3.5.79 \(\int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [479]

Optimal. Leaf size=183 \[ \frac {a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {a^2 \left (a^4+3 a^2 b^2+6 b^4\right ) \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right )^3 d}-\frac {a^2 \tan ^2(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a^3 \left (a^2+3 b^2\right )}{b^3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.22, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3646, 3716, 3707, 3698, 31, 3556} \begin {gather*} -\frac {a^2 \tan ^2(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3}+\frac {a^2 \left (a^4+3 a^2 b^2+6 b^4\right ) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )^3}+\frac {a^3 \left (a^2+3 b^2\right )}{b^3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]

[Out]

(a*(a^2 - 3*b^2)*x)/(a^2 + b^2)^3 + (b*(3*a^2 - b^2)*Log[Cos[c + d*x]])/((a^2 + b^2)^3*d) + (a^2*(a^4 + 3*a^2*
b^2 + 6*b^4)*Log[a + b*Tan[c + d*x]])/(b^3*(a^2 + b^2)^3*d) - (a^2*Tan[c + d*x]^2)/(2*b*(a^2 + b^2)*d*(a + b*T
an[c + d*x])^2) + (a^3*(a^2 + 3*b^2))/(b^3*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3556

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

Rule 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3698

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 3707

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[(a*A + b*B - a*C)*(x/(a^2 + b^2)), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

Rule 3716

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(c^2*C - B*c*d + A*d^2)
*((c + d*Tan[e + f*x])^(n + 1)/(d^2*f*(n + 1)*(c^2 + d^2))), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f
*x])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b*(c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d
+ a*C*d)*Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] &
& NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=-\frac {a^2 \tan ^2(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {\tan (c+d x) \left (2 a^2-2 a b \tan (c+d x)+2 \left (a^2+b^2\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=-\frac {a^2 \tan ^2(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a^3 \left (a^2+3 b^2\right )}{b^3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {2 a^2 \left (a^2+3 b^2\right )-4 a b^3 \tan (c+d x)+2 \left (a^2+b^2\right )^2 \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=\frac {a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac {a^2 \tan ^2(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a^3 \left (a^2+3 b^2\right )}{b^3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (b \left (3 a^2-b^2\right )\right ) \int \tan (c+d x) \, dx}{\left (a^2+b^2\right )^3}+\frac {\left (a^2 \left (a^4+3 a^2 b^2+6 b^4\right )\right ) \int \frac {1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^2 \left (a^2+b^2\right )^3}\\ &=\frac {a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac {a^2 \tan ^2(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a^3 \left (a^2+3 b^2\right )}{b^3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left (a^2 \left (a^4+3 a^2 b^2+6 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^3 \left (a^2+b^2\right )^3 d}\\ &=\frac {a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {a^2 \left (a^4+3 a^2 b^2+6 b^4\right ) \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right )^3 d}-\frac {a^2 \tan ^2(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a^3 \left (a^2+3 b^2\right )}{b^3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.20, size = 351, normalized size = 1.92 \begin {gather*} \frac {\sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (-a^4 b^2 \left (a^2+b^2\right )-2 a^2 b \left (a^2+b^2\right ) \left (a^2+4 b^2\right ) \sin (c+d x) (a \cos (c+d x)+b \sin (c+d x))+2 a b^3 \left (a^2-3 b^2\right ) (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2+2 i a^2 \left (a^4+3 a^2 b^2+6 b^4\right ) (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2-2 i a^2 \left (a^4+3 a^2 b^2+6 b^4\right ) \text {ArcTan}(\tan (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^2-2 \left (a^2+b^2\right )^3 \log (\cos (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^2+a^2 \left (a^4+3 a^2 b^2+6 b^4\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2\right )}{2 b^3 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])*(-(a^4*b^2*(a^2 + b^2)) - 2*a^2*b*(a^2 + b^2)*(a^2 + 4*b^2)*
Sin[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x]) + 2*a*b^3*(a^2 - 3*b^2)*(c + d*x)*(a*Cos[c + d*x] + b*Sin[c + d
*x])^2 + (2*I)*a^2*(a^4 + 3*a^2*b^2 + 6*b^4)*(c + d*x)*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 - (2*I)*a^2*(a^4 +
3*a^2*b^2 + 6*b^4)*ArcTan[Tan[c + d*x]]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 - 2*(a^2 + b^2)^3*Log[Cos[c + d*x]
]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 + a^2*(a^4 + 3*a^2*b^2 + 6*b^4)*Log[(a*Cos[c + d*x] + b*Sin[c + d*x])^2]
*(a*Cos[c + d*x] + b*Sin[c + d*x])^2))/(2*b^3*(a^2 + b^2)^3*d*(a + b*Tan[c + d*x])^3)

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Maple [A]
time = 0.19, size = 169, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (-3 a^{2} b +b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (a^{3}-3 b^{2} a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}-\frac {a^{4}}{2 b^{3} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {a^{2} \left (a^{4}+3 a^{2} b^{2}+6 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3} b^{3}}+\frac {2 a^{3} \left (a^{2}+2 b^{2}\right )}{b^{3} \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(169\)
default	\(\frac {\frac {\frac {\left (-3 a^{2} b +b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (a^{3}-3 b^{2} a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}-\frac {a^{4}}{2 b^{3} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {a^{2} \left (a^{4}+3 a^{2} b^{2}+6 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3} b^{3}}+\frac {2 a^{3} \left (a^{2}+2 b^{2}\right )}{b^{3} \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(169\)
norman	\(\frac {\frac {\left (a^{2}-3 b^{2}\right ) a^{3} x}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}+\frac {b^{2} \left (a^{2}-3 b^{2}\right ) a x \left (\tan ^{2}\left (d x +c \right )\right )}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}+\frac {a^{2} \left (3 a^{4}+7 a^{2} b^{2}\right )}{2 d \,b^{3} \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {2 a \left (a^{4}+2 a^{2} b^{2}\right ) \tan \left (d x +c \right )}{d \,b^{2} \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {2 b \left (a^{2}-3 b^{2}\right ) a^{2} x \tan \left (d x +c \right )}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {a^{2} \left (a^{4}+3 a^{2} b^{2}+6 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) d \,b^{3}}-\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}\)	\(352\)
risch	\(-\frac {x}{3 i b \,a^{2}-i b^{3}-a^{3}+3 b^{2} a}+\frac {2 i x}{b^{3}}+\frac {2 i c}{d \,b^{3}}-\frac {2 i a^{6} x}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) b^{3}}-\frac {2 i a^{6} c}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) b^{3} d}-\frac {6 i a^{4} x}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) b}-\frac {6 i a^{4} c}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) b d}-\frac {12 i a^{2} b x}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}-\frac {12 i a^{2} b c}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) d}+\frac {2 i a^{3} \left (i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+3 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+i a^{3}+4 i a \,b^{2}-a^{2} b -4 b^{3}\right )}{\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} \left (-i a +b \right )^{2} b^{2} d \left (i a +b \right )^{3}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{3} d}+\frac {a^{6} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) b^{3} d}+\frac {3 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) b d}+\frac {6 a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) d}\)	\(623\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 279, normalized size = 1.52 \begin {gather*} \frac {\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 6 \, a^{2} b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}} - \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {3 \, a^{6} + 7 \, a^{4} b^{2} + 4 \, {\left (a^{5} b + 2 \, a^{3} b^{3}\right )} \tan \left (d x + c\right )}{a^{6} b^{3} + 2 \, a^{4} b^{5} + a^{2} b^{7} + {\left (a^{4} b^{5} + 2 \, a^{2} b^{7} + b^{9}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{4} + 2 \, a^{3} b^{6} + a b^{8}\right )} \tan \left (d x + c\right )}}{2 \, 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))^3,x, algorithm="maxima")

[Out]

1/2*(2*(a^3 - 3*a*b^2)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(a^6 + 3*a^4*b^2 + 6*a^2*b^4)*log(b*t
an(d*x + c) + a)/(a^6*b^3 + 3*a^4*b^5 + 3*a^2*b^7 + b^9) - (3*a^2*b - b^3)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^
4*b^2 + 3*a^2*b^4 + b^6) + (3*a^6 + 7*a^4*b^2 + 4*(a^5*b + 2*a^3*b^3)*tan(d*x + c))/(a^6*b^3 + 2*a^4*b^5 + a^2
*b^7 + (a^4*b^5 + 2*a^2*b^7 + b^9)*tan(d*x + c)^2 + 2*(a^5*b^4 + 2*a^3*b^6 + a*b^8)*tan(d*x + c)))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (181) = 362\).
time = 1.35, size = 481, normalized size = 2.63 \begin {gather*} \frac {a^{6} b^{2} + 7 \, a^{4} b^{4} + 2 \, {\left (a^{5} b^{3} - 3 \, a^{3} b^{5}\right )} d x - {\left (3 \, a^{6} b^{2} + 9 \, a^{4} b^{4} - 2 \, {\left (a^{3} b^{5} - 3 \, a b^{7}\right )} d x\right )} \tan \left (d x + c\right )^{2} + {\left (a^{8} + 3 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + {\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 6 \, a^{2} b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 6 \, a^{3} b^{5}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - {\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6} + {\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} - 4 \, a^{3} b^{5} - 2 \, {\left (a^{4} b^{4} - 3 \, a^{2} b^{6}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{5} + 3 \, a^{4} b^{7} + 3 \, a^{2} b^{9} + b^{11}\right )} d \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b^{4} + 3 \, a^{5} b^{6} + 3 \, a^{3} b^{8} + a b^{10}\right )} d \tan \left (d x + c\right ) + {\left (a^{8} b^{3} + 3 \, a^{6} b^{5} + 3 \, a^{4} b^{7} + a^{2} b^{9}\right )} d\right )}} \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))^3,x, algorithm="fricas")

[Out]

1/2*(a^6*b^2 + 7*a^4*b^4 + 2*(a^5*b^3 - 3*a^3*b^5)*d*x - (3*a^6*b^2 + 9*a^4*b^4 - 2*(a^3*b^5 - 3*a*b^7)*d*x)*t
an(d*x + c)^2 + (a^8 + 3*a^6*b^2 + 6*a^4*b^4 + (a^6*b^2 + 3*a^4*b^4 + 6*a^2*b^6)*tan(d*x + c)^2 + 2*(a^7*b + 3
*a^5*b^3 + 6*a^3*b^5)*tan(d*x + c))*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1))
- (a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*tan(d*x + c)^2 + 2*(a^7*b +
3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*tan(d*x + c))*log(1/(tan(d*x + c)^2 + 1)) - 2*(a^7*b + 3*a^5*b^3 - 4*a^3*b^5 -
2*(a^4*b^4 - 3*a^2*b^6)*d*x)*tan(d*x + c))/((a^6*b^5 + 3*a^4*b^7 + 3*a^2*b^9 + b^11)*d*tan(d*x + c)^2 + 2*(a^
7*b^4 + 3*a^5*b^6 + 3*a^3*b^8 + a*b^10)*d*tan(d*x + c) + (a^8*b^3 + 3*a^6*b^5 + 3*a^4*b^7 + a^2*b^9)*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: AttributeError} \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))**3,x)

[Out]

Exception raised: AttributeError >> 'NoneType' object has no attribute 'primitive'

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Giac [A]
time = 1.43, size = 304, normalized size = 1.66 \begin {gather*} \frac {\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 6 \, a^{2} b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}} - \frac {3 \, a^{6} b \tan \left (d x + c\right )^{2} + 9 \, a^{4} b^{3} \tan \left (d x + c\right )^{2} + 18 \, a^{2} b^{5} \tan \left (d x + c\right )^{2} + 2 \, a^{7} \tan \left (d x + c\right ) + 6 \, a^{5} b^{2} \tan \left (d x + c\right ) + 28 \, a^{3} b^{4} \tan \left (d x + c\right ) - a^{6} b + 11 \, a^{4} b^{3}}{{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, 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))^3,x, algorithm="giac")

[Out]

1/2*(2*(a^3 - 3*a*b^2)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (3*a^2*b - b^3)*log(tan(d*x + c)^2 + 1)
/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(a^6 + 3*a^4*b^2 + 6*a^2*b^4)*log(abs(b*tan(d*x + c) + a))/(a^6*b^3 +
3*a^4*b^5 + 3*a^2*b^7 + b^9) - (3*a^6*b*tan(d*x + c)^2 + 9*a^4*b^3*tan(d*x + c)^2 + 18*a^2*b^5*tan(d*x + c)^2
+ 2*a^7*tan(d*x + c) + 6*a^5*b^2*tan(d*x + c) + 28*a^3*b^4*tan(d*x + c) - a^6*b + 11*a^4*b^3)/((a^6*b^2 + 3*a
^4*b^4 + 3*a^2*b^6 + b^8)*(b*tan(d*x + c) + a)^2))/d

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Mupad [B]
time = 4.21, size = 240, normalized size = 1.31 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )}+\frac {\frac {3\,a^6+7\,a^4\,b^2}{2\,b^3\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {2\,a^3\,\mathrm {tan}\left (c+d\,x\right )\,\left (a^2+2\,b^2\right )}{b^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}+\frac {a^2\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^4+3\,a^2\,b^2+6\,b^4\right )}{b^3\,d\,{\left (a^2+b^2\right )}^3}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )} \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))^3,x)

[Out]

log(tan(c + d*x) + 1i)/(2*d*(a*b^2*3i - 3*a^2*b - a^3*1i + b^3)) + (log(tan(c + d*x) - 1i)*1i)/(2*d*(3*a*b^2 -
a^2*b*3i - a^3 + b^3*1i)) + ((3*a^6 + 7*a^4*b^2)/(2*b^3*(a^4 + b^4 + 2*a^2*b^2)) + (2*a^3*tan(c + d*x)*(a^2 +
2*b^2))/(b^2*(a^4 + b^4 + 2*a^2*b^2)))/(d*(a^2 + b^2*tan(c + d*x)^2 + 2*a*b*tan(c + d*x))) + (a^2*log(a + b*t
an(c + d*x))*(a^4 + 6*b^4 + 3*a^2*b^2))/(b^3*d*(a^2 + b^2)^3)

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