3.294 \(\int (c+d x) \sec ^2(a+b x) \tan (a+b x) \, dx\)

Optimal. Leaf size=35 \[ \frac {(c+d x) \sec ^2(a+b x)}{2 b}-\frac {d \tan (a+b x)}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4409, 3767, 8} \[ \frac {(c+d x) \sec ^2(a+b x)}{2 b}-\frac {d \tan (a+b x)}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sec[a + b*x]^2*Tan[a + b*x],x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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

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Mathematica [A]  time = 0.06, size = 48, normalized size = 1.37 \[ -\frac {d \tan (a+b x)}{2 b^2}+\frac {c \sec ^2(a+b x)}{2 b}+\frac {d x \sec ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Sec[a + b*x]^2*Tan[a + b*x],x]

[Out]

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

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fricas [A]  time = 0.41, size = 36, normalized size = 1.03 \[ \frac {b d x - d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + b c}{2 \, b^{2} \cos \left (b x + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sec(b*x+a)^2*tan(b*x+a),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.84, size = 571, normalized size = 16.31 \[ \frac {b d x \tan \left (\frac {1}{2} \, b x\right )^{4} \tan \left (\frac {1}{2} \, a\right )^{4} + b c \tan \left (\frac {1}{2} \, b x\right )^{4} \tan \left (\frac {1}{2} \, a\right )^{4} + 2 \, b d x \tan \left (\frac {1}{2} \, b x\right )^{4} \tan \left (\frac {1}{2} \, a\right )^{2} + 2 \, b d x \tan \left (\frac {1}{2} \, b x\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{4} + 2 \, b c \tan \left (\frac {1}{2} \, b x\right )^{4} \tan \left (\frac {1}{2} \, a\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, b x\right )^{4} \tan \left (\frac {1}{2} \, a\right )^{3} + 2 \, b c \tan \left (\frac {1}{2} \, b x\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{4} + 2 \, d \tan \left (\frac {1}{2} \, b x\right )^{3} \tan \left (\frac {1}{2} \, a\right )^{4} + b d x \tan \left (\frac {1}{2} \, b x\right )^{4} + 4 \, b d x \tan \left (\frac {1}{2} \, b x\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} + b d x \tan \left (\frac {1}{2} \, a\right )^{4} + b c \tan \left (\frac {1}{2} \, b x\right )^{4} - 2 \, d \tan \left (\frac {1}{2} \, b x\right )^{4} \tan \left (\frac {1}{2} \, a\right ) + 4 \, b c \tan \left (\frac {1}{2} \, b x\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} - 12 \, d \tan \left (\frac {1}{2} \, b x\right )^{3} \tan \left (\frac {1}{2} \, a\right )^{2} - 12 \, d \tan \left (\frac {1}{2} \, b x\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{3} + b c \tan \left (\frac {1}{2} \, a\right )^{4} - 2 \, d \tan \left (\frac {1}{2} \, b x\right ) \tan \left (\frac {1}{2} \, a\right )^{4} + 2 \, b d x \tan \left (\frac {1}{2} \, b x\right )^{2} + 2 \, b d x \tan \left (\frac {1}{2} \, a\right )^{2} + 2 \, b c \tan \left (\frac {1}{2} \, b x\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, b x\right )^{3} + 12 \, d \tan \left (\frac {1}{2} \, b x\right )^{2} \tan \left (\frac {1}{2} \, a\right ) + 2 \, b c \tan \left (\frac {1}{2} \, a\right )^{2} + 12 \, d \tan \left (\frac {1}{2} \, b x\right ) \tan \left (\frac {1}{2} \, a\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, a\right )^{3} + b d x + b c - 2 \, d \tan \left (\frac {1}{2} \, b x\right ) - 2 \, d \tan \left (\frac {1}{2} \, a\right )}{2 \, {\left (b^{2} \tan \left (\frac {1}{2} \, b x\right )^{4} \tan \left (\frac {1}{2} \, a\right )^{4} - 2 \, b^{2} \tan \left (\frac {1}{2} \, b x\right )^{4} \tan \left (\frac {1}{2} \, a\right )^{2} - 8 \, b^{2} \tan \left (\frac {1}{2} \, b x\right )^{3} \tan \left (\frac {1}{2} \, a\right )^{3} - 2 \, b^{2} \tan \left (\frac {1}{2} \, b x\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{4} + b^{2} \tan \left (\frac {1}{2} \, b x\right )^{4} + 8 \, b^{2} \tan \left (\frac {1}{2} \, b x\right )^{3} \tan \left (\frac {1}{2} \, a\right ) + 20 \, b^{2} \tan \left (\frac {1}{2} \, b x\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{2} + 8 \, b^{2} \tan \left (\frac {1}{2} \, b x\right ) \tan \left (\frac {1}{2} \, a\right )^{3} + b^{2} \tan \left (\frac {1}{2} \, a\right )^{4} - 2 \, b^{2} \tan \left (\frac {1}{2} \, b x\right )^{2} - 8 \, b^{2} \tan \left (\frac {1}{2} \, b x\right ) \tan \left (\frac {1}{2} \, a\right ) - 2 \, b^{2} \tan \left (\frac {1}{2} \, a\right )^{2} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sec(b*x+a)^2*tan(b*x+a),x, algorithm="giac")

[Out]

1/2*(b*d*x*tan(1/2*b*x)^4*tan(1/2*a)^4 + b*c*tan(1/2*b*x)^4*tan(1/2*a)^4 + 2*b*d*x*tan(1/2*b*x)^4*tan(1/2*a)^2
+ 2*b*d*x*tan(1/2*b*x)^2*tan(1/2*a)^4 + 2*b*c*tan(1/2*b*x)^4*tan(1/2*a)^2 + 2*d*tan(1/2*b*x)^4*tan(1/2*a)^3 +
2*b*c*tan(1/2*b*x)^2*tan(1/2*a)^4 + 2*d*tan(1/2*b*x)^3*tan(1/2*a)^4 + b*d*x*tan(1/2*b*x)^4 + 4*b*d*x*tan(1/2*
b*x)^2*tan(1/2*a)^2 + b*d*x*tan(1/2*a)^4 + b*c*tan(1/2*b*x)^4 - 2*d*tan(1/2*b*x)^4*tan(1/2*a) + 4*b*c*tan(1/2*
b*x)^2*tan(1/2*a)^2 - 12*d*tan(1/2*b*x)^3*tan(1/2*a)^2 - 12*d*tan(1/2*b*x)^2*tan(1/2*a)^3 + b*c*tan(1/2*a)^4 -
2*d*tan(1/2*b*x)*tan(1/2*a)^4 + 2*b*d*x*tan(1/2*b*x)^2 + 2*b*d*x*tan(1/2*a)^2 + 2*b*c*tan(1/2*b*x)^2 + 2*d*ta
n(1/2*b*x)^3 + 12*d*tan(1/2*b*x)^2*tan(1/2*a) + 2*b*c*tan(1/2*a)^2 + 12*d*tan(1/2*b*x)*tan(1/2*a)^2 + 2*d*tan(
1/2*a)^3 + b*d*x + b*c - 2*d*tan(1/2*b*x) - 2*d*tan(1/2*a))/(b^2*tan(1/2*b*x)^4*tan(1/2*a)^4 - 2*b^2*tan(1/2*b
*x)^4*tan(1/2*a)^2 - 8*b^2*tan(1/2*b*x)^3*tan(1/2*a)^3 - 2*b^2*tan(1/2*b*x)^2*tan(1/2*a)^4 + b^2*tan(1/2*b*x)^
4 + 8*b^2*tan(1/2*b*x)^3*tan(1/2*a) + 20*b^2*tan(1/2*b*x)^2*tan(1/2*a)^2 + 8*b^2*tan(1/2*b*x)*tan(1/2*a)^3 + b
^2*tan(1/2*a)^4 - 2*b^2*tan(1/2*b*x)^2 - 8*b^2*tan(1/2*b*x)*tan(1/2*a) - 2*b^2*tan(1/2*a)^2 + b^2)

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maple [A]  time = 0.03, size = 61, normalized size = 1.74 \[ \frac {\frac {d \left (\frac {b x +a}{2 \cos \left (b x +a \right )^{2}}-\frac {\tan \left (b x +a \right )}{2}\right )}{b}-\frac {d a}{2 b \cos \left (b x +a \right )^{2}}+\frac {c}{2 \cos \left (b x +a \right )^{2}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*sec(b*x+a)^2*tan(b*x+a),x)

[Out]

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

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maxima [B]  time = 0.35, size = 283, normalized size = 8.09 \[ \frac {c \tan \left (b x + a\right )^{2} - \frac {a d \tan \left (b x + a\right )^{2}}{b} + \frac {2 \, {\left (4 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + 4 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + {\left (2 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right )\right )} \cos \left (4 \, b x + 4 \, a\right ) + 2 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right ) - 1\right )} \sin \left (4 \, b x + 4 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} d}{{\left (2 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \cos \left (4 \, b x + 4 \, a\right ) + \cos \left (4 \, b x + 4 \, a\right )^{2} + 4 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 4 \, \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, \sin \left (2 \, b x + 2 \, a\right )^{2} + 4 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} b}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sec(b*x+a)^2*tan(b*x+a),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.19, size = 53, normalized size = 1.51 \[ -\frac {d\,1{}\mathrm {i}+{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,\left (-b\,\left (2\,c+2\,d\,x\right )+d\,1{}\mathrm {i}\right )}{b^2\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(a + b*x)*(c + d*x))/cos(a + b*x)^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) \tan {\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sec(b*x+a)**2*tan(b*x+a),x)

[Out]

Integral((c + d*x)*tan(a + b*x)*sec(a + b*x)**2, x)

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