3.229 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=115 \[ \frac{\left (2 c^2+6 c d+7 d^2\right ) \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac{(c-d)^2 \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}+\frac{2 (c+4 d) (c-d) \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2} \]

[Out]

((c - d)^2*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + (2*(c - d)*(c + 4*d)*Tan[e + f*x])/(15*a*f*(a + a*Sec[
e + f*x])^2) + ((2*c^2 + 6*c*d + 7*d^2)*Tan[e + f*x])/(15*f*(a^3 + a^3*Sec[e + f*x]))

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Rubi [A]  time = 0.162603, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3987, 89, 78, 37} \[ \frac{\left (2 c^2+6 c d+7 d^2\right ) \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac{(c-d)^2 \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}+\frac{2 (c+4 d) (c-d) \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x])^2)/(a + a*Sec[e + f*x])^3,x]

[Out]

((c - d)^2*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + (2*(c - d)*(c + 4*d)*Tan[e + f*x])/(15*a*f*(a + a*Sec[
e + f*x])^2) + ((2*c^2 + 6*c*d + 7*d^2)*Tan[e + f*x])/(15*f*(a^3 + a^3*Sec[e + f*x]))

Rule 3987

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[(a^2*g*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x
]]), Subst[Int[((g*x)^(p - 1)*(a + b*x)^(m - 1/2)*(c + d*x)^n)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^2}{\sqrt{a-a x} (a+a x)^{7/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d)^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{a^3 \left (2 c^2+6 c d-3 d^2\right )+5 a^3 d^2 x}{\sqrt{a-a x} (a+a x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d)^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{2 (c-d) (c+4 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}-\frac{\left (\left (2 c^2+6 c d+7 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} (a+a x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d)^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{2 (c-d) (c+4 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{\left (2 c^2+6 c d+7 d^2\right ) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.470205, size = 180, normalized size = 1.57 \[ \frac{\sec \left (\frac{e}{2}\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \left (10 \left (4 c^2+3 c d+2 d^2\right ) \sin \left (\frac{f x}{2}\right )+20 c^2 \sin \left (e+\frac{3 f x}{2}\right )-15 c^2 \sin \left (2 e+\frac{3 f x}{2}\right )+7 c^2 \sin \left (2 e+\frac{5 f x}{2}\right )-30 c (c+d) \sin \left (e+\frac{f x}{2}\right )+30 c d \sin \left (e+\frac{3 f x}{2}\right )+6 c d \sin \left (2 e+\frac{5 f x}{2}\right )+10 d^2 \sin \left (e+\frac{3 f x}{2}\right )+2 d^2 \sin \left (2 e+\frac{5 f x}{2}\right )\right )}{30 a^3 f (\cos (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x])^2)/(a + a*Sec[e + f*x])^3,x]

[Out]

(Cos[(e + f*x)/2]*Sec[e/2]*(10*(4*c^2 + 3*c*d + 2*d^2)*Sin[(f*x)/2] - 30*c*(c + d)*Sin[e + (f*x)/2] + 20*c^2*S
in[e + (3*f*x)/2] + 30*c*d*Sin[e + (3*f*x)/2] + 10*d^2*Sin[e + (3*f*x)/2] - 15*c^2*Sin[2*e + (3*f*x)/2] + 7*c^
2*Sin[2*e + (5*f*x)/2] + 6*c*d*Sin[2*e + (5*f*x)/2] + 2*d^2*Sin[2*e + (5*f*x)/2]))/(30*a^3*f*(1 + Cos[e + f*x]
)^3)

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Maple [A]  time = 0.064, size = 128, normalized size = 1.1 \begin{align*}{\frac{1}{4\,f{a}^{3}} \left ({\frac{{c}^{2}}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}-{\frac{2\,cd}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+{\frac{{d}^{2}}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}-{\frac{2\,{c}^{2}}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+{\frac{2\,{d}^{2}}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ){c}^{2}+2\,cd\tan \left ( 1/2\,fx+e/2 \right ) +\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ){d}^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^3,x)

[Out]

1/4/f/a^3*(1/5*tan(1/2*f*x+1/2*e)^5*c^2-2/5*tan(1/2*f*x+1/2*e)^5*c*d+1/5*tan(1/2*f*x+1/2*e)^5*d^2-2/3*tan(1/2*
f*x+1/2*e)^3*c^2+2/3*tan(1/2*f*x+1/2*e)^3*d^2+tan(1/2*f*x+1/2*e)*c^2+2*c*d*tan(1/2*f*x+1/2*e)+tan(1/2*f*x+1/2*
e)*d^2)

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Maxima [A]  time = 1.0211, size = 248, normalized size = 2.16 \begin{align*} \frac{\frac{d^{2}{\left (\frac{15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac{c^{2}{\left (\frac{15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac{6 \, c d{\left (\frac{5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/60*(d^2*(15*sin(f*x + e)/(cos(f*x + e) + 1) + 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos
(f*x + e) + 1)^5)/a^3 + c^2*(15*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*s
in(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 + 6*c*d*(5*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^5/(cos(f*x +
e) + 1)^5)/a^3)/f

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Fricas [A]  time = 0.45978, size = 270, normalized size = 2.35 \begin{align*} \frac{{\left ({\left (7 \, c^{2} + 6 \, c d + 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, c^{2} + 6 \, c d + 7 \, d^{2} + 6 \,{\left (c^{2} + 3 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*((7*c^2 + 6*c*d + 2*d^2)*cos(f*x + e)^2 + 2*c^2 + 6*c*d + 7*d^2 + 6*(c^2 + 3*c*d + d^2)*cos(f*x + e))*sin
(f*x + e)/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a^3*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c^{2} \sec{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{2 c d \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))**2/(a+a*sec(f*x+e))**3,x)

[Out]

(Integral(c**2*sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(d**2*sec
(e + f*x)**3/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(2*c*d*sec(e + f*x)**2/(
sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x))/a**3

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Giac [A]  time = 1.24547, size = 185, normalized size = 1.61 \begin{align*} \frac{3 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 6 \, c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 3 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 10 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 10 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 30 \, c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 15 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{60 \, a^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/60*(3*c^2*tan(1/2*f*x + 1/2*e)^5 - 6*c*d*tan(1/2*f*x + 1/2*e)^5 + 3*d^2*tan(1/2*f*x + 1/2*e)^5 - 10*c^2*tan(
1/2*f*x + 1/2*e)^3 + 10*d^2*tan(1/2*f*x + 1/2*e)^3 + 15*c^2*tan(1/2*f*x + 1/2*e) + 30*c*d*tan(1/2*f*x + 1/2*e)
+ 15*d^2*tan(1/2*f*x + 1/2*e))/(a^3*f)