+d x)}{2 d}+\frac {b^2 \tan ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]
time = 0.64, size = 110, normalized size = 0.92 \begin {gather*} \frac {30 \left (a^2-b^2\right ) \text {ArcTan}(\tan (c+d x))-60 a b \log (\cos (c+d x))-30 \left (a^2-b^2\right ) \tan (c+d x)-30 a b \tan ^2(c+d x)+10 \left (a^2-b^2\right ) \tan ^3(c+d x)+15 a b \tan ^4(c+d x)+6 b^2 \tan ^5(c+d x)}{30 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4*(a + b*Tan[c + d*x])^2,x]

[Out]

(30*(a^2 - b^2)*ArcTan[Tan[c + d*x]] - 60*a*b*Log[Cos[c + d*x]] - 30*(a^2 - b^2)*Tan[c + d*x] - 30*a*b*Tan[c +
d*x]^2 + 10*(a^2 - b^2)*Tan[c + d*x]^3 + 15*a*b*Tan[c + d*x]^4 + 6*b^2*Tan[c + d*x]^5)/(30*d)

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Maple [A]
time = 0.04, size = 121, normalized size = 1.01

method	result	size
norman	\(\left (a^{2}-b^{2}\right ) x +\frac {b^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}-\frac {\left (a^{2}-b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (a^{2}-b^{2}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a b \left (\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {a b \left (\tan ^{4}\left (d x +c \right )\right )}{2 d}+\frac {a b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(118\)
derivativedivides	\(\frac {\frac {b^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {a b \left (\tan ^{4}\left (d x +c \right )\right )}{2}+\frac {a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {b^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3}-a b \left (\tan ^{2}\left (d x +c \right )\right )-a^{2} \tan \left (d x +c \right )+b^{2} \tan \left (d x +c \right )+a b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+\left (a^{2}-b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(121\)
default	\(\frac {\frac {b^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {a b \left (\tan ^{4}\left (d x +c \right )\right )}{2}+\frac {a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {b^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3}-a b \left (\tan ^{2}\left (d x +c \right )\right )-a^{2} \tan \left (d x +c \right )+b^{2} \tan \left (d x +c \right )+a b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+\left (a^{2}-b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(121\)
risch	\(2 i a b x +a^{2} x -b^{2} x +\frac {4 i a b c}{d}-\frac {2 i \left (-60 i a b \,{\mathrm e}^{8 i \left (d x +c \right )}+30 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-45 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-120 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+90 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-90 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-120 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+110 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-140 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-60 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+70 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-70 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+20 a^{2}-23 b^{2}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/5*b^2*tan(d*x+c)^5+1/2*a*b*tan(d*x+c)^4+1/3*a^2*tan(d*x+c)^3-1/3*b^2*tan(d*x+c)^3-a*b*tan(d*x+c)^2-a^2*
tan(d*x+c)+b^2*tan(d*x+c)+a*b*ln(1+tan(d*x+c)^2)+(a^2-b^2)*arctan(tan(d*x+c)))

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Maxima [A]
time = 0.51, size = 110, normalized size = 0.92 \begin {gather*} \frac {6 \, b^{2} \tan \left (d x + c\right )^{5} + 15 \, a b \tan \left (d x + c\right )^{4} - 30 \, a b \tan \left (d x + c\right )^{2} + 10 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{3} + 30 \, a b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 30 \, {\left (a^{2} - b^{2}\right )} {\left (d x + c\right )} - 30 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/30*(6*b^2*tan(d*x + c)^5 + 15*a*b*tan(d*x + c)^4 - 30*a*b*tan(d*x + c)^2 + 10*(a^2 - b^2)*tan(d*x + c)^3 + 3
0*a*b*log(tan(d*x + c)^2 + 1) + 30*(a^2 - b^2)*(d*x + c) - 30*(a^2 - b^2)*tan(d*x + c))/d

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Fricas [A]
time = 1.00, size = 109, normalized size = 0.91 \begin {gather*} \frac {6 \, b^{2} \tan \left (d x + c\right )^{5} + 15 \, a b \tan \left (d x + c\right )^{4} - 30 \, a b \tan \left (d x + c\right )^{2} + 10 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{3} + 30 \, {\left (a^{2} - b^{2}\right )} d x - 30 \, a b \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 30 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/30*(6*b^2*tan(d*x + c)^5 + 15*a*b*tan(d*x + c)^4 - 30*a*b*tan(d*x + c)^2 + 10*(a^2 - b^2)*tan(d*x + c)^3 + 3
0*(a^2 - b^2)*d*x - 30*a*b*log(1/(tan(d*x + c)^2 + 1)) - 30*(a^2 - b^2)*tan(d*x + c))/d

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Sympy [A]
time = 0.18, size = 139, normalized size = 1.16 \begin {gather*} \begin {cases} a^{2} x + \frac {a^{2} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \tan {\left (c + d x \right )}}{d} + \frac {a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {a b \tan ^{4}{\left (c + d x \right )}}{2 d} - \frac {a b \tan ^{2}{\left (c + d x \right )}}{d} - b^{2} x + \frac {b^{2} \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac {b^{2} \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{2} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right )^{2} \tan ^{4}{\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)**4*(a+b*tan(d*x+c))**2,x)

[Out]

Piecewise((a**2*x + a**2*tan(c + d*x)**3/(3*d) - a**2*tan(c + d*x)/d + a*b*log(tan(c + d*x)**2 + 1)/d + a*b*ta
n(c + d*x)**4/(2*d) - a*b*tan(c + d*x)**2/d - b**2*x + b**2*tan(c + d*x)**5/(5*d) - b**2*tan(c + d*x)**3/(3*d)
+ b**2*tan(c + d*x)/d, Ne(d, 0)), (x*(a + b*tan(c))**2*tan(c)**4, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1315 vs. \(2 (114) = 228\).
time = 2.30, size = 1315, normalized size = 10.96 \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)^4*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/30*(30*a^2*d*x*tan(d*x)^5*tan(c)^5 - 30*b^2*d*x*tan(d*x)^5*tan(c)^5 - 30*a*b*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)^5*tan(c
)^5 - 150*a^2*d*x*tan(d*x)^4*tan(c)^4 + 150*b^2*d*x*tan(d*x)^4*tan(c)^4 - 45*a*b*tan(d*x)^5*tan(c)^5 + 150*a*b
*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 + 30*a^2*tan(d*x)^5*tan(c)^4 - 30*b^2*tan(d*x)^5*tan(c)^4 + 30*a^2*tan(d*x
)^4*tan(c)^5 - 30*b^2*tan(d*x)^4*tan(c)^5 + 300*a^2*d*x*tan(d*x)^3*tan(c)^3 - 300*b^2*d*x*tan(d*x)^3*tan(c)^3
- 30*a*b*tan(d*x)^5*tan(c)^3 + 165*a*b*tan(d*x)^4*tan(c)^4 - 30*a*b*tan(d*x)^3*tan(c)^5 - 10*a^2*tan(d*x)^5*ta
n(c)^2 + 10*b^2*tan(d*x)^5*tan(c)^2 - 300*a*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*ta
n(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 - 150*a^2*tan(d*x)^4*tan(c)^3
+ 150*b^2*tan(d*x)^4*tan(c)^3 - 150*a^2*tan(d*x)^3*tan(c)^4 + 150*b^2*tan(d*x)^3*tan(c)^4 - 10*a^2*tan(d*x)^2
*tan(c)^5 + 10*b^2*tan(d*x)^2*tan(c)^5 + 15*a*b*tan(d*x)^5*tan(c) - 300*a^2*d*x*tan(d*x)^2*tan(c)^2 + 300*b^2*
d*x*tan(d*x)^2*tan(c)^2 + 150*a*b*tan(d*x)^4*tan(c)^2 - 180*a*b*tan(d*x)^3*tan(c)^3 + 150*a*b*tan(d*x)^2*tan(c
)^4 + 15*a*b*tan(d*x)*tan(c)^5 - 6*b^2*tan(d*x)^5 + 20*a^2*tan(d*x)^4*tan(c) - 50*b^2*tan(d*x)^4*tan(c) + 300*
a*b*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)^2*tan(c)^2 + 240*a^2*tan(d*x)^3*tan(c)^2 - 300*b^2*tan(d*x)^3*tan(c)^2 + 240*a^2*t
an(d*x)^2*tan(c)^3 - 300*b^2*tan(d*x)^2*tan(c)^3 + 20*a^2*tan(d*x)*tan(c)^4 - 50*b^2*tan(d*x)*tan(c)^4 - 6*b^2
*tan(c)^5 - 15*a*b*tan(d*x)^4 + 150*a^2*d*x*tan(d*x)*tan(c) - 150*b^2*d*x*tan(d*x)*tan(c) - 150*a*b*tan(d*x)^3
*tan(c) + 180*a*b*tan(d*x)^2*tan(c)^2 - 150*a*b*tan(d*x)*tan(c)^3 - 15*a*b*tan(c)^4 - 10*a^2*tan(d*x)^3 + 10*b
^2*tan(d*x)^3 - 150*a*b*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) - 150*a^2*tan(d*x)^2*tan(c) + 150*b^2*tan(d*x)^2*tan(c)
- 150*a^2*tan(d*x)*tan(c)^2 + 150*b^2*tan(d*x)*tan(c)^2 - 10*a^2*tan(c)^3 + 10*b^2*tan(c)^3 - 30*a^2*d*x + 30
*b^2*d*x + 30*a*b*tan(d*x)^2 - 165*a*b*tan(d*x)*tan(c) + 30*a*b*tan(c)^2 + 30*a*b*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)) + 30*a^2*tan(
d*x) - 30*b^2*tan(d*x) + 30*a^2*tan(c) - 30*b^2*tan(c) + 45*a*b)/(d*tan(d*x)^5*tan(c)^5 - 5*d*tan(d*x)^4*tan(c
)^4 + 10*d*tan(d*x)^3*tan(c)^3 - 10*d*tan(d*x)^2*tan(c)^2 + 5*d*tan(d*x)*tan(c) - d)

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Mupad [B]
time = 4.09, size = 146, normalized size = 1.22 \begin {gather*} \frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {a^2}{3}-\frac {b^2}{3}\right )}{d}+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a^2-b^2\right )}{d}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a+b\right )\,\left (a-b\right )}{a^2-b^2}\right )\,\left (a+b\right )\,\left (a-b\right )}{d}+\frac {a\,b\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{d}-\frac {a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d}+\frac {a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^4}{2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4*(a + b*tan(c + d*x))^2,x)

[Out]

(tan(c + d*x)^3*(a^2/3 - b^2/3))/d + (b^2*tan(c + d*x)^5)/(5*d) - (tan(c + d*x)*(a^2 - b^2))/d + (atan((tan(c
+ d*x)*(a + b)*(a - b))/(a^2 - b^2))*(a + b)*(a - b))/d + (a*b*log(tan(c + d*x)^2 + 1))/d - (a*b*tan(c + d*x)^
2)/d + (a*b*tan(c + d*x)^4)/(2*d)

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