3.539 \(\int \sec (c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=91 \[ \frac{a \left (2 a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b \sec (c+d x) \left (4 \left (4 a^2-b^2\right )+5 a b \tan (c+d x)\right )}{6 d}+\frac{b \sec (c+d x) (a+b \tan (c+d x))^2}{3 d} \]

[Out]

(a*(2*a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + (b*Sec[c + d*x]*(a + b*Tan[c + d*x])^2)/(3*d) + (b*Sec[c + d
*x]*(4*(4*a^2 - b^2) + 5*a*b*Tan[c + d*x]))/(6*d)

________________________________________________________________________________________

Rubi [A]  time = 0.0851151, antiderivative size = 109, normalized size of antiderivative = 1.2, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3512, 743, 780, 215} \[ \frac{b \sec (c+d x) \left (4 \left (4 a^2-b^2\right )+5 a b \tan (c+d x)\right )}{6 d}+\frac{a \left (2 a^2-3 b^2\right ) \sec (c+d x) \sinh ^{-1}(\tan (c+d x))}{2 d \sqrt{\sec ^2(c+d x)}}+\frac{b \sec (c+d x) (a+b \tan (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + b*Tan[c + d*x])^3,x]

[Out]

(a*(2*a^2 - 3*b^2)*ArcSinh[Tan[c + d*x]]*Sec[c + d*x])/(2*d*Sqrt[Sec[c + d*x]^2]) + (b*Sec[c + d*x]*(a + b*Tan
[c + d*x])^2)/(3*d) + (b*Sec[c + d*x]*(4*(4*a^2 - b^2) + 5*a*b*Tan[c + d*x]))/(6*d)

Rule 3512

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(d^(2
*IntPart[m/2])*(d*Sec[e + f*x])^(2*FracPart[m/2]))/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 215

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

Rubi steps

\begin{align*} \int \sec (c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac{\sec (c+d x) \operatorname{Subst}\left (\int \frac{(a+x)^3}{\sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{b d \sqrt{\sec ^2(c+d x)}}\\ &=\frac{b \sec (c+d x) (a+b \tan (c+d x))^2}{3 d}+\frac{(b \sec (c+d x)) \operatorname{Subst}\left (\int \frac{(a+x) \left (-2+\frac{3 a^2}{b^2}+\frac{5 a x}{b^2}\right )}{\sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{3 d \sqrt{\sec ^2(c+d x)}}\\ &=\frac{b \sec (c+d x) (a+b \tan (c+d x))^2}{3 d}+\frac{b \sec (c+d x) \left (4 \left (4 a^2-b^2\right )+5 a b \tan (c+d x)\right )}{6 d}-\frac{\left (a \left (3-\frac{2 a^2}{b^2}\right ) b \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{2 d \sqrt{\sec ^2(c+d x)}}\\ &=\frac{a \left (2 a^2-3 b^2\right ) \sinh ^{-1}(\tan (c+d x)) \sec (c+d x)}{2 d \sqrt{\sec ^2(c+d x)}}+\frac{b \sec (c+d x) (a+b \tan (c+d x))^2}{3 d}+\frac{b \sec (c+d x) \left (4 \left (4 a^2-b^2\right )+5 a b \tan (c+d x)\right )}{6 d}\\ \end{align*}

Mathematica [B]  time = 1.55495, size = 293, normalized size = 3.22 \[ \frac{-6 a \left (2 a^2-3 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 b \sin ^2\left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (\left (18 a^2-5 b^2\right ) \cos (2 (c+d x))+18 a^2+2 b^2 \cos (c+d x)-b^2\right )+36 a^2 b+12 a^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{9 a b^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{9 a b^2}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-18 a b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{b^3}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{b^3}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-10 b^3}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + b*Tan[c + d*x])^3,x]

[Out]

(36*a^2*b - 10*b^3 - 6*a*(2*a^2 - 3*b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*a^3*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]] - 18*a*b^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (9*a*b^2)/(Cos[(c + d*x)/2] - Sin[
(c + d*x)/2])^2 + b^3/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + 2*b*(18*a^2 - b^2 + 2*b^2*Cos[c + d*x] + (18*a
^2 - 5*b^2)*Cos[2*(c + d*x)])*Sec[c + d*x]^3*Sin[(c + d*x)/2]^2 - (9*a*b^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])^2 + b^3/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(12*d)

________________________________________________________________________________________

Maple [B]  time = 0.027, size = 187, normalized size = 2.1 \begin{align*}{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d\cos \left ( dx+c \right ) }}-{\frac{{b}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{2\,{b}^{3}\cos \left ( dx+c \right ) }{3\,d}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,a{b}^{2}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,a{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{b{a}^{2}}{d\cos \left ( dx+c \right ) }}+{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/d*b^3*sin(d*x+c)^4/cos(d*x+c)^3-1/3/d*b^3*sin(d*x+c)^4/cos(d*x+c)-1/3/d*b^3*cos(d*x+c)*sin(d*x+c)^2-2/3/d*
b^3*cos(d*x+c)+3/2/d*a*b^2*sin(d*x+c)^3/cos(d*x+c)^2+3/2/d*a*b^2*sin(d*x+c)-3/2/d*a*b^2*ln(sec(d*x+c)+tan(d*x+
c))+3/d*b*a^2/cos(d*x+c)+1/d*a^3*ln(sec(d*x+c)+tan(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 1.12219, size = 150, normalized size = 1.65 \begin{align*} -\frac{9 \, a b^{2}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - \frac{36 \, a^{2} b}{\cos \left (d x + c\right )} + \frac{4 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} b^{3}}{\cos \left (d x + c\right )^{3}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(9*a*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 12*a^3*
log(sec(d*x + c) + tan(d*x + c)) - 36*a^2*b/cos(d*x + c) + 4*(3*cos(d*x + c)^2 - 1)*b^3/cos(d*x + c)^3)/d

________________________________________________________________________________________

Fricas [A]  time = 1.94229, size = 304, normalized size = 3.34 \begin{align*} \frac{3 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 18 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, b^{3} + 12 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(2*a^3 - 3*a*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*a^3 - 3*a*b^2)*cos(d*x + c)^3*log(-sin(d
*x + c) + 1) + 18*a*b^2*cos(d*x + c)*sin(d*x + c) + 4*b^3 + 12*(3*a^2*b - b^3)*cos(d*x + c)^2)/(d*cos(d*x + c)
^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{3} \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.83717, size = 231, normalized size = 2.54 \begin{align*} \frac{3 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (9 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 18 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 36 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 18 \, a^{2} b + 4 \, b^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(2*a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(9*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 18*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 36*a^2*b*tan(1/2*d*x + 1/2*c)^2
- 12*b^3*tan(1/2*d*x + 1/2*c)^2 - 9*a*b^2*tan(1/2*d*x + 1/2*c) - 18*a^2*b + 4*b^3)/(tan(1/2*d*x + 1/2*c)^2 -
1)^3)/d