\(\int \frac {c+d x}{(a+a \sec (e+f x))^2} \, dx\) [18]

Optimal result

Integrand size = 18, antiderivative size = 140 \[ \int \frac {c+d x}{(a+a \sec (e+f x))^2} \, dx=\frac {(c+d x)^2}{2 a^2 d}-\frac {10 d \log \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3 a^2 f^2}-\frac {d \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 f^2}-\frac {5 (c+d x) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 f} \]

[Out]

1/2*(d*x+c)^2/a^2/d-10/3*d*ln(cos(1/2*f*x+1/2*e))/a^2/f^2-1/6*d*sec(1/2*f*x+1/2*e)^2/a^2/f^2-5/3*(d*x+c)*tan(1
/2*f*x+1/2*e)/a^2/f+1/6*(d*x+c)*sec(1/2*f*x+1/2*e)^2*tan(1/2*f*x+1/2*e)/a^2/f

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {4276, 3399, 4270, 4269, 3556} \[ \int \frac {c+d x}{(a+a \sec (e+f x))^2} \, dx=-\frac {5 (c+d x) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x) \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 f}+\frac {(c+d x)^2}{2 a^2 d}-\frac {d \sec ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{6 a^2 f^2}-\frac {10 d \log \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3 a^2 f^2} \]

[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 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + 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 3556

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

Rule 4269

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 4270

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] &&
GtQ[n, 1] && NeQ[n, 2]

Rule 4276

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 1.98 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.23 \[ \int \frac {c+d x}{(a+a \sec (e+f x))^2} \, dx=\frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x) \left (f (c+d x) \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )-10 f (c+d x) \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )+\cos ^3\left (\frac {1}{2} (e+f x)\right ) \left (3 f^2 x (2 c+d x)-20 d \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )-10 d f x \tan \left (\frac {e}{2}\right )\right )+\cos \left (\frac {1}{2} (e+f x)\right ) \left (-d+f (c+d x) \tan \left (\frac {e}{2}\right )\right )\right )}{3 a^2 f^2 (1+\sec (e+f x))^2} \]

[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)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.61

method	result	size
parallelrisch	\(\frac {10 d \ln \left (\sec \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )+\left (d x +c \right ) f \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) f \left (d x +c \right )+6 f^{2} \left (\frac {d x}{2}+c \right ) x}{6 a^{2} f^{2}}\)	\(86\)
norman	\(\frac {\frac {c x}{a}+\frac {d \,x^{2}}{2 a}-\frac {3 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a f}+\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{6 a f}-\frac {d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{6 a \,f^{2}}-\frac {3 d x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a f}+\frac {d x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{6 a f}}{a}+\frac {5 d \ln \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}{3 a^{2} f^{2}}\)	\(143\)
risch	\(\frac {d \,x^{2}}{2 a^{2}}+\frac {x c}{a^{2}}+\frac {10 i d x}{3 a^{2} f}+\frac {10 i d e}{3 a^{2} f^{2}}-\frac {2 i \left (6 d f x \,{\mathrm e}^{2 i \left (f x +e \right )}-i d \,{\mathrm e}^{2 i \left (f x +e \right )}+6 c f \,{\mathrm e}^{2 i \left (f x +e \right )}+9 d f x \,{\mathrm e}^{i \left (f x +e \right )}-i d \,{\mathrm e}^{i \left (f x +e \right )}+9 c f \,{\mathrm e}^{i \left (f x +e \right )}+5 d f x +5 c f \right )}{3 f^{2} a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3}}-\frac {10 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{3 a^{2} f^{2}}\)	\(172\)

[In]

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

[Out]

1/6*(10*d*ln(sec(1/2*f*x+1/2*e)^2)+(d*x+c)*f*tan(1/2*f*x+1/2*e)^3-d*tan(1/2*f*x+1/2*e)^2-9*tan(1/2*f*x+1/2*e)*
f*(d*x+c)+6*f^2*(1/2*d*x+c)*x)/a^2/f^2

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.31 \[ \int \frac {c+d x}{(a+a \sec (e+f x))^2} \, dx=\frac {3 \, d f^{2} x^{2} + 6 \, c f^{2} x + 3 \, {\left (d f^{2} x^{2} + 2 \, c f^{2} x\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (3 \, d f^{2} x^{2} + 6 \, c f^{2} x - d\right )} \cos \left (f x + e\right ) - 10 \, {\left (d \cos \left (f x + e\right )^{2} + 2 \, d \cos \left (f x + e\right ) + d\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - 2 \, {\left (4 \, d f x + 4 \, c f + 5 \, {\left (d f x + c f\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 2 \, d}{6 \, {\left (a^{2} f^{2} \cos \left (f x + e\right )^{2} + 2 \, a^{2} f^{2} \cos \left (f x + e\right ) + a^{2} f^{2}\right )}} \]

[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)

Sympy [F]

\[ \int \frac {c+d x}{(a+a \sec (e+f x))^2} \, dx=\frac {\int \frac {c}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d x}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1058 vs. \(2 (110) = 220\).

Time = 0.36 (sec) , antiderivative size = 1058, normalized size of antiderivative = 7.56 \[ \int \frac {c+d x}{(a+a \sec (e+f x))^2} \, dx=\text {Too large to display} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 798 vs. \(2 (110) = 220\).

Time = 0.56 (sec) , antiderivative size = 798, normalized size of antiderivative = 5.70 \[ \int \frac {c+d x}{(a+a \sec (e+f x))^2} \, dx=\text {Too large to display} \]

[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*f*x)^2*tan(1/2*e)^2 - 2*tan(1/2*f*x)*tan(1/2*e) + 1)/(tan(1/2*f*
x)^2*tan(1/2*e)^2 + tan(1/2*f*x)^2 + tan(1/2*e)^2 + 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*f*x)^2*tan(1/2*e)^2 - 2*tan(1/2*f*x)*tan(1/2*e) + 1)/(tan(
1/2*f*x)^2*tan(1/2*e)^2 + tan(1/2*f*x)^2 + tan(1/2*e)^2 + 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*f*x)^2*tan(1/2*e)^2 - 2*tan(1/2*f*x
)*tan(1/2*e) + 1)/(tan(1/2*f*x)^2*tan(1/2*e)^2 + tan(1/2*f*x)^2 + tan(1/2*e)^2 + 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*f*x)^2*tan(1/2*e)^2 - 2*tan(1/2*f*x)*tan(1/2*e) + 1)/(tan(1/2*f*x)^2*tan(1/2*e)^2 + tan(1/2*f*x
)^2 + tan(1/2*e)^2 + 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)

Mupad [B] (verification not implemented)

Time = 19.32 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.76 \[ \int \frac {c+d x}{(a+a \sec (e+f x))^2} \, dx=\frac {d\,x^2}{2\,a^2}-\frac {\frac {\left (c+d\,x\right )\,4{}\mathrm {i}}{3\,a^2\,f}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left (c+d\,x\right )\,4{}\mathrm {i}}{3\,a^2\,f}+\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (c+d\,x\right )\,4{}\mathrm {i}}{3\,a^2\,f}}{3\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+3\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}+1}-\frac {10\,d\,\ln \left ({\mathrm {e}}^{e\,1{}\mathrm {i}}\,{\mathrm {e}}^{f\,x\,1{}\mathrm {i}}+1\right )}{3\,a^2\,f^2}-\frac {\left (4\,c\,f+4\,d\,f\,x-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{3\,a^2\,f^2\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1\right )}+\frac {\left (c\,f+d\,f\,x-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{3\,a^2\,f^2\,\left (2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}+\frac {x\,\left (3\,c\,f+d\,10{}\mathrm {i}\right )}{3\,a^2\,f} \]

[In]

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

[Out]

(d*x^2)/(2*a^2) - (((c + d*x)*4i)/(3*a^2*f) + (exp(e*1i + f*x*1i)*(c + d*x)*4i)/(3*a^2*f) + (exp(e*2i + f*x*2i
)*(c + d*x)*4i)/(3*a^2*f))/(3*exp(e*1i + f*x*1i) + 3*exp(e*2i + f*x*2i) + exp(e*3i + f*x*3i) + 1) - (10*d*log(
exp(e*1i)*exp(f*x*1i) + 1))/(3*a^2*f^2) - ((4*c*f - d*1i + 4*d*f*x)*2i)/(3*a^2*f^2*(exp(e*1i + f*x*1i) + 1)) +
((c*f - d*1i + d*f*x)*2i)/(3*a^2*f^2*(2*exp(e*1i + f*x*1i) + exp(e*2i + f*x*2i) + 1)) + (x*(d*10i + 3*c*f))/(
3*a^2*f)