3.1.19 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^5} \, dx\) [19]

Optimal. Leaf size=121 \[ -\frac {(a+a \sec (e+f x))^2 \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}-\frac {2 (a+a \sec (e+f x))^2 \tan (e+f x)}{63 c f (c-c \sec (e+f x))^4}-\frac {2 (a+a \sec (e+f x))^2 \tan (e+f x)}{315 c^2 f (c-c \sec (e+f x))^3} \]

[Out]

-1/9*(a+a*sec(f*x+e))^2*tan(f*x+e)/f/(c-c*sec(f*x+e))^5-2/63*(a+a*sec(f*x+e))^2*tan(f*x+e)/c/f/(c-c*sec(f*x+e)
)^4-2/315*(a+a*sec(f*x+e))^2*tan(f*x+e)/c^2/f/(c-c*sec(f*x+e))^3

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Rubi [A]
time = 0.17, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4036, 4035} \begin {gather*} -\frac {2 \tan (e+f x) (a \sec (e+f x)+a)^2}{315 c^2 f (c-c \sec (e+f x))^3}-\frac {2 \tan (e+f x) (a \sec (e+f x)+a)^2}{63 c f (c-c \sec (e+f x))^4}-\frac {\tan (e+f x) (a \sec (e+f x)+a)^2}{9 f (c-c \sec (e+f x))^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c - c*Sec[e + f*x])^5,x]

[Out]

-1/9*((a + a*Sec[e + f*x])^2*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^5) - (2*(a + a*Sec[e + f*x])^2*Tan[e + f*x]
)/(63*c*f*(c - c*Sec[e + f*x])^4) - (2*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(315*c^2*f*(c - c*Sec[e + f*x])^3)

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))^2}{(c-c \sec (e+f x))^5} \, dx &=-\frac {(a+a \sec (e+f x))^2 \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}+\frac {2 \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^4} \, dx}{9 c}\\ &=-\frac {(a+a \sec (e+f x))^2 \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}-\frac {2 (a+a \sec (e+f x))^2 \tan (e+f x)}{63 c f (c-c \sec (e+f x))^4}+\frac {2 \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^3} \, dx}{63 c^2}\\ &=-\frac {(a+a \sec (e+f x))^2 \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}-\frac {2 (a+a \sec (e+f x))^2 \tan (e+f x)}{63 c f (c-c \sec (e+f x))^4}-\frac {2 (a+a \sec (e+f x))^2 \tan (e+f x)}{315 c^2 f (c-c \sec (e+f x))^3}\\ \end {align*}

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Mathematica [A]
time = 0.43, size = 141, normalized size = 1.17 \begin {gather*} -\frac {a^2 \csc \left (\frac {e}{2}\right ) \csc ^9\left (\frac {1}{2} (e+f x)\right ) \left (3402 \sin \left (\frac {f x}{2}\right )+2520 \sin \left (e+\frac {f x}{2}\right )-1638 \sin \left (e+\frac {3 f x}{2}\right )-2310 \sin \left (2 e+\frac {3 f x}{2}\right )+1062 \sin \left (2 e+\frac {5 f x}{2}\right )+630 \sin \left (3 e+\frac {5 f x}{2}\right )-108 \sin \left (3 e+\frac {7 f x}{2}\right )-315 \sin \left (4 e+\frac {7 f x}{2}\right )+47 \sin \left (4 e+\frac {9 f x}{2}\right )\right )}{80640 c^5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c - c*Sec[e + f*x])^5,x]

[Out]

-1/80640*(a^2*Csc[e/2]*Csc[(e + f*x)/2]^9*(3402*Sin[(f*x)/2] + 2520*Sin[e + (f*x)/2] - 1638*Sin[e + (3*f*x)/2]
- 2310*Sin[2*e + (3*f*x)/2] + 1062*Sin[2*e + (5*f*x)/2] + 630*Sin[3*e + (5*f*x)/2] - 108*Sin[3*e + (7*f*x)/2]
- 315*Sin[4*e + (7*f*x)/2] + 47*Sin[4*e + (9*f*x)/2]))/(c^5*f)

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Maple [A]
time = 0.21, size = 52, normalized size = 0.43

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {1}{9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {2}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}\right )}{4 f \,c^{5}}\)	\(52\)
default	\(\frac {a^{2} \left (\frac {1}{9 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {2}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}\right )}{4 f \,c^{5}}\)	\(52\)
risch	\(\frac {2 i a^{2} \left (315 \,{\mathrm e}^{8 i \left (f x +e \right )}-630 \,{\mathrm e}^{7 i \left (f x +e \right )}+2310 \,{\mathrm e}^{6 i \left (f x +e \right )}-2520 \,{\mathrm e}^{5 i \left (f x +e \right )}+3402 \,{\mathrm e}^{4 i \left (f x +e \right )}-1638 \,{\mathrm e}^{3 i \left (f x +e \right )}+1062 \,{\mathrm e}^{2 i \left (f x +e \right )}-108 \,{\mathrm e}^{i \left (f x +e \right )}+47\right )}{315 f \,c^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{9}}\)	\(116\)
norman	\(\frac {\frac {a^{2}}{36 c f}-\frac {8 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{63 c f}+\frac {139 a^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{630 c f}-\frac {6 a^{2} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{35 c f}+\frac {a^{2} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{20 c f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2} c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^5,x,method=_RETURNVERBOSE)

[Out]

1/4/f*a^2/c^5*(1/9/tan(1/2*f*x+1/2*e)^9+1/5/tan(1/2*f*x+1/2*e)^5-2/7/tan(1/2*f*x+1/2*e)^7)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (127) = 254\).
time = 0.31, size = 293, normalized size = 2.42 \begin {gather*} -\frac {\frac {a^{2} {\left (\frac {180 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {378 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {420 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {315 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 35\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}} + \frac {10 \, a^{2} {\left (\frac {18 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {42 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {63 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 7\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}} + \frac {7 \, a^{2} {\left (\frac {18 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {45 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 5\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}}}{5040 \, 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))^2/(c-c*sec(f*x+e))^5,x, algorithm="maxima")

[Out]

-1/5040*(a^2*(180*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 378*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 420*sin(f*x
+ e)^6/(cos(f*x + e) + 1)^6 - 315*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 35)*(cos(f*x + e) + 1)^9/(c^5*sin(f*x
+ e)^9) + 10*a^2*(18*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 42*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 63*sin(f*x
+ e)^8/(cos(f*x + e) + 1)^8 - 7)*(cos(f*x + e) + 1)^9/(c^5*sin(f*x + e)^9) + 7*a^2*(18*sin(f*x + e)^4/(cos(f*
x + e) + 1)^4 - 45*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 5)*(cos(f*x + e) + 1)^9/(c^5*sin(f*x + e)^9))/f

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Fricas [A]
time = 3.23, size = 150, normalized size = 1.24 \begin {gather*} \frac {47 \, a^{2} \cos \left (f x + e\right )^{5} + 127 \, a^{2} \cos \left (f x + e\right )^{4} + 101 \, a^{2} \cos \left (f x + e\right )^{3} + 11 \, a^{2} \cos \left (f x + e\right )^{2} - 8 \, a^{2} \cos \left (f x + e\right ) + 2 \, a^{2}}{315 \, {\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} + 6 \, c^{5} f \cos \left (f x + e\right )^{2} - 4 \, c^{5} f \cos \left (f x + e\right ) + c^{5} 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))^2/(c-c*sec(f*x+e))^5,x, algorithm="fricas")

[Out]

1/315*(47*a^2*cos(f*x + e)^5 + 127*a^2*cos(f*x + e)^4 + 101*a^2*cos(f*x + e)^3 + 11*a^2*cos(f*x + e)^2 - 8*a^2
*cos(f*x + e) + 2*a^2)/((c^5*f*cos(f*x + e)^4 - 4*c^5*f*cos(f*x + e)^3 + 6*c^5*f*cos(f*x + e)^2 - 4*c^5*f*cos(
f*x + e) + c^5*f)*sin(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {a^{2} \left (\int \frac {\sec {\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {2 \sec ^{2}{\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec {\left (e + f x \right )} - 1}\, dx\right )}{c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**2/(c-c*sec(f*x+e))**5,x)

[Out]

-a**2*(Integral(sec(e + f*x)/(sec(e + f*x)**5 - 5*sec(e + f*x)**4 + 10*sec(e + f*x)**3 - 10*sec(e + f*x)**2 +
5*sec(e + f*x) - 1), x) + Integral(2*sec(e + f*x)**2/(sec(e + f*x)**5 - 5*sec(e + f*x)**4 + 10*sec(e + f*x)**3
- 10*sec(e + f*x)**2 + 5*sec(e + f*x) - 1), x) + Integral(sec(e + f*x)**3/(sec(e + f*x)**5 - 5*sec(e + f*x)**
4 + 10*sec(e + f*x)**3 - 10*sec(e + f*x)**2 + 5*sec(e + f*x) - 1), x))/c**5

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Giac [A]
time = 0.60, size = 57, normalized size = 0.47 \begin {gather*} \frac {63 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 90 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 35 \, a^{2}}{1260 \, c^{5} f \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^5,x, algorithm="giac")

[Out]

1/1260*(63*a^2*tan(1/2*f*x + 1/2*e)^4 - 90*a^2*tan(1/2*f*x + 1/2*e)^2 + 35*a^2)/(c^5*f*tan(1/2*f*x + 1/2*e)^9)

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Mupad [B]
time = 1.63, size = 67, normalized size = 0.55 \begin {gather*} \frac {a^2\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{20\,c^5\,f}-\frac {a^2\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{14\,c^5\,f}+\frac {a^2\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{36\,c^5\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^2/(cos(e + f*x)*(c - c/cos(e + f*x))^5),x)

[Out]

(a^2*cot(e/2 + (f*x)/2)^5)/(20*c^5*f) - (a^2*cot(e/2 + (f*x)/2)^7)/(14*c^5*f) + (a^2*cot(e/2 + (f*x)/2)^9)/(36
*c^5*f)

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