3.298 \(\int \frac {\tan ^3(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {B \tan ^2(c+d x)}{2 d}+\frac {B \log (\cos (c+d x))}{d} \]

[Out]

B*ln(cos(d*x+c))/d+1/2*B*tan(d*x+c)^2/d

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Rubi [A]  time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {21, 3473, 3475} \[ \frac {B \tan ^2(c+d x)}{2 d}+\frac {B \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*Log[Cos[c + d*x]])/d + (B*Tan[c + d*x]^2)/(2*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3475

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 26, normalized size = 0.90 \[ \frac {B \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(2*Log[Cos[c + d*x]] + Tan[c + d*x]^2))/(2*d)

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fricas [A]  time = 0.76, size = 31, normalized size = 1.07 \[ \frac {B \tan \left (d x + c\right )^{2} + B \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*tan(d*x + c)^2 + B*log(1/(tan(d*x + c)^2 + 1)))/d

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giac [B]  time = 0.76, size = 187, normalized size = 6.45 \[ -\frac {B \log \left ({\left | -\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} - \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 2 \right |}\right ) - B \log \left ({\left | -\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} - \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 2 \right |}\right ) + \frac {B {\left (\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )} + 6 \, B}{\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 2}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*log(abs(-(cos(d*x + c) + 1)/(cos(d*x + c) - 1) - (cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2)) - B*log(a
bs(-(cos(d*x + c) + 1)/(cos(d*x + c) - 1) - (cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2)) + (B*((cos(d*x + c) +
1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1)) + 6*B)/((cos(d*x + c) + 1)/(cos(d*x + c) - 1) +
(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2))/d

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maple [A]  time = 0.18, size = 33, normalized size = 1.14 \[ \frac {B \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*tan(d*x+c)^2/d-1/2/d*B*ln(1+tan(d*x+c)^2)

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maxima [A]  time = 1.09, size = 30, normalized size = 1.03 \[ \frac {B \tan \left (d x + c\right )^{2} - B \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*tan(d*x + c)^2 - B*log(tan(d*x + c)^2 + 1))/d

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mupad [B]  time = 6.21, size = 28, normalized size = 0.97 \[ -\frac {B\,\left (\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )-{\mathrm {tan}\left (c+d\,x\right )}^2\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*(log(tan(c + d*x)^2 + 1) - tan(c + d*x)^2))/(2*d)

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sympy [A]  time = 0.84, size = 53, normalized size = 1.83 \[ \begin {cases} - \frac {B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\\frac {x \left (B a + B b \tan {\relax (c )}\right ) \tan ^{3}{\relax (c )}}{a + b \tan {\relax (c )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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