3.1.74 \(\int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^4} \, dx\) [74]

Optimal. Leaf size=112 \[ \frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {11 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {13 \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {13 \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )} \]

[Out]

1/7*tan(d*x+c)/d/(a+a*sec(d*x+c))^4-11/35*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3+13/105*tan(d*x+c)/d/(a^2+a^2*sec(d
*x+c))^2+13/105*tan(d*x+c)/d/(a^4+a^4*sec(d*x+c))

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Rubi [A]
time = 0.11, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3884, 4085, 3881, 3879} \begin {gather*} \frac {13 \tan (c+d x)}{105 d \left (a^4 \sec (c+d x)+a^4\right )}+\frac {13 \tan (c+d x)}{105 d \left (a^2 \sec (c+d x)+a^2\right )^2}-\frac {11 \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + a*Sec[c + d*x])^4,x]

[Out]

Tan[c + d*x]/(7*d*(a + a*Sec[c + d*x])^4) - (11*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3) + (13*Tan[c + d*
x])/(105*d*(a^2 + a^2*Sec[c + d*x])^2) + (13*Tan[c + d*x])/(105*d*(a^4 + a^4*Sec[c + d*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 3884

Int[csc[(e_.) + (f_.)*(x_)]^3*(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[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)
]

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 ^3(c+d x)}{(a+a \sec (c+d x))^4} \, dx &=\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\sec (c+d x) (-4 a+7 a \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {11 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {13 \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{35 a^2}\\ &=\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {11 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {13 \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {13 \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}\\ &=\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {11 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {13 \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {13 \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 87, normalized size = 0.78 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (35 \sin \left (\frac {d x}{2}\right )-35 \sin \left (c+\frac {d x}{2}\right )+2 \left (21 \sin \left (c+\frac {3 d x}{2}\right )+7 \sin \left (2 c+\frac {5 d x}{2}\right )+\sin \left (3 c+\frac {7 d x}{2}\right )\right )\right )}{1680 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + a*Sec[c + d*x])^4,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(35*Sin[(d*x)/2] - 35*Sin[c + (d*x)/2] + 2*(21*Sin[c + (3*d*x)/2] + 7*Sin[2*c + (
5*d*x)/2] + Sin[3*c + (7*d*x)/2])))/(1680*a^4*d)

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Maple [A]
time = 0.09, size = 58, normalized size = 0.52

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(58\)
default	\(\frac {-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(58\)
risch	\(\frac {8 i \left (35 \,{\mathrm e}^{4 i \left (d x +c \right )}+35 \,{\mathrm e}^{3 i \left (d x +c \right )}+42 \,{\mathrm e}^{2 i \left (d x +c \right )}+14 \,{\mathrm e}^{i \left (d x +c \right )}+2\right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(69\)
norman	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}+\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{60 a d}+\frac {31 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{420 a d}+\frac {3 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{280 a d}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a^{3}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/8/d/a^4*(-1/7*tan(1/2*d*x+1/2*c)^7-1/5*tan(1/2*d*x+1/2*c)^5+1/3*tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1/2*c))

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Maxima [A]
time = 0.28, size = 87, normalized size = 0.78 \begin {gather*} \frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{840 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 4.04, size = 99, normalized size = 0.88 \begin {gather*} \frac {{\left (8 \, \cos \left (d x + c\right )^{3} + 32 \, \cos \left (d x + c\right )^{2} + 52 \, \cos \left (d x + c\right ) + 13\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/105*(8*cos(d*x + c)^3 + 32*cos(d*x + c)^2 + 52*cos(d*x + c) + 13)*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4
*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+a*sec(d*x+c))**4,x)

[Out]

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

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Giac [A]
time = 0.48, size = 59, normalized size = 0.53 \begin {gather*} -\frac {15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

-1/840*(15*tan(1/2*d*x + 1/2*c)^7 + 21*tan(1/2*d*x + 1/2*c)^5 - 35*tan(1/2*d*x + 1/2*c)^3 - 105*tan(1/2*d*x +
1/2*c))/(a^4*d)

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Mupad [B]
time = 0.66, size = 58, normalized size = 0.52 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+105\right )}{840\,a^4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a + a/cos(c + d*x))^4),x)

[Out]

(tan(c/2 + (d*x)/2)*(35*tan(c/2 + (d*x)/2)^2 - 21*tan(c/2 + (d*x)/2)^4 - 15*tan(c/2 + (d*x)/2)^6 + 105))/(840*
a^4*d)

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