3.4.50 \(\int \frac {\sec ^2(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [350]

Optimal. Leaf size=125 \[ \frac {C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(B-C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(4 B-29 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

C*arctanh(sin(d*x+c))/a^3/d+1/5*(B-C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(2*B-7*C)*tan(d*x+c)/a
/d/(a+a*sec(d*x+c))^2+1/15*(4*B-29*C)*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]
time = 0.27, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4157, 4104, 4093, 4083, 3855, 3879} \begin {gather*} \frac {(4 B-29 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {(2 B-7 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(C*ArcTanh[Sin[c + d*x]])/(a^3*d) + ((B - C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((2*B
- 7*C)*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) + ((4*B - 29*C)*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d
*x]))

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{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 4083

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4093

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

Rule 4104

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

Rule 4157

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

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=\int \frac {\sec ^3(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\\ &=\frac {(B-C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^2(c+d x) (2 a (B-C)+5 a C \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac {(B-C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec (c+d x) \left (-2 a^2 (2 B-7 C)-15 a^2 C \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac {(B-C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(4 B-29 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}+\frac {C \int \sec (c+d x) \, dx}{a^3}\\ &=\frac {C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(B-C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(4 B-29 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.71, size = 136, normalized size = 1.09 \begin {gather*} \frac {15 C \left (-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+4 (2 B-7 C) \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+24 (B-C) \csc ^5(c+d x) \sin ^6\left (\frac {1}{2} (c+d x)\right )+2 (B-11 C) \tan \left (\frac {1}{2} (c+d x)\right )}{15 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*C*(-Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 4*(2*B - 7*C)*C
sc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 24*(B - C)*Csc[c + d*x]^5*Sin[(c + d*x)/2]^6 + 2*(B - 11*C)*Tan[(c + d*x)/2
])/(15*a^3*d)

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Maple [A]
time = 0.48, size = 119, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}}{4 d \,a^{3}}\)	\(119\)
default	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}}{4 d \,a^{3}}\)	\(119\)
risch	\(-\frac {2 i \left (15 C \,{\mathrm e}^{4 i \left (d x +c \right )}+75 C \,{\mathrm e}^{3 i \left (d x +c \right )}-20 B \,{\mathrm e}^{2 i \left (d x +c \right )}+145 C \,{\mathrm e}^{2 i \left (d x +c \right )}-10 B \,{\mathrm e}^{i \left (d x +c \right )}+95 C \,{\mathrm e}^{i \left (d x +c \right )}-2 B +22 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}\)	\(146\)
norman	\(\frac {\frac {\left (B -11 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}-\frac {\left (B -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (B -C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {\left (B +9 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 a d}-\frac {\left (3 B -43 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 a d}+\frac {\left (-59 C +7 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} a^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\)	\(209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 187, normalized size = 1.50 \begin {gather*} -\frac {C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(C*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) +
1) - 1)/a^3) - B*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c
)^5/(cos(d*x + c) + 1)^5)/a^3)/d

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Fricas [A]
time = 3.96, size = 183, normalized size = 1.46 \begin {gather*} \frac {15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (B - 11 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B - 17 \, C\right )} \cos \left (d x + c\right ) + 7 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(15*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*log(sin(d*x + c) + 1) - 15*(C*cos(d*x
+ c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*log(-sin(d*x + c) + 1) + 2*(2*(B - 11*C)*cos(d*x + c)^2 +
3*(2*B - 17*C)*cos(d*x + c) + 7*B - 32*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3
*d*cos(d*x + c) + a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**3,x)

[Out]

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

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Giac [A]
time = 0.53, size = 147, normalized size = 1.18 \begin {gather*} \frac {\frac {60 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + (3*B*a^12*ta
n(1/2*d*x + 1/2*c)^5 - 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 10*B*a^12*tan(1/2*d*x + 1/2*c)^3 - 20*C*a^12*tan(1/2*
d*x + 1/2*c)^3 + 15*B*a^12*tan(1/2*d*x + 1/2*c) - 105*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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Mupad [B]
time = 2.90, size = 124, normalized size = 0.99 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {B-C}{12\,a^3}+\frac {B-3\,C}{12\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B-C}{4\,a^3}+\frac {B-3\,C}{4\,a^3}-\frac {B+3\,C}{4\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-C\right )}{20\,a^3\,d}+\frac {2\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a/cos(c + d*x))^3),x)

[Out]

(tan(c/2 + (d*x)/2)^3*((B - C)/(12*a^3) + (B - 3*C)/(12*a^3)))/d + (tan(c/2 + (d*x)/2)*((B - C)/(4*a^3) + (B -
3*C)/(4*a^3) - (B + 3*C)/(4*a^3)))/d + (tan(c/2 + (d*x)/2)^5*(B - C))/(20*a^3*d) + (2*C*atanh(tan(c/2 + (d*x)
/2)))/(a^3*d)

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