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

Optimal. Leaf size=28 \[ \frac {d \log (\cos (a+b x))}{b^2}+\frac {(c+d x) \tan (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4184, 3475} \[ \frac {d \log (\cos (a+b x))}{b^2}+\frac {(c+d x) \tan (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sec[a + b*x]^2,x]

[Out]

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

Rule 3475

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

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 36, normalized size = 1.29 \[ \frac {d \log (\cos (a+b x))}{b^2}+\frac {c \tan (a+b x)}{b}+\frac {d x \tan (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Sec[a + b*x]^2,x]

[Out]

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

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fricas [A]  time = 0.68, size = 45, normalized size = 1.61 \[ \frac {d \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right )\right ) + {\left (b d x + b c\right )} \sin \left (b x + a\right )}{b^{2} \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 5.83, size = 1459, normalized size = 52.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*b*d*x*tan(1/2*b*x)^2*tan(1/2*a) + 4*b*d*x*tan(1/2*b*x)*tan(1/2*a)^2 - d*log(4*(tan(1/2*b*x)^8*tan(1/2*
a)^4 - 2*tan(1/2*b*x)^8*tan(1/2*a)^2 - 8*tan(1/2*b*x)^7*tan(1/2*a)^3 + tan(1/2*b*x)^8 + 8*tan(1/2*b*x)^7*tan(1
/2*a) + 16*tan(1/2*b*x)^6*tan(1/2*a)^2 - 8*tan(1/2*b*x)^5*tan(1/2*a)^3 - 2*tan(1/2*b*x)^4*tan(1/2*a)^4 + 8*tan
(1/2*b*x)^5*tan(1/2*a) + 36*tan(1/2*b*x)^4*tan(1/2*a)^2 + 8*tan(1/2*b*x)^3*tan(1/2*a)^3 - 2*tan(1/2*b*x)^4 - 8
*tan(1/2*b*x)^3*tan(1/2*a) + 16*tan(1/2*b*x)^2*tan(1/2*a)^2 + 8*tan(1/2*b*x)*tan(1/2*a)^3 + tan(1/2*a)^4 - 8*t
an(1/2*b*x)*tan(1/2*a) - 2*tan(1/2*a)^2 + 1)/(tan(1/2*a)^4 + 2*tan(1/2*a)^2 + 1))*tan(1/2*b*x)^2*tan(1/2*a)^2
+ 4*b*c*tan(1/2*b*x)^2*tan(1/2*a) + 4*b*c*tan(1/2*b*x)*tan(1/2*a)^2 - 4*b*d*x*tan(1/2*b*x) + d*log(4*(tan(1/2*
b*x)^8*tan(1/2*a)^4 - 2*tan(1/2*b*x)^8*tan(1/2*a)^2 - 8*tan(1/2*b*x)^7*tan(1/2*a)^3 + tan(1/2*b*x)^8 + 8*tan(1
/2*b*x)^7*tan(1/2*a) + 16*tan(1/2*b*x)^6*tan(1/2*a)^2 - 8*tan(1/2*b*x)^5*tan(1/2*a)^3 - 2*tan(1/2*b*x)^4*tan(1
/2*a)^4 + 8*tan(1/2*b*x)^5*tan(1/2*a) + 36*tan(1/2*b*x)^4*tan(1/2*a)^2 + 8*tan(1/2*b*x)^3*tan(1/2*a)^3 - 2*tan
(1/2*b*x)^4 - 8*tan(1/2*b*x)^3*tan(1/2*a) + 16*tan(1/2*b*x)^2*tan(1/2*a)^2 + 8*tan(1/2*b*x)*tan(1/2*a)^3 + tan
(1/2*a)^4 - 8*tan(1/2*b*x)*tan(1/2*a) - 2*tan(1/2*a)^2 + 1)/(tan(1/2*a)^4 + 2*tan(1/2*a)^2 + 1))*tan(1/2*b*x)^
2 - 4*b*d*x*tan(1/2*a) + 4*d*log(4*(tan(1/2*b*x)^8*tan(1/2*a)^4 - 2*tan(1/2*b*x)^8*tan(1/2*a)^2 - 8*tan(1/2*b*
x)^7*tan(1/2*a)^3 + tan(1/2*b*x)^8 + 8*tan(1/2*b*x)^7*tan(1/2*a) + 16*tan(1/2*b*x)^6*tan(1/2*a)^2 - 8*tan(1/2*
b*x)^5*tan(1/2*a)^3 - 2*tan(1/2*b*x)^4*tan(1/2*a)^4 + 8*tan(1/2*b*x)^5*tan(1/2*a) + 36*tan(1/2*b*x)^4*tan(1/2*
a)^2 + 8*tan(1/2*b*x)^3*tan(1/2*a)^3 - 2*tan(1/2*b*x)^4 - 8*tan(1/2*b*x)^3*tan(1/2*a) + 16*tan(1/2*b*x)^2*tan(
1/2*a)^2 + 8*tan(1/2*b*x)*tan(1/2*a)^3 + tan(1/2*a)^4 - 8*tan(1/2*b*x)*tan(1/2*a) - 2*tan(1/2*a)^2 + 1)/(tan(1
/2*a)^4 + 2*tan(1/2*a)^2 + 1))*tan(1/2*b*x)*tan(1/2*a) + d*log(4*(tan(1/2*b*x)^8*tan(1/2*a)^4 - 2*tan(1/2*b*x)
^8*tan(1/2*a)^2 - 8*tan(1/2*b*x)^7*tan(1/2*a)^3 + tan(1/2*b*x)^8 + 8*tan(1/2*b*x)^7*tan(1/2*a) + 16*tan(1/2*b*
x)^6*tan(1/2*a)^2 - 8*tan(1/2*b*x)^5*tan(1/2*a)^3 - 2*tan(1/2*b*x)^4*tan(1/2*a)^4 + 8*tan(1/2*b*x)^5*tan(1/2*a
) + 36*tan(1/2*b*x)^4*tan(1/2*a)^2 + 8*tan(1/2*b*x)^3*tan(1/2*a)^3 - 2*tan(1/2*b*x)^4 - 8*tan(1/2*b*x)^3*tan(1
/2*a) + 16*tan(1/2*b*x)^2*tan(1/2*a)^2 + 8*tan(1/2*b*x)*tan(1/2*a)^3 + tan(1/2*a)^4 - 8*tan(1/2*b*x)*tan(1/2*a
) - 2*tan(1/2*a)^2 + 1)/(tan(1/2*a)^4 + 2*tan(1/2*a)^2 + 1))*tan(1/2*a)^2 - 4*b*c*tan(1/2*b*x) - 4*b*c*tan(1/2
*a) - d*log(4*(tan(1/2*b*x)^8*tan(1/2*a)^4 - 2*tan(1/2*b*x)^8*tan(1/2*a)^2 - 8*tan(1/2*b*x)^7*tan(1/2*a)^3 + t
an(1/2*b*x)^8 + 8*tan(1/2*b*x)^7*tan(1/2*a) + 16*tan(1/2*b*x)^6*tan(1/2*a)^2 - 8*tan(1/2*b*x)^5*tan(1/2*a)^3 -
2*tan(1/2*b*x)^4*tan(1/2*a)^4 + 8*tan(1/2*b*x)^5*tan(1/2*a) + 36*tan(1/2*b*x)^4*tan(1/2*a)^2 + 8*tan(1/2*b*x)
^3*tan(1/2*a)^3 - 2*tan(1/2*b*x)^4 - 8*tan(1/2*b*x)^3*tan(1/2*a) + 16*tan(1/2*b*x)^2*tan(1/2*a)^2 + 8*tan(1/2*
b*x)*tan(1/2*a)^3 + tan(1/2*a)^4 - 8*tan(1/2*b*x)*tan(1/2*a) - 2*tan(1/2*a)^2 + 1)/(tan(1/2*a)^4 + 2*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 = 37, normalized size = 1.32 \[ \frac {d \tan \left (b x +a \right ) x}{b}+\frac {d \ln \left (\cos \left (b x +a \right )\right )}{b^{2}}+\frac {c \tan \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.01, size = 159, normalized size = 5.68 \[ \frac {2 \, c \tan \left (b x + a\right ) - \frac {2 \, a d \tan \left (b x + a\right )}{b} + \frac {{\left ({\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 4 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d}{{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} b}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.83, size = 55, normalized size = 1.96 \[ \frac {d\,\ln \left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}+1\right )}{b^2}+\frac {\left (c+d\,x\right )\,2{}\mathrm {i}}{b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}-\frac {d\,x\,2{}\mathrm {i}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/cos(a + b*x)^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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