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

Optimal. Leaf size=116 \[ \frac {\left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))}{b^5 d}-\frac {a \left (a^2+2 b^2\right ) \tan (c+d x)}{b^4 d}+\frac {\left (a^2+2 b^2\right ) \tan ^2(c+d x)}{2 b^3 d}-\frac {a \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^4(c+d x)}{4 b d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 116, 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} \[ \frac {\left (a^2+2 b^2\right ) \tan ^2(c+d x)}{2 b^3 d}-\frac {a \left (a^2+2 b^2\right ) \tan (c+d x)}{b^4 d}+\frac {\left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))}{b^5 d}-\frac {a \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^4(c+d x)}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 1.18, size = 99, normalized size = 0.85 \[ \frac {6 b^2 \left (a^2+b^2\right ) \tan ^2(c+d x)-12 a b \left (a^2+2 b^2\right ) \tan (c+d x)+12 \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))-4 a b^3 \tan ^3(c+d x)+3 b^4 \sec ^4(c+d x)}{12 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(a^2 + b^2)^2*Log[a + b*Tan[c + d*x]] + 3*b^4*Sec[c + d*x]^4 - 12*a*b*(a^2 + 2*b^2)*Tan[c + d*x] + 6*b^2*(
a^2 + b^2)*Tan[c + d*x]^2 - 4*a*b^3*Tan[c + d*x]^3)/(12*b^5*d)

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fricas [A]  time = 0.60, size = 183, normalized size = 1.58 \[ \frac {6 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 6 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\cos \left (d x + c\right )^{2}\right ) + 3 \, b^{4} + 6 \, {\left (a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (a b^{3} \cos \left (d x + c\right ) + {\left (3 \, a^{3} b + 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \, b^{5} d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*(a^4 + 2*a^2*b^2 + b^4)*cos(d*x + c)^4*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^
2 + b^2) - 6*(a^4 + 2*a^2*b^2 + b^4)*cos(d*x + c)^4*log(cos(d*x + c)^2) + 3*b^4 + 6*(a^2*b^2 + b^4)*cos(d*x +
c)^2 - 4*(a*b^3*cos(d*x + c) + (3*a^3*b + 5*a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/(b^5*d*cos(d*x + c)^4)

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giac [A]  time = 0.83, size = 120, normalized size = 1.03 \[ \frac {\frac {3 \, b^{3} \tan \left (d x + c\right )^{4} - 4 \, a b^{2} \tan \left (d x + c\right )^{3} + 6 \, a^{2} b \tan \left (d x + c\right )^{2} + 12 \, b^{3} \tan \left (d x + c\right )^{2} - 12 \, a^{3} \tan \left (d x + c\right ) - 24 \, a b^{2} \tan \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{5}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((3*b^3*tan(d*x + c)^4 - 4*a*b^2*tan(d*x + c)^3 + 6*a^2*b*tan(d*x + c)^2 + 12*b^3*tan(d*x + c)^2 - 12*a^3
*tan(d*x + c) - 24*a*b^2*tan(d*x + c))/b^4 + 12*(a^4 + 2*a^2*b^2 + b^4)*log(abs(b*tan(d*x + c) + a))/b^5)/d

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maple [A]  time = 0.46, size = 162, normalized size = 1.40 \[ \frac {\tan ^{4}\left (d x +c \right )}{4 b d}-\frac {a \left (\tan ^{3}\left (d x +c \right )\right )}{3 b^{2} d}+\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d \,b^{3}}+\frac {\tan ^{2}\left (d x +c \right )}{b d}-\frac {a^{3} \tan \left (d x +c \right )}{d \,b^{4}}-\frac {2 a \tan \left (d x +c \right )}{b^{2} d}+\frac {\ln \left (a +b \tan \left (d x +c \right )\right ) a^{4}}{d \,b^{5}}+\frac {2 \ln \left (a +b \tan \left (d x +c \right )\right ) a^{2}}{d \,b^{3}}+\frac {\ln \left (a +b \tan \left (d x +c \right )\right )}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 108, normalized size = 0.93 \[ \frac {\frac {3 \, b^{3} \tan \left (d x + c\right )^{4} - 4 \, a b^{2} \tan \left (d x + c\right )^{3} + 6 \, {\left (a^{2} b + 2 \, b^{3}\right )} \tan \left (d x + c\right )^{2} - 12 \, {\left (a^{3} + 2 \, a b^{2}\right )} \tan \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{5}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((3*b^3*tan(d*x + c)^4 - 4*a*b^2*tan(d*x + c)^3 + 6*(a^2*b + 2*b^3)*tan(d*x + c)^2 - 12*(a^3 + 2*a*b^2)*t
an(d*x + c))/b^4 + 12*(a^4 + 2*a^2*b^2 + b^4)*log(b*tan(d*x + c) + a)/b^5)/d

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mupad [B]  time = 3.73, size = 119, normalized size = 1.03 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{4\,b\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {1}{b}+\frac {a^2}{2\,b^3}\right )}{d}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{b^5\,d}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,b^2\,d}-\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {2}{b}+\frac {a^2}{b^3}\right )}{b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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