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

Optimal. Leaf size=65 \[ \frac {(5 A+4 C) \tan (c+d x)}{5 d}+\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 A+4 C) \tan ^3(c+d x)}{15 d} \]

[Out]

1/5*(5*A+4*C)*tan(d*x+c)/d+1/5*C*sec(d*x+c)^4*tan(d*x+c)/d+1/15*(5*A+4*C)*tan(d*x+c)^3/d

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Rubi [A]
time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {4131, 3852} \begin {gather*} \frac {(5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {(5 A+4 C) \tan (c+d x)}{5 d}+\frac {C \tan (c+d x) \sec ^4(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(A + C*Sec[c + d*x]^2),x]

[Out]

((5*A + 4*C)*Tan[c + d*x])/(5*d) + (C*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((5*A + 4*C)*Tan[c + d*x]^3)/(15*d)

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 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 ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} (5 A+4 C) \int \sec ^4(c+d x) \, dx\\ &=\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {(5 A+4 C) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {(5 A+4 C) \tan (c+d x)}{5 d}+\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 A+4 C) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 61, normalized size = 0.94 \begin {gather*} \frac {A \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d}+\frac {C \left (\tan (c+d x)+\frac {2}{3} \tan ^3(c+d x)+\frac {1}{5} \tan ^5(c+d x)\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(A*(Tan[c + d*x] + Tan[c + d*x]^3/3))/d + (C*(Tan[c + d*x] + (2*Tan[c + d*x]^3)/3 + Tan[c + d*x]^5/5))/d

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Maple [A]
time = 0.41, size = 58, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {-A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-C \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(58\)
default	\(\frac {-A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-C \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(58\)
risch	\(\frac {4 i \left (15 A \,{\mathrm e}^{6 i \left (d x +c \right )}+35 A \,{\mathrm e}^{4 i \left (d x +c \right )}+40 C \,{\mathrm e}^{4 i \left (d x +c \right )}+25 A \,{\mathrm e}^{2 i \left (d x +c \right )}+20 C \,{\mathrm e}^{2 i \left (d x +c \right )}+5 A +4 C \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\)	\(87\)
norman	\(\frac {-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (A +C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 \left (2 A +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 \left (2 A +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 \left (25 A +29 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(-A*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)-C*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]
time = 0.28, size = 43, normalized size = 0.66 \begin {gather*} \frac {3 \, C \tan \left (d x + c\right )^{5} + 5 \, {\left (A + 2 \, C\right )} \tan \left (d x + c\right )^{3} + 15 \, {\left (A + C\right )} \tan \left (d x + c\right )}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/15*(3*C*tan(d*x + c)^5 + 5*(A + 2*C)*tan(d*x + c)^3 + 15*(A + C)*tan(d*x + c))/d

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Fricas [A]
time = 2.31, size = 56, normalized size = 0.86 \begin {gather*} \frac {{\left (2 \, {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/15*(2*(5*A + 4*C)*cos(d*x + c)^4 + (5*A + 4*C)*cos(d*x + c)^2 + 3*C)*sin(d*x + c)/(d*cos(d*x + c)^5)

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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 ^{4}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**4, x)

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Giac [A]
time = 0.44, size = 57, normalized size = 0.88 \begin {gather*} \frac {3 \, C \tan \left (d x + c\right )^{5} + 5 \, A \tan \left (d x + c\right )^{3} + 10 \, C \tan \left (d x + c\right )^{3} + 15 \, A \tan \left (d x + c\right ) + 15 \, C \tan \left (d x + c\right )}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/15*(3*C*tan(d*x + c)^5 + 5*A*tan(d*x + c)^3 + 10*C*tan(d*x + c)^3 + 15*A*tan(d*x + c) + 15*C*tan(d*x + c))/d

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Mupad [B]
time = 2.32, size = 42, normalized size = 0.65 \begin {gather*} \frac {\frac {C\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+\left (\frac {A}{3}+\frac {2\,C}{3}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (A+C\right )\,\mathrm {tan}\left (c+d\,x\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)^4,x)

[Out]

((C*tan(c + d*x)^5)/5 + tan(c + d*x)*(A + C) + tan(c + d*x)^3*(A/3 + (2*C)/3))/d

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