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]

1/4*(a^2+b^2)^2*(a+b*tan(d*x+c))^4/b^5/d-4/5*a*(a^2+b^2)*(a+b*tan(d*x+c))^5/b^5/d+1/3*(3*a^2+b^2)*(a+b*tan(d*x
+c))^6/b^5/d-4/7*a*(a+b*tan(d*x+c))^7/b^5/d+1/8*(a+b*tan(d*x+c))^8/b^5/d

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Rubi [A]  time = 0.13, antiderivative size = 138, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.57, 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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fricas [A]  time = 0.78, size = 128, normalized size = 0.93 \[ \frac {105 \, b^{3} + 140 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{7} + 4 \, {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 45 \, a b^{2} \cos \left (d x + c\right ) + 3 \, {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{840 \, d \cos \left (d x + c\right )^{8}} \]

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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giac [A]  time = 2.72, size = 166, normalized size = 1.20 \[ \frac {105 \, b^{3} \tan \left (d x + c\right )^{8} + 360 \, a b^{2} \tan \left (d x + c\right )^{7} + 420 \, a^{2} b \tan \left (d x + c\right )^{6} + 280 \, b^{3} \tan \left (d x + c\right )^{6} + 168 \, a^{3} \tan \left (d x + c\right )^{5} + 1008 \, a b^{2} \tan \left (d x + c\right )^{5} + 1260 \, a^{2} b \tan \left (d x + c\right )^{4} + 210 \, b^{3} \tan \left (d x + c\right )^{4} + 560 \, a^{3} \tan \left (d x + c\right )^{3} + 840 \, a b^{2} \tan \left (d x + c\right )^{3} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 840 \, a^{3} \tan \left (d x + c\right )}{840 \, d} \]

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

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maple [A]  time = 0.46, size = 173, normalized size = 1.25 \[ \frac {-a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {a^{2} b}{2 \cos \left (d x +c \right )^{6}}+3 b^{2} a \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}\right )}{d} \]

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*(-a^3*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+1/2*a^2*b/cos(d*x+c)^6+3*b^2*a*(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)+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))

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maxima [A]  time = 0.33, size = 142, normalized size = 1.03 \[ \frac {105 \, b^{3} \tan \left (d x + c\right )^{8} + 360 \, a b^{2} \tan \left (d x + c\right )^{7} + 140 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \tan \left (d x + c\right )^{6} + 168 \, {\left (a^{3} + 6 \, a b^{2}\right )} \tan \left (d x + c\right )^{5} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 210 \, {\left (6 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{4} + 840 \, a^{3} \tan \left (d x + c\right ) + 280 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{3}}{840 \, d} \]

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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mupad [B]  time = 3.60, size = 139, normalized size = 1.01 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {2\,a^3}{3}+a\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {a^3}{5}+\frac {6\,a\,b^2}{5}\right )+{\mathrm {tan}\left (c+d\,x\right )}^6\,\left (\frac {a^2\,b}{2}+\frac {b^3}{3}\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {3\,a^2\,b}{2}+\frac {b^3}{4}\right )+a^3\,\mathrm {tan}\left (c+d\,x\right )+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^8}{8}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {3\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^7}{7}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c + d*x)^3*(a*b^2 + (2*a^3)/3) + tan(c + d*x)^5*((6*a*b^2)/5 + a^3/5) + tan(c + d*x)^6*((a^2*b)/2 + b^3/3
) + tan(c + d*x)^4*((3*a^2*b)/2 + b^3/4) + a^3*tan(c + d*x) + (b^3*tan(c + d*x)^8)/8 + (3*a^2*b*tan(c + d*x)^2
)/2 + (3*a*b^2*tan(c + d*x)^7)/7)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sec ^{6}{\left (c + d x \right )}\, dx \]

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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