3.4.42 \(\int \frac {\sec (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [342]

Optimal. Leaf size=79 \[ \frac {C \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {(2 B-5 C) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {4157, 4093, 4083, 3855, 3879} \begin {gather*} \frac {(2 B-5 C) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {C \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(B-C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(C*ArcTanh[Sin[c + d*x]])/(a^2*d) + ((2*B - 5*C)*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((B - C)*Tan[c +
d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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

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Mathematica [A]
time = 0.79, size = 106, normalized size = 1.34 \begin {gather*} \frac {3 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 (B-C) \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+(B-4 C) \tan \left (\frac {1}{2} (c+d x)\right )}{3 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*(-Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 4*(B - C)*Csc[c
+ d*x]^3*Sin[(c + d*x)/2]^4 + (B - 4*C)*Tan[(c + d*x)/2])/(3*a^2*d)

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Maple [A]
time = 0.46, size = 91, normalized size = 1.15

method	result	size
derivativedivides	\(\frac {\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\)	\(91\)
default	\(\frac {\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\)	\(91\)
risch	\(-\frac {2 i \left (3 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}+9 C \,{\mathrm e}^{i \left (d x +c \right )}-B +4 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{2} d}\)	\(110\)
norman	\(\frac {\frac {\left (B -7 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\left (B -C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (5 B -17 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 145, normalized size = 1.84 \begin {gather*} -\frac {C {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - \frac {B {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(C*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(c
os(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) - B*(3*sin(d*x + c)/(cos(d*x + c)
+ 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d

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Fricas [A]
time = 2.27, size = 129, normalized size = 1.63 \begin {gather*} \frac {3 \, {\left (C \cos \left (d x + c\right )^{2} + 2 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (C \cos \left (d x + c\right )^{2} + 2 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (B - 4 \, C\right )} \cos \left (d x + c\right ) + 2 \, B - 5 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(C*cos(d*x + c)^2 + 2*C*cos(d*x + c) + C)*log(sin(d*x + c) + 1) - 3*(C*cos(d*x + c)^2 + 2*C*cos(d*x + c
) + C)*log(-sin(d*x + c) + 1) + 2*((B - 4*C)*cos(d*x + c) + 2*B - 5*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2
*a^2*d*cos(d*x + c) + a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.46, size = 112, normalized size = 1.42 \begin {gather*} \frac {\frac {6 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + (B*a^4*tan(1/2*
d*x + 1/2*c)^3 - C*a^4*tan(1/2*d*x + 1/2*c)^3 + 3*B*a^4*tan(1/2*d*x + 1/2*c) - 9*C*a^4*tan(1/2*d*x + 1/2*c))/a
^6)/d

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Mupad [B]
time = 2.80, size = 74, normalized size = 0.94 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {C}{a^2}-\frac {B-C}{2\,a^2}\right )}{d}+\frac {2\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,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)*(a + a/cos(c + d*x))^2),x)

[Out]

(tan(c/2 + (d*x)/2)^3*(B - C))/(6*a^2*d) - (tan(c/2 + (d*x)/2)*(C/a^2 - (B - C)/(2*a^2)))/d + (2*C*atanh(tan(c
/2 + (d*x)/2)))/(a^2*d)

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