3.18 \(\int \frac{c+d x}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=140 \[ -\frac{5 (c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}+\frac{(c+d x)^2}{2 a^2 d}-\frac{d \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f^2}-\frac{10 d \log \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{3 a^2 f^2} \]

[Out]

(c + d*x)^2/(2*a^2*d) - (10*d*Log[Cos[e/2 + (f*x)/2]])/(3*a^2*f^2) - (d*Sec[e/2 + (f*x)/2]^2)/(6*a^2*f^2) - (5
*(c + d*x)*Tan[e/2 + (f*x)/2])/(3*a^2*f) + ((c + d*x)*Sec[e/2 + (f*x)/2]^2*Tan[e/2 + (f*x)/2])/(6*a^2*f)

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Rubi [A]  time = 0.19643, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {4191, 3318, 4185, 4184, 3475} \[ -\frac{5 (c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}+\frac{(c+d x)^2}{2 a^2 d}-\frac{d \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f^2}-\frac{10 d \log \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{3 a^2 f^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + a*Sec[e + f*x])^2,x]

[Out]

(c + d*x)^2/(2*a^2*d) - (10*d*Log[Cos[e/2 + (f*x)/2]])/(3*a^2*f^2) - (d*Sec[e/2 + (f*x)/2]^2)/(6*a^2*f^2) - (5
*(c + d*x)*Tan[e/2 + (f*x)/2])/(3*a^2*f) + ((c + d*x)*Sec[e/2 + (f*x)/2]^2*Tan[e/2 + (f*x)/2])/(6*a^2*f)

Rule 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

Rule 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

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]

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{c+d x}{(a+a \sec (e+f x))^2} \, dx &=\int \left (\frac{c+d x}{a^2}+\frac{c+d x}{a^2 (1+\cos (e+f x))^2}-\frac{2 (c+d x)}{a^2 (1+\cos (e+f x))}\right ) \, dx\\ &=\frac{(c+d x)^2}{2 a^2 d}+\frac{\int \frac{c+d x}{(1+\cos (e+f x))^2} \, dx}{a^2}-\frac{2 \int \frac{c+d x}{1+\cos (e+f x)} \, dx}{a^2}\\ &=\frac{(c+d x)^2}{2 a^2 d}+\frac{\int (c+d x) \csc ^4\left (\frac{e+\pi }{2}+\frac{f x}{2}\right ) \, dx}{4 a^2}-\frac{\int (c+d x) \csc ^2\left (\frac{e+\pi }{2}+\frac{f x}{2}\right ) \, dx}{a^2}\\ &=\frac{(c+d x)^2}{2 a^2 d}-\frac{d \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f^2}-\frac{2 (c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{a^2 f}+\frac{(c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}+\frac{\int (c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{6 a^2}+\frac{(2 d) \int \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{a^2 f}\\ &=\frac{(c+d x)^2}{2 a^2 d}-\frac{4 d \log \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{a^2 f^2}-\frac{d \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f^2}-\frac{5 (c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}-\frac{d \int \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{3 a^2 f}\\ &=\frac{(c+d x)^2}{2 a^2 d}-\frac{10 d \log \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{3 a^2 f^2}-\frac{d \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f^2}-\frac{5 (c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}\\ \end{align*}

Mathematica [A]  time = 1.60322, size = 172, normalized size = 1.23 \[ \frac{2 \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \left (\cos ^3\left (\frac{1}{2} (e+f x)\right ) \left (3 f^2 x (2 c+d x)-10 d f x \tan \left (\frac{e}{2}\right )-20 d \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )+\cos \left (\frac{1}{2} (e+f x)\right ) \left (f \tan \left (\frac{e}{2}\right ) (c+d x)-d\right )+f \sec \left (\frac{e}{2}\right ) (c+d x) \sin \left (\frac{f x}{2}\right )-10 f \sec \left (\frac{e}{2}\right ) (c+d x) \sin \left (\frac{f x}{2}\right ) \cos ^2\left (\frac{1}{2} (e+f x)\right )\right )}{3 a^2 f^2 (\sec (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)/(a + a*Sec[e + f*x])^2,x]

[Out]

(2*Cos[(e + f*x)/2]*Sec[e + f*x]^2*(f*(c + d*x)*Sec[e/2]*Sin[(f*x)/2] - 10*f*(c + d*x)*Cos[(e + f*x)/2]^2*Sec[
e/2]*Sin[(f*x)/2] + Cos[(e + f*x)/2]^3*(3*f^2*x*(2*c + d*x) - 20*d*Log[Cos[(e + f*x)/2]] - 10*d*f*x*Tan[e/2])
+ Cos[(e + f*x)/2]*(-d + f*(c + d*x)*Tan[e/2])))/(3*a^2*f^2*(1 + Sec[e + f*x])^2)

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Maple [A]  time = 0.074, size = 138, normalized size = 1. \begin{align*}{\frac{cx}{{a}^{2}}}+{\frac{d{x}^{2}}{2\,{a}^{2}}}-{\frac{3\,c}{2\,f{a}^{2}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }+{\frac{c}{6\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-{\frac{d}{6\,{a}^{2}{f}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2}}+{\frac{dx}{6\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-{\frac{3\,dx}{2\,f{a}^{2}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }+{\frac{5\,d}{3\,{a}^{2}{f}^{2}}\ln \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/(a+a*sec(f*x+e))^2,x)

[Out]

1/a^2*c*x+1/2/a^2*d*x^2-3/2/a^2*c/f*tan(1/2*f*x+1/2*e)+1/6/a^2*c/f*tan(1/2*f*x+1/2*e)^3-1/6/a^2*d/f^2*tan(1/2*
f*x+1/2*e)^2+1/6/a^2*d/f*x*tan(1/2*f*x+1/2*e)^3-3/2/a^2/f*x*d*tan(1/2*f*x+1/2*e)+5/3*d/a^2/f^2*ln(1+tan(1/2*f*
x+1/2*e)^2)

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Maxima [B]  time = 1.94266, size = 1428, normalized size = 10.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(d*e*((9*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2*f) - 12*arctan(sin(f*
x + e)/(cos(f*x + e) + 1))/(a^2*f)) - c*((9*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1
)^3)/a^2 - 12*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + (3*(f*x + e)^2*cos(3*f*x + 3*e)^2 + 3*(f*x + e)^2
*sin(3*f*x + 3*e)^2 + 3*(9*(f*x + e)^2 - 4)*cos(2*f*x + 2*e)^2 + 3*(9*(f*x + e)^2 - 4)*cos(f*x + e)^2 + 3*(9*(
f*x + e)^2 - 4)*sin(2*f*x + 2*e)^2 + 3*(9*(f*x + e)^2 - 4)*sin(f*x + e)^2 + 3*(f*x + e)^2 + 2*(3*(f*x + e)^2 +
(9*(f*x + e)^2 - 2)*cos(2*f*x + 2*e) + (9*(f*x + e)^2 - 2)*cos(f*x + e) + 12*(f*x + e)*sin(2*f*x + 2*e) + 18*
(f*x + e)*sin(f*x + e))*cos(3*f*x + 3*e) + 2*(9*(f*x + e)^2 + 3*(9*(f*x + e)^2 - 4)*cos(f*x + e) + 18*(f*x + e
)*sin(f*x + e) - 2)*cos(2*f*x + 2*e) + 2*(9*(f*x + e)^2 - 2)*cos(f*x + e) - 10*(2*(3*cos(2*f*x + 2*e) + 3*cos(
f*x + e) + 1)*cos(3*f*x + 3*e) + cos(3*f*x + 3*e)^2 + 6*(3*cos(f*x + e) + 1)*cos(2*f*x + 2*e) + 9*cos(2*f*x +
2*e)^2 + 9*cos(f*x + e)^2 + 6*(sin(2*f*x + 2*e) + sin(f*x + e))*sin(3*f*x + 3*e) + sin(3*f*x + 3*e)^2 + 9*sin(
2*f*x + 2*e)^2 + 18*sin(2*f*x + 2*e)*sin(f*x + e) + 9*sin(f*x + e)^2 + 6*cos(f*x + e) + 1)*log(cos(f*x + e)^2
+ sin(f*x + e)^2 + 2*cos(f*x + e) + 1) - 2*(10*f*x + 12*(f*x + e)*cos(2*f*x + 2*e) + 18*(f*x + e)*cos(f*x + e)
- (9*(f*x + e)^2 - 2)*sin(2*f*x + 2*e) - (9*(f*x + e)^2 - 2)*sin(f*x + e) + 10*e)*sin(3*f*x + 3*e) - 6*(6*f*x
+ 6*(f*x + e)*cos(f*x + e) - (9*(f*x + e)^2 - 4)*sin(f*x + e) + 6*e)*sin(2*f*x + 2*e) - 24*(f*x + e)*sin(f*x
+ e))*d/(a^2*f*cos(3*f*x + 3*e)^2 + 9*a^2*f*cos(2*f*x + 2*e)^2 + 9*a^2*f*cos(f*x + e)^2 + a^2*f*sin(3*f*x + 3*
e)^2 + 9*a^2*f*sin(2*f*x + 2*e)^2 + 18*a^2*f*sin(2*f*x + 2*e)*sin(f*x + e) + 9*a^2*f*sin(f*x + e)^2 + 6*a^2*f*
cos(f*x + e) + a^2*f + 2*(3*a^2*f*cos(2*f*x + 2*e) + 3*a^2*f*cos(f*x + e) + a^2*f)*cos(3*f*x + 3*e) + 6*(3*a^2
*f*cos(f*x + e) + a^2*f)*cos(2*f*x + 2*e) + 6*(a^2*f*sin(2*f*x + 2*e) + a^2*f*sin(f*x + e))*sin(3*f*x + 3*e)))
/f

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Fricas [A]  time = 1.67214, size = 443, normalized size = 3.16 \begin{align*} \frac{3 \, d f^{2} x^{2} + 6 \, c f^{2} x + 3 \,{\left (d f^{2} x^{2} + 2 \, c f^{2} x\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (3 \, d f^{2} x^{2} + 6 \, c f^{2} x - d\right )} \cos \left (f x + e\right ) - 10 \,{\left (d \cos \left (f x + e\right )^{2} + 2 \, d \cos \left (f x + e\right ) + d\right )} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) - 2 \,{\left (4 \, d f x + 4 \, c f + 5 \,{\left (d f x + c f\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 2 \, d}{6 \,{\left (a^{2} f^{2} \cos \left (f x + e\right )^{2} + 2 \, a^{2} f^{2} \cos \left (f x + e\right ) + a^{2} f^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*d*f^2*x^2 + 6*c*f^2*x + 3*(d*f^2*x^2 + 2*c*f^2*x)*cos(f*x + e)^2 + 2*(3*d*f^2*x^2 + 6*c*f^2*x - d)*cos(
f*x + e) - 10*(d*cos(f*x + e)^2 + 2*d*cos(f*x + e) + d)*log(1/2*cos(f*x + e) + 1/2) - 2*(4*d*f*x + 4*c*f + 5*(
d*f*x + c*f)*cos(f*x + e))*sin(f*x + e) - 2*d)/(a^2*f^2*cos(f*x + e)^2 + 2*a^2*f^2*cos(f*x + e) + a^2*f^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d x}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+a*sec(f*x+e))**2,x)

[Out]

(Integral(c/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(d*x/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1),
x))/a**2

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Giac [B]  time = 1.6313, size = 1215, normalized size = 8.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*d*f^2*x^2*tan(1/2*f*x)^3*tan(1/2*e)^3 + 6*c*f^2*x*tan(1/2*f*x)^3*tan(1/2*e)^3 - 9*d*f^2*x^2*tan(1/2*f*x
)^2*tan(1/2*e)^2 - 18*c*f^2*x*tan(1/2*f*x)^2*tan(1/2*e)^2 + 9*d*f*x*tan(1/2*f*x)^3*tan(1/2*e)^2 + 9*d*f*x*tan(
1/2*f*x)^2*tan(1/2*e)^3 - 10*d*log(4*(tan(1/2*e)^2 + 1)/(tan(1/2*f*x)^4*tan(1/2*e)^2 - 2*tan(1/2*f*x)^3*tan(1/
2*e) + tan(1/2*f*x)^2*tan(1/2*e)^2 + tan(1/2*f*x)^2 - 2*tan(1/2*f*x)*tan(1/2*e) + 1))*tan(1/2*f*x)^3*tan(1/2*e
)^3 + 9*d*f^2*x^2*tan(1/2*f*x)*tan(1/2*e) + 9*c*f*tan(1/2*f*x)^3*tan(1/2*e)^2 + 9*c*f*tan(1/2*f*x)^2*tan(1/2*e
)^3 - d*tan(1/2*f*x)^3*tan(1/2*e)^3 - d*f*x*tan(1/2*f*x)^3 + 18*c*f^2*x*tan(1/2*f*x)*tan(1/2*e) - 21*d*f*x*tan
(1/2*f*x)^2*tan(1/2*e) - 21*d*f*x*tan(1/2*f*x)*tan(1/2*e)^2 + 30*d*log(4*(tan(1/2*e)^2 + 1)/(tan(1/2*f*x)^4*ta
n(1/2*e)^2 - 2*tan(1/2*f*x)^3*tan(1/2*e) + tan(1/2*f*x)^2*tan(1/2*e)^2 + tan(1/2*f*x)^2 - 2*tan(1/2*f*x)*tan(1
/2*e) + 1))*tan(1/2*f*x)^2*tan(1/2*e)^2 - d*f*x*tan(1/2*e)^3 - 3*d*f^2*x^2 - c*f*tan(1/2*f*x)^3 - 21*c*f*tan(1
/2*f*x)^2*tan(1/2*e) - d*tan(1/2*f*x)^3*tan(1/2*e) - 21*c*f*tan(1/2*f*x)*tan(1/2*e)^2 + d*tan(1/2*f*x)^2*tan(1
/2*e)^2 - c*f*tan(1/2*e)^3 - d*tan(1/2*f*x)*tan(1/2*e)^3 - 6*c*f^2*x + 9*d*f*x*tan(1/2*f*x) + 9*d*f*x*tan(1/2*
e) - 30*d*log(4*(tan(1/2*e)^2 + 1)/(tan(1/2*f*x)^4*tan(1/2*e)^2 - 2*tan(1/2*f*x)^3*tan(1/2*e) + tan(1/2*f*x)^2
*tan(1/2*e)^2 + tan(1/2*f*x)^2 - 2*tan(1/2*f*x)*tan(1/2*e) + 1))*tan(1/2*f*x)*tan(1/2*e) + 9*c*f*tan(1/2*f*x)
+ d*tan(1/2*f*x)^2 + 9*c*f*tan(1/2*e) - d*tan(1/2*f*x)*tan(1/2*e) + d*tan(1/2*e)^2 + 10*d*log(4*(tan(1/2*e)^2
+ 1)/(tan(1/2*f*x)^4*tan(1/2*e)^2 - 2*tan(1/2*f*x)^3*tan(1/2*e) + tan(1/2*f*x)^2*tan(1/2*e)^2 + tan(1/2*f*x)^2
- 2*tan(1/2*f*x)*tan(1/2*e) + 1)) + d)/(a^2*f^2*tan(1/2*f*x)^3*tan(1/2*e)^3 - 3*a^2*f^2*tan(1/2*f*x)^2*tan(1/
2*e)^2 + 3*a^2*f^2*tan(1/2*f*x)*tan(1/2*e) - a^2*f^2)