3.4.40 \(\int \frac {\sec ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [340]

Optimal. Leaf size=156 \[ -\frac {(4 B-7 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {2 (5 B-8 C) \tan (c+d x)}{3 a^2 d}-\frac {(4 B-7 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {(5 B-8 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

-1/2*(4*B-7*C)*arctanh(sin(d*x+c))/a^2/d+2/3*(5*B-8*C)*tan(d*x+c)/a^2/d-1/2*(4*B-7*C)*sec(d*x+c)*tan(d*x+c)/a^
2/d+1/3*(5*B-8*C)*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(1+sec(d*x+c))+1/3*(B-C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*
x+c))^2

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Rubi [A]
time = 0.27, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4157, 4104, 3872, 3852, 8, 3853, 3855} \begin {gather*} \frac {2 (5 B-8 C) \tan (c+d x)}{3 a^2 d}-\frac {(4 B-7 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {(5 B-8 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {(4 B-7 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac {(B-C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((4*B - 7*C)*ArcTanh[Sin[c + d*x]])/(a^2*d) + (2*(5*B - 8*C)*Tan[c + d*x])/(3*a^2*d) - ((4*B - 7*C)*Sec[c
+ d*x]*Tan[c + d*x])/(2*a^2*d) + ((5*B - 8*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + ((B
- C)*Sec[c + d*x]^3*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

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

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

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 ^3(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 ^4(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^3(c+d x) (3 a (B-C)-a (2 B-5 C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac {(5 B-8 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \sec ^2(c+d x) \left (2 a^2 (5 B-8 C)-3 a^2 (4 B-7 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac {(5 B-8 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(2 (5 B-8 C)) \int \sec ^2(c+d x) \, dx}{3 a^2}-\frac {(4 B-7 C) \int \sec ^3(c+d x) \, dx}{a^2}\\ &=-\frac {(4 B-7 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {(5 B-8 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(4 B-7 C) \int \sec (c+d x) \, dx}{2 a^2}-\frac {(2 (5 B-8 C)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d}\\ &=-\frac {(4 B-7 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {2 (5 B-8 C) \tan (c+d x)}{3 a^2 d}-\frac {(4 B-7 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {(5 B-8 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(379\) vs. \(2(156)=312\).
time = 1.59, size = 379, normalized size = 2.43 \begin {gather*} \frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (3 (4 B-7 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 )+8 (B+5 C) \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-64 (B-C) \csc ^5(c+d x) \sin ^8\left (\frac {1}{2} (c+d x)\right )-128 C \csc ^7(c+d x) \sin ^{12}\left (\frac {1}{2} (c+d x)\right )+(26 B-44 C) \tan \left (\frac {1}{2} (c+d x)\right )-6 (4 B-7 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 ) \tan ^2\left (\frac {1}{2} (c+d x)\right )-8 (5 B-8 C) \tan ^3\left (\frac {1}{2} (c+d x)\right )+3 (4 B-7 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 ) \tan ^4\left (\frac {1}{2} (c+d x)\right )+\left (14 B-20 C+B \sec ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^5\left (\frac {1}{2} (c+d x)\right )\right )}{6 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(3*(4*B - 7*C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x
)/2] + Sin[(c + d*x)/2]]) + 8*(B + 5*C)*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 64*(B - C)*Csc[c + d*x]^5*Sin[(c +
d*x)/2]^8 - 128*C*Csc[c + d*x]^7*Sin[(c + d*x)/2]^12 + (26*B - 44*C)*Tan[(c + d*x)/2] - 6*(4*B - 7*C)*(Log[Co
s[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*Tan[(c + d*x)/2]^2 - 8*(5*B - 8
*C)*Tan[(c + d*x)/2]^3 + 3*(4*B - 7*C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[
(c + d*x)/2]])*Tan[(c + d*x)/2]^4 + (14*B - 20*C + B*Sec[(c + d*x)/2]^2)*Tan[(c + d*x)/2]^5))/(6*a^2*d)

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Maple [A]
time = 0.48, size = 177, normalized size = 1.13

method	result	size
derivativedivides	\(\frac {\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (4 B -7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (7 C -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{2 d \,a^{2}}\)	\(177\)
default	\(\frac {\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (4 B -7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (7 C -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{2 d \,a^{2}}\)	\(177\)
norman	\(\frac {\frac {\left (B -C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (9 B -13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (11 B -18 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (11 B -17 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (61 B -100 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\left (95 B -149 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} a}+\frac {\left (4 B -7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2} d}-\frac {\left (4 B -7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2} d}\)	\(228\)
risch	\(\frac {i \left (12 B \,{\mathrm e}^{6 i \left (d x +c \right )}-21 C \,{\mathrm e}^{6 i \left (d x +c \right )}+36 B \,{\mathrm e}^{5 i \left (d x +c \right )}-63 C \,{\mathrm e}^{5 i \left (d x +c \right )}+56 B \,{\mathrm e}^{4 i \left (d x +c \right )}-98 C \,{\mathrm e}^{4 i \left (d x +c \right )}+84 B \,{\mathrm e}^{3 i \left (d x +c \right )}-126 C \,{\mathrm e}^{3 i \left (d x +c \right )}+64 B \,{\mathrm e}^{2 i \left (d x +c \right )}-97 C \,{\mathrm e}^{2 i \left (d x +c \right )}+48 B \,{\mathrm e}^{i \left (d x +c \right )}-75 C \,{\mathrm e}^{i \left (d x +c \right )}+20 B -32 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{2} d}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a^{2} d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{2} d}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a^{2} d}\)	\(276\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (146) = 292\).
time = 0.28, size = 336, normalized size = 2.15 \begin {gather*} -\frac {C {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \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 {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - B {\left (\frac {\frac {15 \, \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 {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(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*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) + s
in(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x + c
)/(cos(d*x + c) + 1) - 1)/a^2) - B*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)
/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 12*
sin(d*x + c)/((a^2 - a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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Fricas [A]
time = 2.88, size = 228, normalized size = 1.46 \begin {gather*} -\frac {3 \, {\left ({\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (5 \, B - 8 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (28 \, B - 43 \, C\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (B - C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*((4*B - 7*C)*cos(d*x + c)^4 + 2*(4*B - 7*C)*cos(d*x + c)^3 + (4*B - 7*C)*cos(d*x + c)^2)*log(sin(d*x
+ c) + 1) - 3*((4*B - 7*C)*cos(d*x + c)^4 + 2*(4*B - 7*C)*cos(d*x + c)^3 + (4*B - 7*C)*cos(d*x + c)^2)*log(-si
n(d*x + c) + 1) - 2*(4*(5*B - 8*C)*cos(d*x + c)^3 + (28*B - 43*C)*cos(d*x + c)^2 + 6*(B - C)*cos(d*x + c) + 3*
C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {B \sec ^{4}{\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 ^{5}{\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)**3*(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)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**5/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

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Giac [A]
time = 0.49, size = 198, normalized size = 1.27 \begin {gather*} -\frac {\frac {3 \, {\left (4 \, B - 7 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (4 \, B - 7 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} 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} + 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 21 \, 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)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(3*(4*B - 7*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(4*B - 7*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/
a^2 + 6*(2*B*tan(1/2*d*x + 1/2*c)^3 - 5*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) + 3*C*tan(1/2*d*x
+ 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*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
+ 15*B*a^4*tan(1/2*d*x + 1/2*c) - 21*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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Mupad [B]
time = 2.90, size = 166, normalized size = 1.06 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{2\,a^2}+\frac {2\,B-4\,C}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-5\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-3\,C\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,B-7\,C\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)^3*(a + a/cos(c + d*x))^2),x)

[Out]

(tan(c/2 + (d*x)/2)*((3*(B - C))/(2*a^2) + (2*B - 4*C)/(2*a^2)))/d - (tan(c/2 + (d*x)/2)^3*(2*B - 5*C) - tan(c
/2 + (d*x)/2)*(2*B - 3*C))/(d*(a^2*tan(c/2 + (d*x)/2)^4 - 2*a^2*tan(c/2 + (d*x)/2)^2 + a^2)) + (tan(c/2 + (d*x
)/2)^3*(B - C))/(6*a^2*d) - (atanh(tan(c/2 + (d*x)/2))*(4*B - 7*C))/(a^2*d)

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