3.3.30 \(\int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx\) [230]

Optimal. Leaf size=102 \[ \frac {(c-d) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {(2 c+3 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {(2 c+3 d) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )} \]

[Out]

1/5*(c-d)*tan(f*x+e)/f/(a+a*sec(f*x+e))^3+1/15*(2*c+3*d)*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^2+1/15*(2*c+3*d)*tan(
f*x+e)/f/(a^3+a^3*sec(f*x+e))

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Rubi [A]
time = 0.08, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {4085, 3881, 3879} \begin {gather*} \frac {(2 c+3 d) \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac {(2 c+3 d) \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2}+\frac {(c-d) \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3881

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

Rule 4085

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(A*b - a*B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Dist[(a*B*m + A*b*
(m + 1))/(a*b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x
] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]

Rubi steps

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

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Mathematica [A]
time = 0.35, size = 135, normalized size = 1.32 \begin {gather*} \frac {\cos \left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {e}{2}\right ) \left (5 (8 c+3 d) \sin \left (\frac {f x}{2}\right )-15 (2 c+d) \sin \left (e+\frac {f x}{2}\right )+20 c \sin \left (e+\frac {3 f x}{2}\right )+15 d \sin \left (e+\frac {3 f x}{2}\right )-15 c \sin \left (2 e+\frac {3 f x}{2}\right )+7 c \sin \left (2 e+\frac {5 f x}{2}\right )+3 d \sin \left (2 e+\frac {5 f x}{2}\right )\right )}{30 a^3 f (1+\cos (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(e + f*x)/2]*Sec[e/2]*(5*(8*c + 3*d)*Sin[(f*x)/2] - 15*(2*c + d)*Sin[e + (f*x)/2] + 20*c*Sin[e + (3*f*x)/
2] + 15*d*Sin[e + (3*f*x)/2] - 15*c*Sin[2*e + (3*f*x)/2] + 7*c*Sin[2*e + (5*f*x)/2] + 3*d*Sin[2*e + (5*f*x)/2]
))/(30*a^3*f*(1 + Cos[e + f*x])^3)

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Maple [A]
time = 0.22, size = 64, normalized size = 0.63

method	result	size
derivativedivides	\(\frac {\frac {\left (c -d \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}-\frac {2 c \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f \,a^{3}}\)	\(64\)
default	\(\frac {\frac {\left (c -d \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}-\frac {2 c \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f \,a^{3}}\)	\(64\)
risch	\(\frac {2 i \left (15 c \,{\mathrm e}^{4 i \left (f x +e \right )}+30 c \,{\mathrm e}^{3 i \left (f x +e \right )}+15 d \,{\mathrm e}^{3 i \left (f x +e \right )}+40 \,{\mathrm e}^{2 i \left (f x +e \right )} c +15 d \,{\mathrm e}^{2 i \left (f x +e \right )}+20 \,{\mathrm e}^{i \left (f x +e \right )} c +15 d \,{\mathrm e}^{i \left (f x +e \right )}+7 c +3 d \right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}\)	\(114\)
norman	\(\frac {\frac {\left (c -d \right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{20 a f}-\frac {\left (c +d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 a f}+\frac {\left (5 c +3 d \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{12 f a}-\frac {\left (13 c -3 d \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{60 f a}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) a^{2}}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 125, normalized size = 1.23 \begin {gather*} \frac {\frac {c {\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 {3 \, 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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(c*(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 + 3*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 = 3.15, size = 99, normalized size = 0.97 \begin {gather*} \frac {{\left ({\left (7 \, c + 3 \, d\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (2 \, c + 3 \, d\right )} \cos \left (f x + e\right ) + 2 \, c + 3 \, d\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((7*c + 3*d)*cos(f*x + e)^2 + 3*(2*c + 3*d)*cos(f*x + e) + 2*c + 3*d)*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {c \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 \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(c*sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(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 = 0.48, size = 75, normalized size = 0.74 \begin {gather*} \frac {3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 3 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{60 \, a^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.74, size = 66, normalized size = 0.65 \begin {gather*} \frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (15\,c+15\,d-10\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-3\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{60\,a^3\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(e/2 + (f*x)/2)*(15*c + 15*d - 10*c*tan(e/2 + (f*x)/2)^2 + 3*c*tan(e/2 + (f*x)/2)^4 - 3*d*tan(e/2 + (f*x)/
2)^4))/(60*a^3*f)

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