3.532 \(\int \sec ^6(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=138 \[ \frac{\left (3 a^2+b^2\right ) (a+b \tan (c+d x))^6}{3 b^5 d}-\frac{4 a \left (a^2+b^2\right ) (a+b \tan (c+d x))^5}{5 b^5 d}+\frac{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^4}{4 b^5 d}+\frac{(a+b \tan (c+d x))^8}{8 b^5 d}-\frac{4 a (a+b \tan (c+d x))^7}{7 b^5 d} \]

[Out]

((a^2 + b^2)^2*(a + b*Tan[c + d*x])^4)/(4*b^5*d) - (4*a*(a^2 + b^2)*(a + b*Tan[c + d*x])^5)/(5*b^5*d) + ((3*a^
2 + b^2)*(a + b*Tan[c + d*x])^6)/(3*b^5*d) - (4*a*(a + b*Tan[c + d*x])^7)/(7*b^5*d) + (a + b*Tan[c + d*x])^8/(
8*b^5*d)

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Rubi [A]  time = 0.125019, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3506, 697} \[ \frac{\left (3 a^2+b^2\right ) (a+b \tan (c+d x))^6}{3 b^5 d}-\frac{4 a \left (a^2+b^2\right ) (a+b \tan (c+d x))^5}{5 b^5 d}+\frac{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^4}{4 b^5 d}+\frac{(a+b \tan (c+d x))^8}{8 b^5 d}-\frac{4 a (a+b \tan (c+d x))^7}{7 b^5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + b*Tan[c + d*x])^3,x]

[Out]

((a^2 + b^2)^2*(a + b*Tan[c + d*x])^4)/(4*b^5*d) - (4*a*(a^2 + b^2)*(a + b*Tan[c + d*x])^5)/(5*b^5*d) + ((3*a^
2 + b^2)*(a + b*Tan[c + d*x])^6)/(3*b^5*d) - (4*a*(a + b*Tan[c + d*x])^7)/(7*b^5*d) + (a + b*Tan[c + d*x])^8/(
8*b^5*d)

Rule 3506

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

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \sec ^6(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac{\operatorname{Subst}\left (\int (a+x)^3 \left (1+\frac{x^2}{b^2}\right )^2 \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{\left (a^2+b^2\right )^2 (a+x)^3}{b^4}-\frac{4 a \left (a^2+b^2\right ) (a+x)^4}{b^4}+\frac{2 \left (3 a^2+b^2\right ) (a+x)^5}{b^4}-\frac{4 a (a+x)^6}{b^4}+\frac{(a+x)^7}{b^4}\right ) \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^4}{4 b^5 d}-\frac{4 a \left (a^2+b^2\right ) (a+b \tan (c+d x))^5}{5 b^5 d}+\frac{\left (3 a^2+b^2\right ) (a+b \tan (c+d x))^6}{3 b^5 d}-\frac{4 a (a+b \tan (c+d x))^7}{7 b^5 d}+\frac{(a+b \tan (c+d x))^8}{8 b^5 d}\\ \end{align*}

Mathematica [A]  time = 0.570698, size = 115, normalized size = 0.83 \[ \frac{\frac{1}{3} \left (3 a^2+b^2\right ) (a+b \tan (c+d x))^6-\frac{4}{5} a \left (a^2+b^2\right ) (a+b \tan (c+d x))^5+\frac{1}{4} \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^4+\frac{1}{8} (a+b \tan (c+d x))^8-\frac{4}{7} a (a+b \tan (c+d x))^7}{b^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + b*Tan[c + d*x])^3,x]

[Out]

(((a^2 + b^2)^2*(a + b*Tan[c + d*x])^4)/4 - (4*a*(a^2 + b^2)*(a + b*Tan[c + d*x])^5)/5 + ((3*a^2 + b^2)*(a + b
*Tan[c + d*x])^6)/3 - (4*a*(a + b*Tan[c + d*x])^7)/7 + (a + b*Tan[c + d*x])^8/8)/(b^5*d)

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Maple [A]  time = 0.069, size = 173, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({b}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{12\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{24\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \right ) +3\,a{b}^{2} \left ( 1/7\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{b{a}^{2}}{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}-{a}^{3} \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+b*tan(d*x+c))^3,x)

[Out]

1/d*(b^3*(1/8*sin(d*x+c)^4/cos(d*x+c)^8+1/12*sin(d*x+c)^4/cos(d*x+c)^6+1/24*sin(d*x+c)^4/cos(d*x+c)^4)+3*a*b^2
*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3)+1/2*b*a^2/cos(
d*x+c)^6-a^3*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.16301, size = 192, normalized size = 1.39 \begin{align*} \frac{105 \, b^{3} \tan \left (d x + c\right )^{8} + 360 \, a b^{2} \tan \left (d x + c\right )^{7} + 140 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \tan \left (d x + c\right )^{6} + 168 \,{\left (a^{3} + 6 \, a b^{2}\right )} \tan \left (d x + c\right )^{5} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 210 \,{\left (6 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{4} + 840 \, a^{3} \tan \left (d x + c\right ) + 280 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{3}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/840*(105*b^3*tan(d*x + c)^8 + 360*a*b^2*tan(d*x + c)^7 + 140*(3*a^2*b + 2*b^3)*tan(d*x + c)^6 + 168*(a^3 + 6
*a*b^2)*tan(d*x + c)^5 + 1260*a^2*b*tan(d*x + c)^2 + 210*(6*a^2*b + b^3)*tan(d*x + c)^4 + 840*a^3*tan(d*x + c)
+ 280*(2*a^3 + 3*a*b^2)*tan(d*x + c)^3)/d

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Fricas [A]  time = 2.12018, size = 304, normalized size = 2.2 \begin{align*} \frac{105 \, b^{3} + 140 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (8 \,{\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{7} + 4 \,{\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 45 \, a b^{2} \cos \left (d x + c\right ) + 3 \,{\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{840 \, d \cos \left (d x + c\right )^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/840*(105*b^3 + 140*(3*a^2*b - b^3)*cos(d*x + c)^2 + 8*(8*(7*a^3 - 3*a*b^2)*cos(d*x + c)^7 + 4*(7*a^3 - 3*a*b
^2)*cos(d*x + c)^5 + 45*a*b^2*cos(d*x + c) + 3*(7*a^3 - 3*a*b^2)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)
^8)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{3} \sec ^{6}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*sec(c + d*x)**6, x)

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Giac [A]  time = 1.76878, size = 224, normalized size = 1.62 \begin{align*} \frac{105 \, b^{3} \tan \left (d x + c\right )^{8} + 360 \, a b^{2} \tan \left (d x + c\right )^{7} + 420 \, a^{2} b \tan \left (d x + c\right )^{6} + 280 \, b^{3} \tan \left (d x + c\right )^{6} + 168 \, a^{3} \tan \left (d x + c\right )^{5} + 1008 \, a b^{2} \tan \left (d x + c\right )^{5} + 1260 \, a^{2} b \tan \left (d x + c\right )^{4} + 210 \, b^{3} \tan \left (d x + c\right )^{4} + 560 \, a^{3} \tan \left (d x + c\right )^{3} + 840 \, a b^{2} \tan \left (d x + c\right )^{3} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 840 \, a^{3} \tan \left (d x + c\right )}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/840*(105*b^3*tan(d*x + c)^8 + 360*a*b^2*tan(d*x + c)^7 + 420*a^2*b*tan(d*x + c)^6 + 280*b^3*tan(d*x + c)^6 +
168*a^3*tan(d*x + c)^5 + 1008*a*b^2*tan(d*x + c)^5 + 1260*a^2*b*tan(d*x + c)^4 + 210*b^3*tan(d*x + c)^4 + 560
*a^3*tan(d*x + c)^3 + 840*a*b^2*tan(d*x + c)^3 + 1260*a^2*b*tan(d*x + c)^2 + 840*a^3*tan(d*x + c))/d