3.1.5 \(\int \sec ^2(c+d x) (A+C \sec ^2(c+d x)) \, dx\) [5]

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

[Out]

1/3*(3*A+2*C)*tan(d*x+c)/d+1/3*C*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 = {4131, 3852, 8} \begin {gather*} \frac {(3 A+2 C) \tan (c+d x)}{3 d}+\frac {C \tan (c+d x) \sec ^2(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.07, size = 36, normalized size = 0.84 \begin {gather*} \frac {A \tan (c+d x)}{d}+\frac {C \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.32, size = 35, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {A \tan \left (d x +c \right )-C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(35\)
default	\(\frac {A \tan \left (d x +c \right )-C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(35\)
risch	\(\frac {2 i \left (3 A \,{\mathrm e}^{4 i \left (d x +c \right )}+6 A \,{\mathrm e}^{2 i \left (d x +c \right )}+6 C \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A +2 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(63\)
norman	\(\frac {-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (A +C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (3 A +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 34, normalized size = 0.79 \begin {gather*} \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C + 3 \, A \tan \left (d x + c\right )}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.50, size = 37, normalized size = 0.86 \begin {gather*} \frac {{\left ({\left (3 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2} + C\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(sec(d*x+c)^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/3*((3*A + 2*C)*cos(d*x + c)^2 + C)*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 \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.44, size = 34, normalized size = 0.79 \begin {gather*} \frac {C \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(sec(d*x+c)^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/3*(C*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 = 2.37, size = 28, normalized size = 0.65 \begin {gather*} \frac {C\,{\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)^2,x)

[Out]

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

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