3.288 \(\int \frac{(a B+b B \cos (c+d x)) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=28 \[ \frac{B \tan ^3(c+d x)}{3 d}+\frac{B \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0155774, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {21, 3767} \[ \frac{B \tan ^3(c+d x)}{3 d}+\frac{B \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a*B + b*B*Cos[c + d*x])*Sec[c + d*x]^4)/(a + b*Cos[c + d*x]),x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3767

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 \frac{(a B+b B \cos (c+d x)) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx &=B \int \sec ^4(c+d x) \, dx\\ &=-\frac{B \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac{B \tan (c+d x)}{d}+\frac{B \tan ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.0371981, size = 24, normalized size = 0.86 \[ \frac{B \left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a*B + b*B*Cos[c + d*x])*Sec[c + d*x]^4)/(a + b*Cos[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.063, size = 25, normalized size = 0.9 \begin{align*} -{\frac{B\tan \left ( dx+c \right ) }{d} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*B+b*B*cos(d*x+c))*sec(d*x+c)^4/(a+b*cos(d*x+c)),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36135, size = 84, normalized size = 3. \begin{align*} \frac{{\left (2 \, B \cos \left (d x + c\right )^{2} + B\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 114.743, size = 42, normalized size = 1.5 \begin{align*} \begin{cases} \frac{B \left (\frac{\tan ^{3}{\left (c + d x \right )}}{3} + \tan{\left (c + d x \right )}\right )}{d} & \text{for}\: d \neq 0 \\\frac{x \left (B a + B b \cos{\left (c \right )}\right ) \sec ^{4}{\left (c \right )}}{a + b \cos{\left (c \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)**4/(a+b*cos(d*x+c)),x)

[Out]

Piecewise((B*(tan(c + d*x)**3/3 + tan(c + d*x))/d, Ne(d, 0)), (x*(B*a + B*b*cos(c))*sec(c)**4/(a + b*cos(c)),
True))

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Giac [A]  time = 1.56388, size = 34, normalized size = 1.21 \begin{align*} \frac{B \tan \left (d x + c\right )^{3} + 3 \, B \tan \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

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