3.1 \(\int \tan (c+d x) (a+b \tan (c+d x)) (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\)
Optimal. Leaf size=87 \[ \frac {(a C+b B) \tan ^2(c+d x)}{2 d}+\frac {(a B-b C) \tan (c+d x)}{d}+\frac {(a C+b B) \log (\cos (c+d x))}{d}-x (a B-b C)+\frac {b C \tan ^3(c+d x)}{3 d} \]
[Out]
-(B*a-C*b)*x+(B*b+C*a)*ln(cos(d*x+c))/d+(B*a-C*b)*tan(d*x+c)/d+1/2*(B*b+C*a)*tan(d*x+c)^2/d+1/3*b*C*tan(d*x+c)
^3/d
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Rubi [A] time = 0.13, antiderivative size = 87, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used =
{3632, 3592, 3528, 3525, 3475} \[ \frac {(a C+b B) \tan ^2(c+d x)}{2 d}+\frac {(a B-b C) \tan (c+d x)}{d}+\frac {(a C+b B) \log (\cos (c+d x))}{d}-x (a B-b C)+\frac {b C \tan ^3(c+d x)}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[Tan[c + d*x]*(a + b*Tan[c + d*x])*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]
[Out]
-((a*B - b*C)*x) + ((b*B + a*C)*Log[Cos[c + d*x]])/d + ((a*B - b*C)*Tan[c + d*x])/d + ((b*B + a*C)*Tan[c + d*x
]^2)/(2*d) + (b*C*Tan[c + d*x]^3)/(3*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]
Rule 3592
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(B*d*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] && !LeQ[m, -1]
Rule 3632
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Tan[e + f*x])
^(m + 1)*(c + d*Tan[e + f*x])^n*(b*B - a*C + b*C*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Rubi steps
\begin {align*} \int \tan (c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx &=\int \tan ^2(c+d x) (a+b \tan (c+d x)) (B+C \tan (c+d x)) \, dx\\ &=\frac {b C \tan ^3(c+d x)}{3 d}+\int \tan ^2(c+d x) (a B-b C+(b B+a C) \tan (c+d x)) \, dx\\ &=\frac {(b B+a C) \tan ^2(c+d x)}{2 d}+\frac {b C \tan ^3(c+d x)}{3 d}+\int \tan (c+d x) (-b B-a C+(a B-b C) \tan (c+d x)) \, dx\\ &=-(a B-b C) x+\frac {(a B-b C) \tan (c+d x)}{d}+\frac {(b B+a C) \tan ^2(c+d x)}{2 d}+\frac {b C \tan ^3(c+d x)}{3 d}+(-b B-a C) \int \tan (c+d x) \, dx\\ &=-(a B-b C) x+\frac {(b B+a C) \log (\cos (c+d x))}{d}+\frac {(a B-b C) \tan (c+d x)}{d}+\frac {(b B+a C) \tan ^2(c+d x)}{2 d}+\frac {b C \tan ^3(c+d x)}{3 d}\\ \end {align*}
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Mathematica [A] time = 0.60, size = 86, normalized size = 0.99 \[ \frac {(6 b C-6 a B) \tan ^{-1}(\tan (c+d x))+3 (a C+b B) \tan ^2(c+d x)+6 (a B-b C) \tan (c+d x)+6 (a C+b B) \log (\cos (c+d x))+2 b C \tan ^3(c+d x)}{6 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[c + d*x]*(a + b*Tan[c + d*x])*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]
[Out]
((-6*a*B + 6*b*C)*ArcTan[Tan[c + d*x]] + 6*(b*B + a*C)*Log[Cos[c + d*x]] + 6*(a*B - b*C)*Tan[c + d*x] + 3*(b*B
+ a*C)*Tan[c + d*x]^2 + 2*b*C*Tan[c + d*x]^3)/(6*d)
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fricas [A] time = 0.60, size = 85, normalized size = 0.98 \[ \frac {2 \, C b \tan \left (d x + c\right )^{3} - 6 \, {\left (B a - C b\right )} d x + 3 \, {\left (C a + B b\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (C a + B b\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 6 \, {\left (B a - C b\right )} \tan \left (d x + c\right )}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")
[Out]
1/6*(2*C*b*tan(d*x + c)^3 - 6*(B*a - C*b)*d*x + 3*(C*a + B*b)*tan(d*x + c)^2 + 3*(C*a + B*b)*log(1/(tan(d*x +
c)^2 + 1)) + 6*(B*a - C*b)*tan(d*x + c))/d
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giac [B] time = 3.96, size = 1017, normalized size = 11.69 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")
[Out]
-1/6*(6*B*a*d*x*tan(d*x)^3*tan(c)^3 - 6*C*b*d*x*tan(d*x)^3*tan(c)^3 - 3*C*a*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
- 3*B*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)^3*tan(c)^3 - 18*B*a*d*x*tan(d*x)^2*tan(c)^2 + 18*C*b*d*x*tan(d*x)^2*tan(c)^2
- 3*C*a*tan(d*x)^3*tan(c)^3 - 3*B*b*tan(d*x)^3*tan(c)^3 + 9*C*a*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 + 9*B*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 + 6*B*a*tan(d*x)^3*tan(c)^2 - 6*C*b*tan(d*x)^3*tan(c)^2 + 6*B*a*tan(d*x)^2*tan
(c)^3 - 6*C*b*tan(d*x)^2*tan(c)^3 + 18*B*a*d*x*tan(d*x)*tan(c) - 18*C*b*d*x*tan(d*x)*tan(c) - 3*C*a*tan(d*x)^3
*tan(c) - 3*B*b*tan(d*x)^3*tan(c) + 3*C*a*tan(d*x)^2*tan(c)^2 + 3*B*b*tan(d*x)^2*tan(c)^2 - 3*C*a*tan(d*x)*tan
(c)^3 - 3*B*b*tan(d*x)*tan(c)^3 + 2*C*b*tan(d*x)^3 - 9*C*a*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) - 9*B*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) - 12*B*a*tan(d*x)^2*tan(c) + 18*C*b*tan(d*x)^2*tan(c) - 12*B*a*tan(d*x)*tan(c)^2 + 18*C*b*t
an(d*x)*tan(c)^2 + 2*C*b*tan(c)^3 - 6*B*a*d*x + 6*C*b*d*x + 3*C*a*tan(d*x)^2 + 3*B*b*tan(d*x)^2 - 3*C*a*tan(d*
x)*tan(c) - 3*B*b*tan(d*x)*tan(c) + 3*C*a*tan(c)^2 + 3*B*b*tan(c)^2 + 3*C*a*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)) + 3*B*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)) + 6*B*a*tan(d*x) - 6*C*b*tan(d*x) + 6*B*a*tan(c) - 6*C*b*tan(c) + 3*C*a + 3*B*b)/(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)
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maple [A] time = 0.03, size = 135, normalized size = 1.55 \[ \frac {b C \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {b B \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {C \left (\tan ^{2}\left (d x +c \right )\right ) a}{2 d}+\frac {a B \tan \left (d x +c \right )}{d}-\frac {b C \tan \left (d x +c \right )}{d}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B b}{2 d}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a C}{2 d}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) a}{d}+\frac {C \arctan \left (\tan \left (d x +c \right )\right ) b}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(d*x+c)*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x)
[Out]
1/3*b*C*tan(d*x+c)^3/d+1/2*b*B*tan(d*x+c)^2/d+1/2/d*C*tan(d*x+c)^2*a+1/d*a*B*tan(d*x+c)-b*C*tan(d*x+c)/d-1/2/d
*ln(1+tan(d*x+c)^2)*B*b-1/2/d*ln(1+tan(d*x+c)^2)*a*C-1/d*B*arctan(tan(d*x+c))*a+1/d*C*arctan(tan(d*x+c))*b
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maxima [A] time = 0.54, size = 86, normalized size = 0.99 \[ \frac {2 \, C b \tan \left (d x + c\right )^{3} + 3 \, {\left (C a + B b\right )} \tan \left (d x + c\right )^{2} - 6 \, {\left (B a - C b\right )} {\left (d x + c\right )} - 3 \, {\left (C a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 \, {\left (B a - C b\right )} \tan \left (d x + c\right )}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")
[Out]
1/6*(2*C*b*tan(d*x + c)^3 + 3*(C*a + B*b)*tan(d*x + c)^2 - 6*(B*a - C*b)*(d*x + c) - 3*(C*a + B*b)*log(tan(d*x
+ c)^2 + 1) + 6*(B*a - C*b)*tan(d*x + c))/d
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mupad [B] time = 8.83, size = 84, normalized size = 0.97 \[ \frac {\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a-C\,b\right )-\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {B\,b}{2}+\frac {C\,a}{2}\right )+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {B\,b}{2}+\frac {C\,a}{2}\right )-d\,x\,\left (B\,a-C\,b\right )+\frac {C\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(c + d*x)*(B*tan(c + d*x) + C*tan(c + d*x)^2)*(a + b*tan(c + d*x)),x)
[Out]
(tan(c + d*x)*(B*a - C*b) - log(tan(c + d*x)^2 + 1)*((B*b)/2 + (C*a)/2) + tan(c + d*x)^2*((B*b)/2 + (C*a)/2) -
d*x*(B*a - C*b) + (C*b*tan(c + d*x)^3)/3)/d
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sympy [A] time = 0.37, size = 139, normalized size = 1.60 \[ \begin {cases} - B a x + \frac {B a \tan {\left (c + d x \right )}}{d} - \frac {B b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b \tan ^{2}{\left (c + d x \right )}}{2 d} - \frac {C a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C a \tan ^{2}{\left (c + d x \right )}}{2 d} + C b x + \frac {C b \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {C b \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\relax (c )}\right ) \left (B \tan {\relax (c )} + C \tan ^{2}{\relax (c )}\right ) \tan {\relax (c )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)**2),x)
[Out]
Piecewise((-B*a*x + B*a*tan(c + d*x)/d - B*b*log(tan(c + d*x)**2 + 1)/(2*d) + B*b*tan(c + d*x)**2/(2*d) - C*a*
log(tan(c + d*x)**2 + 1)/(2*d) + C*a*tan(c + d*x)**2/(2*d) + C*b*x + C*b*tan(c + d*x)**3/(3*d) - C*b*tan(c + d
*x)/d, Ne(d, 0)), (x*(a + b*tan(c))*(B*tan(c) + C*tan(c)**2)*tan(c), True))
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