3.1.13 \(\int (A+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [13]

Optimal. Leaf size=43 \[ \frac {(2 A+3 C) \tan (c+d x)}{3 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 43, 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 = {3091, 3852, 8} \begin {gather*} \frac {(2 A+3 C) \tan (c+d x)}{3 d}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 3091

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A*Cos[e +
f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

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]

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 36, normalized size = 0.84 \begin {gather*} \frac {C \tan (c+d x)}{d}+\frac {A \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.18, size = 35, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {-A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \tan \left (d x +c \right )}{d}\)	\(35\)
default	\(\frac {-A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \tan \left (d x +c \right )}{d}\)	\(35\)
risch	\(\frac {2 i \left (3 C \,{\mathrm e}^{4 i \left (d x +c \right )}+6 A \,{\mathrm e}^{2 i \left (d x +c \right )}+6 C \,{\mathrm e}^{2 i \left (d x +c \right )}+2 A +3 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(63\)
norman	\(\frac {-\frac {8 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 A \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 \left (A -3 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (A +C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 27, normalized size = 0.63 \begin {gather*} \frac {A \tan \left (d x + c\right )^{3} + 3 \, {\left (A + C\right )} \tan \left (d x + c\right )}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 37, normalized size = 0.86 \begin {gather*} \frac {{\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + A\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (A + C \cos ^{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((A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

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

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Giac [A]
time = 0.40, size = 34, normalized size = 0.79 \begin {gather*} \frac {A \tan \left (d x + c\right )^{3} + 3 \, A \tan \left (d x + c\right ) + 3 \, C \tan \left (d x + c\right )}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.64, size = 28, normalized size = 0.65 \begin {gather*} \frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A+C\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]

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

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