3.4.8 \(\int (a+a \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [308]

Optimal. Leaf size=56 \[ \frac {a (2 B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (B+C) \tan (c+d x)}{d}+\frac {a C \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4133, 3855, 3852, 8} \begin {gather*} \frac {a (B+C) \tan (c+d x)}{d}+\frac {a (2 B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a C \tan (c+d x) \sec (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*B + C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(B + C)*Tan[c + d*x])/d + (a*C*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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 3855

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

Rule 4133

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 75, normalized size = 1.34 \begin {gather*} \frac {a B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a B \tan (c+d x)}{d}+\frac {a C \tan (c+d x)}{d}+\frac {a C \sec (c+d x) \tan (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*B*ArcTanh[Sin[c + d*x]])/d + (a*C*ArcTanh[Sin[c + d*x]])/(2*d) + (a*B*Tan[c + d*x])/d + (a*C*Tan[c + d*x])/
d + (a*C*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Maple [A]
time = 0.85, size = 75, normalized size = 1.34

method	result	size
derivativedivides	\(\frac {B a \tan \left (d x +c \right )+a C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a C \tan \left (d x +c \right )}{d}\)	\(75\)
default	\(\frac {B a \tan \left (d x +c \right )+a C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a C \tan \left (d x +c \right )}{d}\)	\(75\)
norman	\(\frac {\frac {a \left (2 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a \left (2 B +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a \left (2 B +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (2 B +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(108\)
risch	\(-\frac {i a \left (C \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C \,{\mathrm e}^{2 i \left (d x +c \right )}-C \,{\mathrm e}^{i \left (d x +c \right )}-2 B -2 C \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(B*a*tan(d*x+c)+a*C*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+B*a*ln(sec(d*x+c)+tan(d*x+c)
)+a*C*tan(d*x+c))

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Maxima [A]
time = 0.28, size = 88, normalized size = 1.57 \begin {gather*} -\frac {C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, B a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 4 \, B a \tan \left (d x + c\right ) - 4 \, C a \tan \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(C*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*B*a*log(se
c(d*x + c) + tan(d*x + c)) - 4*B*a*tan(d*x + c) - 4*C*a*tan(d*x + c))/d

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Fricas [A]
time = 3.62, size = 89, normalized size = 1.59 \begin {gather*} \frac {{\left (2 \, B + C\right )} a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, B + C\right )} a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (B + C\right )} a \cos \left (d x + c\right ) + C a\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (52) = 104\).
time = 0.47, size = 124, normalized size = 2.21 \begin {gather*} \frac {{\left (2 \, B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a \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}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.54, size = 94, normalized size = 1.68 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B\,a+3\,C\,a\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B\,a+C\,a\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,B+C\right )}{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)*(a + a/cos(c + d*x)),x)

[Out]

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

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