3.1.8 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^4} \, dx\) [8]

Optimal. Leaf size=116 \[ -\frac {(a+a \sec (e+f x)) \tan (e+f x)}{7 f (c-c \sec (e+f x))^4}-\frac {2 (a+a \sec (e+f x)) \tan (e+f x)}{35 c f (c-c \sec (e+f x))^3}-\frac {2 (a+a \sec (e+f x)) \tan (e+f x)}{105 f \left (c^2-c^2 \sec (e+f x)\right )^2} \]

[Out]

-1/7*(a+a*sec(f*x+e))*tan(f*x+e)/f/(c-c*sec(f*x+e))^4-2/35*(a+a*sec(f*x+e))*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^3-
2/105*(a+a*sec(f*x+e))*tan(f*x+e)/f/(c^2-c^2*sec(f*x+e))^2

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Rubi [A]
time = 0.11, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {4036, 4035} \begin {gather*} -\frac {2 \tan (e+f x) (a \sec (e+f x)+a)}{105 f \left (c^2-c^2 \sec (e+f x)\right )^2}-\frac {2 \tan (e+f x) (a \sec (e+f x)+a)}{35 c f (c-c \sec (e+f x))^3}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{7 f (c-c \sec (e+f x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c - c*Sec[e + f*x])^4,x]

[Out]

-1/7*((a + a*Sec[e + f*x])*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^4) - (2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(3
5*c*f*(c - c*Sec[e + f*x])^3) - (2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(105*f*(c^2 - c^2*Sec[e + f*x])^2)

Rule 4035

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

Rule 4036

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[b*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] +
Dist[(m + n + 1)/(a*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[m + n + 1, 0] && NeQ[2
*m + 1, 0] &&  !LtQ[n, 0] &&  !(IGtQ[n + 1/2, 0] && LtQ[n + 1/2, -(m + n)])

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^4} \, dx &=-\frac {(a+a \sec (e+f x)) \tan (e+f x)}{7 f (c-c \sec (e+f x))^4}+\frac {2 \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^3} \, dx}{7 c}\\ &=-\frac {(a+a \sec (e+f x)) \tan (e+f x)}{7 f (c-c \sec (e+f x))^4}-\frac {2 (a+a \sec (e+f x)) \tan (e+f x)}{35 c f (c-c \sec (e+f x))^3}+\frac {2 \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^2} \, dx}{35 c^2}\\ &=-\frac {(a+a \sec (e+f x)) \tan (e+f x)}{7 f (c-c \sec (e+f x))^4}-\frac {2 (a+a \sec (e+f x)) \tan (e+f x)}{35 c f (c-c \sec (e+f x))^3}-\frac {2 (a+a \sec (e+f x)) \tan (e+f x)}{105 f \left (c^2-c^2 \sec (e+f x)\right )^2}\\ \end {align*}

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Mathematica [A]
time = 0.41, size = 113, normalized size = 0.97 \begin {gather*} -\frac {a \csc \left (\frac {e}{2}\right ) \csc ^7\left (\frac {1}{2} (e+f x)\right ) \left (350 \sin \left (\frac {f x}{2}\right )+455 \sin \left (e+\frac {f x}{2}\right )-273 \sin \left (e+\frac {3 f x}{2}\right )-210 \sin \left (2 e+\frac {3 f x}{2}\right )+56 \sin \left (2 e+\frac {5 f x}{2}\right )+105 \sin \left (3 e+\frac {5 f x}{2}\right )-23 \sin \left (3 e+\frac {7 f x}{2}\right )\right )}{6720 c^4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c - c*Sec[e + f*x])^4,x]

[Out]

-1/6720*(a*Csc[e/2]*Csc[(e + f*x)/2]^7*(350*Sin[(f*x)/2] + 455*Sin[e + (f*x)/2] - 273*Sin[e + (3*f*x)/2] - 210
*Sin[2*e + (3*f*x)/2] + 56*Sin[2*e + (5*f*x)/2] + 105*Sin[3*e + (5*f*x)/2] - 23*Sin[3*e + (7*f*x)/2]))/(c^4*f)

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Maple [A]
time = 0.17, size = 50, normalized size = 0.43

method	result	size
derivativedivides	\(\frac {a \left (-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {2}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}\right )}{4 f \,c^{4}}\)	\(50\)
default	\(\frac {a \left (-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {2}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}\right )}{4 f \,c^{4}}\)	\(50\)
risch	\(\frac {2 i a \left (105 \,{\mathrm e}^{6 i \left (f x +e \right )}-210 \,{\mathrm e}^{5 i \left (f x +e \right )}+455 \,{\mathrm e}^{4 i \left (f x +e \right )}-350 \,{\mathrm e}^{3 i \left (f x +e \right )}+273 \,{\mathrm e}^{2 i \left (f x +e \right )}-56 \,{\mathrm e}^{i \left (f x +e \right )}+23\right )}{105 f \,c^{4} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{7}}\)	\(92\)
norman	\(\frac {\frac {a}{28 c f}-\frac {19 a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{140 c f}+\frac {11 a \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{60 c f}-\frac {a \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{12 c f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

1/4/f*a/c^4*(-1/7/tan(1/2*f*x+1/2*e)^7-1/3/tan(1/2*f*x+1/2*e)^3+2/5/tan(1/2*f*x+1/2*e)^5)

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Maxima [A]
time = 0.28, size = 193, normalized size = 1.66 \begin {gather*} \frac {\frac {a {\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {105 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 15\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}} + \frac {3 \, a {\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {35 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 5\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}}}{840 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^4,x, algorithm="maxima")

[Out]

1/840*(a*(21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 105*sin(f*x + e)^6
/(cos(f*x + e) + 1)^6 - 15)*(cos(f*x + e) + 1)^7/(c^4*sin(f*x + e)^7) + 3*a*(21*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 - 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 35*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 5)*(cos(f*x + e) + 1
)^7/(c^4*sin(f*x + e)^7))/f

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Fricas [A]
time = 4.04, size = 112, normalized size = 0.97 \begin {gather*} \frac {23 \, a \cos \left (f x + e\right )^{4} + 36 \, a \cos \left (f x + e\right )^{3} + 5 \, a \cos \left (f x + e\right )^{2} - 6 \, a \cos \left (f x + e\right ) + 2 \, a}{105 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

1/105*(23*a*cos(f*x + e)^4 + 36*a*cos(f*x + e)^3 + 5*a*cos(f*x + e)^2 - 6*a*cos(f*x + e) + 2*a)/((c^4*f*cos(f*
x + e)^3 - 3*c^4*f*cos(f*x + e)^2 + 3*c^4*f*cos(f*x + e) - c^4*f)*sin(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {a \left (\int \frac {\sec {\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec {\left (e + f x \right )} + 1}\, dx\right )}{c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))**4,x)

[Out]

a*(Integral(sec(e + f*x)/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1), x) +
Integral(sec(e + f*x)**2/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1), x))/c
**4

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Giac [A]
time = 0.71, size = 51, normalized size = 0.44 \begin {gather*} -\frac {35 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 42 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, a}{420 \, c^{4} f \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^4,x, algorithm="giac")

[Out]

-1/420*(35*a*tan(1/2*f*x + 1/2*e)^4 - 42*a*tan(1/2*f*x + 1/2*e)^2 + 15*a)/(c^4*f*tan(1/2*f*x + 1/2*e)^7)

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Mupad [B]
time = 1.75, size = 61, normalized size = 0.53 \begin {gather*} \frac {a\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{10\,c^4\,f}-\frac {a\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{12\,c^4\,f}-\frac {a\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{28\,c^4\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))/(cos(e + f*x)*(c - c/cos(e + f*x))^4),x)

[Out]

(a*cot(e/2 + (f*x)/2)^5)/(10*c^4*f) - (a*cot(e/2 + (f*x)/2)^3)/(12*c^4*f) - (a*cot(e/2 + (f*x)/2)^7)/(28*c^4*f
)

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