3.447 \(\int \tan (c+d x) (a+b \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=130 \[ \frac {\left (a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac {a b \left (a^2-3 b^2\right ) \tan (c+d x)}{d}-4 a b x \left (a^2-b^2\right )-\frac {\left (a^4-6 a^2 b^2+b^4\right ) \log (\cos (c+d x))}{d}+\frac {(a+b \tan (c+d x))^4}{4 d}+\frac {a (a+b \tan (c+d x))^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3528, 3525, 3475} \[ \frac {\left (a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac {a b \left (a^2-3 b^2\right ) \tan (c+d x)}{d}-\frac {\left (-6 a^2 b^2+a^4+b^4\right ) \log (\cos (c+d x))}{d}-4 a b x \left (a^2-b^2\right )+\frac {(a+b \tan (c+d x))^4}{4 d}+\frac {a (a+b \tan (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3475

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

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 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]

Rubi steps

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

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Mathematica [C]  time = 1.66, size = 123, normalized size = 0.95 \[ \frac {-6 b^2 \left (b^2-6 a^2\right ) \tan ^2(c+d x)+48 a b \left (a^2-b^2\right ) \tan (c+d x)+16 a b^3 \tan ^3(c+d x)+6 \left ((a-i b)^4 \log (\tan (c+d x)+i)+(a+i b)^4 \log (-\tan (c+d x)+i)\right )+3 b^4 \tan ^4(c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*((a + I*b)^4*Log[I - Tan[c + d*x]] + (a - I*b)^4*Log[I + Tan[c + d*x]]) + 48*a*b*(a^2 - b^2)*Tan[c + d*x] -
6*b^2*(-6*a^2 + b^2)*Tan[c + d*x]^2 + 16*a*b^3*Tan[c + d*x]^3 + 3*b^4*Tan[c + d*x]^4)/(12*d)

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fricas [A]  time = 0.44, size = 123, normalized size = 0.95 \[ \frac {3 \, b^{4} \tan \left (d x + c\right )^{4} + 16 \, a b^{3} \tan \left (d x + c\right )^{3} - 48 \, {\left (a^{3} b - a b^{3}\right )} d x + 6 \, {\left (6 \, a^{2} b^{2} - b^{4}\right )} \tan \left (d x + c\right )^{2} - 6 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 48 \, {\left (a^{3} b - a b^{3}\right )} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^4*tan(d*x + c)^4 + 16*a*b^3*tan(d*x + c)^3 - 48*(a^3*b - a*b^3)*d*x + 6*(6*a^2*b^2 - b^4)*tan(d*x +
c)^2 - 6*(a^4 - 6*a^2*b^2 + b^4)*log(1/(tan(d*x + c)^2 + 1)) + 48*(a^3*b - a*b^3)*tan(d*x + c))/d

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giac [B]  time = 10.72, size = 1886, normalized size = 14.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(48*a^3*b*d*x*tan(d*x)^4*tan(c)^4 - 48*a*b^3*d*x*tan(d*x)^4*tan(c)^4 + 6*a^4*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*t
an(c)^4 - 36*a^2*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*t
an(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^4*tan(c)^4 + 6*b^4*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 - 192*a^3*b
*d*x*tan(d*x)^3*tan(c)^3 + 192*a*b^3*d*x*tan(d*x)^3*tan(c)^3 - 36*a^2*b^2*tan(d*x)^4*tan(c)^4 + 9*b^4*tan(d*x)
^4*tan(c)^4 - 24*a^4*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*t
an(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 + 144*a^2*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*
b^4*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 + 48*a^3*b*tan(d*x)^4*tan(c)^3 - 48*a*b^3*tan(d*x)^4*tan(c)^3 + 48*a^3*
b*tan(d*x)^3*tan(c)^4 - 48*a*b^3*tan(d*x)^3*tan(c)^4 + 288*a^3*b*d*x*tan(d*x)^2*tan(c)^2 - 288*a*b^3*d*x*tan(d
*x)^2*tan(c)^2 - 36*a^2*b^2*tan(d*x)^4*tan(c)^2 + 6*b^4*tan(d*x)^4*tan(c)^2 + 72*a^2*b^2*tan(d*x)^3*tan(c)^3 -
24*b^4*tan(d*x)^3*tan(c)^3 - 36*a^2*b^2*tan(d*x)^2*tan(c)^4 + 6*b^4*tan(d*x)^2*tan(c)^4 + 16*a*b^3*tan(d*x)^4
*tan(c) + 36*a^4*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 - 216*a^2*b^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*ta
n(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 + 36*b^4*
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 - 144*a^3*b*tan(d*x)^3*tan(c)^2 + 192*a*b^3*tan(d*x)^3*tan(c)^2 - 144*a^3*b
*tan(d*x)^2*tan(c)^3 + 192*a*b^3*tan(d*x)^2*tan(c)^3 + 16*a*b^3*tan(d*x)*tan(c)^4 - 3*b^4*tan(d*x)^4 - 192*a^3
*b*d*x*tan(d*x)*tan(c) + 192*a*b^3*d*x*tan(d*x)*tan(c) + 72*a^2*b^2*tan(d*x)^3*tan(c) - 24*b^4*tan(d*x)^3*tan(
c) - 72*a^2*b^2*tan(d*x)^2*tan(c)^2 + 12*b^4*tan(d*x)^2*tan(c)^2 + 72*a^2*b^2*tan(d*x)*tan(c)^3 - 24*b^4*tan(d
*x)*tan(c)^3 - 3*b^4*tan(c)^4 - 16*a*b^3*tan(d*x)^3 - 24*a^4*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) + 144*a^2*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) - 24*b^4*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + t
an(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)*tan(c) + 144*a^3*b*tan(d*x)^2*tan(c) - 192*a*b^3*t
an(d*x)^2*tan(c) + 144*a^3*b*tan(d*x)*tan(c)^2 - 192*a*b^3*tan(d*x)*tan(c)^2 - 16*a*b^3*tan(c)^3 + 48*a^3*b*d*
x - 48*a*b^3*d*x - 36*a^2*b^2*tan(d*x)^2 + 6*b^4*tan(d*x)^2 + 72*a^2*b^2*tan(d*x)*tan(c) - 24*b^4*tan(d*x)*tan
(c) - 36*a^2*b^2*tan(c)^2 + 6*b^4*tan(c)^2 + 6*a^4*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)) - 36*a^2*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)) + 6*b^4*log(4*(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(c)^2
+ 1)) - 48*a^3*b*tan(d*x) + 48*a*b^3*tan(d*x) - 48*a^3*b*tan(c) + 48*a*b^3*tan(c) - 36*a^2*b^2 + 9*b^4)/(d*ta
n(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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maple [A]  time = 0.02, size = 192, normalized size = 1.48 \[ \frac {\left (\tan ^{4}\left (d x +c \right )\right ) b^{4}}{4 d}+\frac {4 \left (\tan ^{3}\left (d x +c \right )\right ) a \,b^{3}}{3 d}+\frac {3 a^{2} b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {b^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {4 a^{3} b \tan \left (d x +c \right )}{d}-\frac {4 a \,b^{3} \tan \left (d x +c \right )}{d}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{4}}{2 d}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} b^{2}}{d}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{4}}{2 d}-\frac {4 \arctan \left (\tan \left (d x +c \right )\right ) a^{3} b}{d}+\frac {4 \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.77, size = 124, normalized size = 0.95 \[ \frac {3 \, b^{4} \tan \left (d x + c\right )^{4} + 16 \, a b^{3} \tan \left (d x + c\right )^{3} + 6 \, {\left (6 \, a^{2} b^{2} - b^{4}\right )} \tan \left (d x + c\right )^{2} - 48 \, {\left (a^{3} b - a b^{3}\right )} {\left (d x + c\right )} + 6 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 48 \, {\left (a^{3} b - a b^{3}\right )} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^4*tan(d*x + c)^4 + 16*a*b^3*tan(d*x + c)^3 + 6*(6*a^2*b^2 - b^4)*tan(d*x + c)^2 - 48*(a^3*b - a*b^3)
*(d*x + c) + 6*(a^4 - 6*a^2*b^2 + b^4)*log(tan(d*x + c)^2 + 1) + 48*(a^3*b - a*b^3)*tan(d*x + c))/d

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mupad [B]  time = 3.92, size = 168, normalized size = 1.29 \[ \frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {a^4}{2}-3\,a^2\,b^2+\frac {b^4}{2}\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {b^4}{2}-3\,a^2\,b^2\right )}{d}+\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a\,b^3-4\,a^3\,b\right )}{d}+\frac {4\,a\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}+\frac {4\,a\,b\,\mathrm {atan}\left (\frac {4\,a\,b\,\mathrm {tan}\left (c+d\,x\right )\,\left (a+b\right )\,\left (a-b\right )}{4\,a\,b^3-4\,a^3\,b}\right )\,\left (a+b\right )\,\left (a-b\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.56, size = 187, normalized size = 1.44 \[ \begin {cases} \frac {a^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 4 a^{3} b x + \frac {4 a^{3} b \tan {\left (c + d x \right )}}{d} - \frac {3 a^{2} b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {3 a^{2} b^{2} \tan ^{2}{\left (c + d x \right )}}{d} + 4 a b^{3} x + \frac {4 a b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 a b^{3} \tan {\left (c + d x \right )}}{d} + \frac {b^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b^{4} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{4} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\relax (c )}\right )^{4} \tan {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a+b*tan(d*x+c))**4,x)

[Out]

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

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