3.567 \(\int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=121 \[ -\frac {\left (a^2+b^2\right )^2}{2 b^5 d (a+b \tan (c+d x))^2}+\frac {4 a \left (a^2+b^2\right )}{b^5 d (a+b \tan (c+d x))}+\frac {2 \left (3 a^2+b^2\right ) \log (a+b \tan (c+d x))}{b^5 d}-\frac {3 a \tan (c+d x)}{b^4 d}+\frac {\tan ^2(c+d x)}{2 b^3 d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 121, 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} \[ -\frac {\left (a^2+b^2\right )^2}{2 b^5 d (a+b \tan (c+d x))^2}+\frac {4 a \left (a^2+b^2\right )}{b^5 d (a+b \tan (c+d x))}+\frac {2 \left (3 a^2+b^2\right ) \log (a+b \tan (c+d x))}{b^5 d}-\frac {3 a \tan (c+d x)}{b^4 d}+\frac {\tan ^2(c+d x)}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 3.48, size = 140, normalized size = 1.16 \[ \frac {-2 a \left (-\frac {a^2+b^2}{a+b \tan (c+d x)}-2 a \log (a+b \tan (c+d x))+b \tan (c+d x)\right )+2 \left (a^2+b^2\right ) \left (\frac {3 a^2+4 a b \tan (c+d x)-b^2}{2 (a+b \tan (c+d x))^2}+\log (a+b \tan (c+d x))\right )+\frac {b^4 \sec ^4(c+d x)}{2 (a+b \tan (c+d x))^2}}{b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.81, size = 354, normalized size = 2.93 \[ \frac {24 \, a^{2} b^{2} \cos \left (d x + c\right )^{4} + b^{4} - 2 \, {\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left ({\left (3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 2 \, {\left ({\left (3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) - 4 \, {\left (a b^{3} \cos \left (d x + c\right ) + 3 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (2 \, a b^{6} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + b^{7} d \cos \left (d x + c\right )^{2} + {\left (a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{4}\right )}} \]

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]

1/2*(24*a^2*b^2*cos(d*x + c)^4 + b^4 - 2*(9*a^2*b^2 + b^4)*cos(d*x + c)^2 + 2*((3*a^4 - 2*a^2*b^2 - b^4)*cos(d
*x + c)^4 + 2*(3*a^3*b + a*b^3)*cos(d*x + c)^3*sin(d*x + c) + (3*a^2*b^2 + b^4)*cos(d*x + c)^2)*log(2*a*b*cos(
d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 2*((3*a^4 - 2*a^2*b^2 - b^4)*cos(d*x + c)^4 + 2*(3
*a^3*b + a*b^3)*cos(d*x + c)^3*sin(d*x + c) + (3*a^2*b^2 + b^4)*cos(d*x + c)^2)*log(cos(d*x + c)^2) - 4*(a*b^3
*cos(d*x + c) + 3*(a^3*b - a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/(2*a*b^6*d*cos(d*x + c)^3*sin(d*x + c) + b^7*d
*cos(d*x + c)^2 + (a^2*b^5 - b^7)*d*cos(d*x + c)^4)

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giac [A]  time = 2.04, size = 140, normalized size = 1.16 \[ \frac {\frac {4 \, {\left (3 \, a^{2} + b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{5}} + \frac {b^{3} \tan \left (d x + c\right )^{2} - 6 \, a b^{2} \tan \left (d x + c\right )}{b^{6}} - \frac {18 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 6 \, b^{4} \tan \left (d x + c\right )^{2} + 28 \, a^{3} b \tan \left (d x + c\right ) + 4 \, a b^{3} \tan \left (d x + c\right ) + 11 \, a^{4} + b^{4}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} b^{5}}}{2 \, d} \]

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]

1/2*(4*(3*a^2 + b^2)*log(abs(b*tan(d*x + c) + a))/b^5 + (b^3*tan(d*x + c)^2 - 6*a*b^2*tan(d*x + c))/b^6 - (18*
a^2*b^2*tan(d*x + c)^2 + 6*b^4*tan(d*x + c)^2 + 28*a^3*b*tan(d*x + c) + 4*a*b^3*tan(d*x + c) + 11*a^4 + b^4)/(
(b*tan(d*x + c) + a)^2*b^5))/d

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maple [A]  time = 0.50, size = 184, normalized size = 1.52 \[ \frac {\tan ^{2}\left (d x +c \right )}{2 b^{3} d}-\frac {3 a \tan \left (d x +c \right )}{b^{4} d}+\frac {6 \ln \left (a +b \tan \left (d x +c \right )\right ) a^{2}}{d \,b^{5}}+\frac {2 \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3} d}+\frac {4 a^{3}}{d \,b^{5} \left (a +b \tan \left (d x +c \right )\right )}+\frac {4 a}{b^{3} d \left (a +b \tan \left (d x +c \right )\right )}-\frac {a^{4}}{2 d \,b^{5} \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {a^{2}}{d \,b^{3} \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {1}{2 b d \left (a +b \tan \left (d x +c \right )\right )^{2}} \]

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]

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

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maxima [A]  time = 0.44, size = 128, normalized size = 1.06 \[ \frac {\frac {7 \, a^{4} + 6 \, a^{2} b^{2} - b^{4} + 8 \, {\left (a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )}{b^{7} \tan \left (d x + c\right )^{2} + 2 \, a b^{6} \tan \left (d x + c\right ) + a^{2} b^{5}} + \frac {b \tan \left (d x + c\right )^{2} - 6 \, a \tan \left (d x + c\right )}{b^{4}} + \frac {4 \, {\left (3 \, a^{2} + b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{5}}}{2 \, d} \]

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]

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

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mupad [B]  time = 3.73, size = 143, normalized size = 1.18 \[ \frac {\frac {7\,a^4+6\,a^2\,b^2-b^4}{2\,b}+\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3+4\,a\,b^2\right )}{d\,\left (a^2\,b^4+2\,a\,b^5\,\mathrm {tan}\left (c+d\,x\right )+b^6\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,b^3\,d}-\frac {3\,a\,\mathrm {tan}\left (c+d\,x\right )}{b^4\,d}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (6\,a^2+2\,b^2\right )}{b^5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^6*(a + b*tan(c + d*x))^3),x)

[Out]

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

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

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

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