3.250 \(\int (c+d x) \sec (a+b x) \tan (a+b x) \, dx\)

Optimal. Leaf size=29 \[ \frac {(c+d x) \sec (a+b x)}{b}-\frac {d \tanh ^{-1}(\sin (a+b x))}{b^2} \]

[Out]

-d*arctanh(sin(b*x+a))/b^2+(d*x+c)*sec(b*x+a)/b

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Rubi [A]  time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4409, 3770} \[ \frac {(c+d x) \sec (a+b x)}{b}-\frac {d \tanh ^{-1}(\sin (a+b x))}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*ArcTanh[Sin[a + b*x]])/b^2) + ((c + d*x)*Sec[a + b*x])/b

Rule 3770

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

Rule 4409

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
((c + d*x)^m*Sec[a + b*x]^n)/(b*n), x] - Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rubi steps

\begin {align*} \int (c+d x) \sec (a+b x) \tan (a+b x) \, dx &=\frac {(c+d x) \sec (a+b x)}{b}-\frac {d \int \sec (a+b x) \, dx}{b}\\ &=-\frac {d \tanh ^{-1}(\sin (a+b x))}{b^2}+\frac {(c+d x) \sec (a+b x)}{b}\\ \end {align*}

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Mathematica [B]  time = 0.04, size = 93, normalized size = 3.21 \[ \frac {d \log \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right )-\sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{b^2}-\frac {d \log \left (\sin \left (\frac {a}{2}+\frac {b x}{2}\right )+\cos \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{b^2}+\frac {c \sec (a+b x)}{b}+\frac {d x \sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Log[Cos[a/2 + (b*x)/2] - Sin[a/2 + (b*x)/2]])/b^2 - (d*Log[Cos[a/2 + (b*x)/2] + Sin[a/2 + (b*x)/2]])/b^2 +
(c*Sec[a + b*x])/b + (d*x*Sec[a + b*x])/b

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fricas [B]  time = 0.47, size = 60, normalized size = 2.07 \[ \frac {2 \, b d x - d \cos \left (b x + a\right ) \log \left (\sin \left (b x + a\right ) + 1\right ) + d \cos \left (b x + a\right ) \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, b c}{2 \, b^{2} \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*d*x - d*cos(b*x + a)*log(sin(b*x + a) + 1) + d*cos(b*x + a)*log(-sin(b*x + a) + 1) + 2*b*c)/(b^2*cos(
b*x + a))

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giac [B]  time = 1.31, size = 1537, normalized size = 53.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 49, normalized size = 1.69 \[ \frac {d x}{b \cos \left (b x +a \right )}-\frac {d \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b^{2}}+\frac {c}{b \cos \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 259, normalized size = 8.93 \[ \frac {\frac {{\left (4 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) \cos \left (b x + a\right ) + 4 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) + 4 \, {\left (b x + a\right )} \cos \left (b x + a\right ) - {\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \sin \left (b x + a\right ) + 1\right ) + {\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \sin \left (b x + a\right ) + 1\right )\right )} d}{{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} b} + \frac {2 \, c}{\cos \left (b x + a\right )} - \frac {2 \, a d}{b \cos \left (b x + a\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((4*(b*x + a)*cos(2*b*x + 2*a)*cos(b*x + a) + 4*(b*x + a)*sin(2*b*x + 2*a)*sin(b*x + a) + 4*(b*x + a)*cos(
b*x + a) - (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a
)^2 + 2*sin(b*x + a) + 1) + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a
)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1))*d/((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) +
1)*b) + 2*c/cos(b*x + a) - 2*a*d/(b*cos(b*x + a)))/b

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mupad [B]  time = 2.58, size = 78, normalized size = 2.69 \[ \frac {d\,\ln \left ({\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}-\mathrm {i}\right )}{b^2}-\frac {d\,\ln \left ({\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}{b^2}+\frac {2\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (c+d\,x\right )}{b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(a + b*x)*(c + d*x))/cos(a + b*x),x)

[Out]

(d*log(exp(a*1i + b*x*1i) - 1i))/b^2 - (d*log(exp(a*1i + b*x*1i) + 1i))/b^2 + (2*exp(a*1i + b*x*1i)*(c + d*x))
/(b*(exp(a*2i + b*x*2i) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) \tan {\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)*tan(a + b*x)*sec(a + b*x), x)

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