3.178 \(\int (a+b \sec ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=73 \[ a^3 x+\frac {b \left (3 a^2+3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b^2 (3 a+2 b) \tan ^3(c+d x)}{3 d}+\frac {b^3 \tan ^5(c+d x)}{5 d} \]

[Out]

a^3*x+b*(3*a^2+3*a*b+b^2)*tan(d*x+c)/d+1/3*b^2*(3*a+2*b)*tan(d*x+c)^3/d+1/5*b^3*tan(d*x+c)^5/d

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Rubi [A]  time = 0.04, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4128, 390, 203} \[ \frac {b \left (3 a^2+3 a b+b^2\right ) \tan (c+d x)}{d}+a^3 x+\frac {b^2 (3 a+2 b) \tan ^3(c+d x)}{3 d}+\frac {b^3 \tan ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x]^2)^3,x]

[Out]

a^3*x + (b*(3*a^2 + 3*a*b + b^2)*Tan[c + d*x])/d + (b^2*(3*a + 2*b)*Tan[c + d*x]^3)/(3*d) + (b^3*Tan[c + d*x]^
5)/(5*d)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \left (a+b \sec ^2(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b+b x^2\right )^3}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (b \left (3 a^2+3 a b+b^2\right )+b^2 (3 a+2 b) x^2+b^3 x^4+\frac {a^3}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {b \left (3 a^2+3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b^2 (3 a+2 b) \tan ^3(c+d x)}{3 d}+\frac {b^3 \tan ^5(c+d x)}{5 d}+\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=a^3 x+\frac {b \left (3 a^2+3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b^2 (3 a+2 b) \tan ^3(c+d x)}{3 d}+\frac {b^3 \tan ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [B]  time = 1.02, size = 268, normalized size = 3.67 \[ \frac {\sec (c) \sec ^5(c+d x) \left (150 a^3 d x \cos (2 c+d x)+75 a^3 d x \cos (2 c+3 d x)+75 a^3 d x \cos (4 c+3 d x)+15 a^3 d x \cos (4 c+5 d x)+15 a^3 d x \cos (6 c+5 d x)+150 a^3 d x \cos (d x)-360 a^2 b \sin (2 c+d x)+360 a^2 b \sin (2 c+3 d x)-90 a^2 b \sin (4 c+3 d x)+90 a^2 b \sin (4 c+5 d x)+540 a^2 b \sin (d x)-180 a b^2 \sin (2 c+d x)+300 a b^2 \sin (2 c+3 d x)+60 a b^2 \sin (4 c+5 d x)+420 a b^2 \sin (d x)+80 b^3 \sin (2 c+3 d x)+16 b^3 \sin (4 c+5 d x)+160 b^3 \sin (d x)\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x]^2)^3,x]

[Out]

(Sec[c]*Sec[c + d*x]^5*(150*a^3*d*x*Cos[d*x] + 150*a^3*d*x*Cos[2*c + d*x] + 75*a^3*d*x*Cos[2*c + 3*d*x] + 75*a
^3*d*x*Cos[4*c + 3*d*x] + 15*a^3*d*x*Cos[4*c + 5*d*x] + 15*a^3*d*x*Cos[6*c + 5*d*x] + 540*a^2*b*Sin[d*x] + 420
*a*b^2*Sin[d*x] + 160*b^3*Sin[d*x] - 360*a^2*b*Sin[2*c + d*x] - 180*a*b^2*Sin[2*c + d*x] + 360*a^2*b*Sin[2*c +
3*d*x] + 300*a*b^2*Sin[2*c + 3*d*x] + 80*b^3*Sin[2*c + 3*d*x] - 90*a^2*b*Sin[4*c + 3*d*x] + 90*a^2*b*Sin[4*c
+ 5*d*x] + 60*a*b^2*Sin[4*c + 5*d*x] + 16*b^3*Sin[4*c + 5*d*x]))/(480*d)

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fricas [A]  time = 0.75, size = 90, normalized size = 1.23 \[ \frac {15 \, a^{3} d x \cos \left (d x + c\right )^{5} + {\left ({\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, b^{3} + {\left (15 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*a^3*d*x*cos(d*x + c)^5 + ((45*a^2*b + 30*a*b^2 + 8*b^3)*cos(d*x + c)^4 + 3*b^3 + (15*a*b^2 + 4*b^3)*c
os(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [A]  time = 0.33, size = 91, normalized size = 1.25 \[ \frac {3 \, b^{3} \tan \left (d x + c\right )^{5} + 15 \, a b^{2} \tan \left (d x + c\right )^{3} + 10 \, b^{3} \tan \left (d x + c\right )^{3} + 15 \, {\left (d x + c\right )} a^{3} + 45 \, a^{2} b \tan \left (d x + c\right ) + 45 \, a b^{2} \tan \left (d x + c\right ) + 15 \, b^{3} \tan \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^3*tan(d*x + c)^5 + 15*a*b^2*tan(d*x + c)^3 + 10*b^3*tan(d*x + c)^3 + 15*(d*x + c)*a^3 + 45*a^2*b*tan
(d*x + c) + 45*a*b^2*tan(d*x + c) + 15*b^3*tan(d*x + c))/d

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maple [A]  time = 1.25, size = 84, normalized size = 1.15 \[ \frac {a^{3} \left (d x +c \right )+3 a^{2} b \tan \left (d x +c \right )-3 b^{2} a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-b^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c)^2)^3,x)

[Out]

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

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maxima [A]  time = 0.35, size = 83, normalized size = 1.14 \[ a^{3} x + \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{2}}{d} + \frac {{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} b^{3}}{15 \, d} + \frac {3 \, a^{2} b \tan \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x + (tan(d*x + c)^3 + 3*tan(d*x + c))*a*b^2/d + 1/15*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x +
c))*b^3/d + 3*a^2*b*tan(d*x + c)/d

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mupad [B]  time = 4.51, size = 73, normalized size = 1.00 \[ \frac {\mathrm {tan}\left (c+d\,x\right )\,\left (3\,b\,{\left (a+b\right )}^2-3\,b^2\,\left (a+b\right )+b^3\right )+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (b^2\,\left (a+b\right )-\frac {b^3}{3}\right )+a^3\,d\,x}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x)^2)^3,x)

[Out]

(tan(c + d*x)*(3*b*(a + b)^2 - 3*b^2*(a + b) + b^3) + (b^3*tan(c + d*x)^5)/5 + tan(c + d*x)^3*(b^2*(a + b) - b
^3/3) + a^3*d*x)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec ^{2}{\left (c + d x \right )}\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c)**2)**3,x)

[Out]

Integral((a + b*sec(c + d*x)**2)**3, x)

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