3.1.4 \(\int \sec ^3(c+d x) (A+C \sec ^2(c+d x)) \, dx\) [4]

Optimal. Leaf size=70 \[ \frac {(4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[Out]

1/8*(4*A+3*C)*arctanh(sin(d*x+c))/d+1/8*(4*A+3*C)*sec(d*x+c)*tan(d*x+c)/d+1/4*C*sec(d*x+c)^3*tan(d*x+c)/d

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Rubi [A]
time = 0.03, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4131, 3853, 3855} \begin {gather*} \frac {(4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A + 3*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((4*A + 3*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (C*Sec[c + d*x]^3*
Tan[c + d*x])/(4*d)

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 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 54, normalized size = 0.77 \begin {gather*} \frac {(4 A+3 C) \tanh ^{-1}(\sin (c+d x))+\sec (c+d x) \left (4 A+3 C+2 C \sec ^2(c+d x)\right ) \tan (c+d x)}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A + 3*C)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(4*A + 3*C + 2*C*Sec[c + d*x]^2)*Tan[c + d*x])/(8*d)

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Maple [A]
time = 0.46, size = 85, normalized size = 1.21

method	result	size
derivativedivides	\(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(85\)
default	\(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(85\)
norman	\(\frac {-\frac {\left (4 A -3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {\left (4 A -3 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (4 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (4 A +5 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(157\)
risch	\(-\frac {i \left (4 A \,{\mathrm e}^{7 i \left (d x +c \right )}+3 C \,{\mathrm e}^{7 i \left (d x +c \right )}+4 A \,{\mathrm e}^{5 i \left (d x +c \right )}+11 C \,{\mathrm e}^{5 i \left (d x +c \right )}-4 A \,{\mathrm e}^{3 i \left (d x +c \right )}-11 C \,{\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{i \left (d x +c \right )} A -3 C \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(194\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+C*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*
x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c))))

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Maxima [A]
time = 0.28, size = 97, normalized size = 1.39 \begin {gather*} \frac {{\left (4 \, A + 3 \, C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, A + 3 \, C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (4 \, A + 3 \, C\right )} \sin \left (d x + c\right )^{3} - {\left (4 \, A + 5 \, C\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((4*A + 3*C)*log(sin(d*x + c) + 1) - (4*A + 3*C)*log(sin(d*x + c) - 1) - 2*((4*A + 3*C)*sin(d*x + c)^3 -
(4*A + 5*C)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

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Fricas [A]
time = 2.41, size = 95, normalized size = 1.36 \begin {gather*} \frac {{\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, C\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((4*A + 3*C)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (4*A + 3*C)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2
*((4*A + 3*C)*cos(d*x + c)^2 + 2*C)*sin(d*x + c))/(d*cos(d*x + c)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2),x)

[Out]

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

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Giac [A]
time = 0.46, size = 98, normalized size = 1.40 \begin {gather*} \frac {{\left (4 \, A + 3 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (4 \, A + 3 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (4 \, A \sin \left (d x + c\right )^{3} + 3 \, C \sin \left (d x + c\right )^{3} - 4 \, A \sin \left (d x + c\right ) - 5 \, C \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((4*A + 3*C)*log(abs(sin(d*x + c) + 1)) - (4*A + 3*C)*log(abs(sin(d*x + c) - 1)) - 2*(4*A*sin(d*x + c)^3
+ 3*C*sin(d*x + c)^3 - 4*A*sin(d*x + c) - 5*C*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d

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Mupad [B]
time = 2.45, size = 77, normalized size = 1.10 \begin {gather*} \frac {\sin \left (c+d\,x\right )\,\left (\frac {A}{2}+\frac {5\,C}{8}\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {A}{2}+\frac {3\,C}{8}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {A}{2}+\frac {3\,C}{8}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(c + d*x)*(A/2 + (5*C)/8) - sin(c + d*x)^3*(A/2 + (3*C)/8))/(d*(sin(c + d*x)^4 - 2*sin(c + d*x)^2 + 1)) +
(atanh(sin(c + d*x))*(A/2 + (3*C)/8))/d

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