3.300 \(\int \frac{\tan ^7(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=179 \[ \frac{3 \left (a^2-b^2\right ) \sec ^2(c+d x)}{2 b^4 d}-\frac{2 a \left (2 a^2-3 b^2\right ) \sec (c+d x)}{b^5 d}+\frac{\left (a^2-b^2\right )^3}{a b^6 d (a+b \sec (c+d x))}+\frac{\left (a^2-b^2\right )^2 \left (5 a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^6 d}+\frac{\log (\cos (c+d x))}{a^2 d}-\frac{2 a \sec ^3(c+d x)}{3 b^3 d}+\frac{\sec ^4(c+d x)}{4 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.145637, antiderivative size = 179, 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 = {3885, 894} \[ \frac{3 \left (a^2-b^2\right ) \sec ^2(c+d x)}{2 b^4 d}-\frac{2 a \left (2 a^2-3 b^2\right ) \sec (c+d x)}{b^5 d}+\frac{\left (a^2-b^2\right )^3}{a b^6 d (a+b \sec (c+d x))}+\frac{\left (a^2-b^2\right )^2 \left (5 a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^6 d}+\frac{\log (\cos (c+d x))}{a^2 d}-\frac{2 a \sec ^3(c+d x)}{3 b^3 d}+\frac{\sec ^4(c+d x)}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^7/(a + b*Sec[c + d*x])^2,x]

[Out]

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

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [B]  time = 6.19821, size = 383, normalized size = 2.14 \[ \frac{2 a \left (3 b^2-2 a^2\right ) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{b^5 d (a+b \sec (c+d x))^2}+\frac{(b-a)^3 (a+b)^3 \sec ^2(c+d x) (a \cos (c+d x)+b)}{a^2 b^5 d (a+b \sec (c+d x))^2}+\frac{\left (9 a^2 b^2-5 a^4-3 b^4\right ) \sec ^2(c+d x) \log (\cos (c+d x)) (a \cos (c+d x)+b)^2}{b^6 d (a+b \sec (c+d x))^2}+\frac{\left (-9 a^4 b^2+3 a^2 b^4+5 a^6+b^6\right ) \sec ^2(c+d x) (a \cos (c+d x)+b)^2 \log (a \cos (c+d x)+b)}{a^2 b^6 d (a+b \sec (c+d x))^2}+\frac{\sec ^6(c+d x) (a \cos (c+d x)+b)^2}{4 b^2 d (a+b \sec (c+d x))^2}-\frac{2 a \sec ^5(c+d x) (a \cos (c+d x)+b)^2}{3 b^3 d (a+b \sec (c+d x))^2}-\frac{3 (b-a) (a+b) \sec ^4(c+d x) (a \cos (c+d x)+b)^2}{2 b^4 d (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^7/(a + b*Sec[c + d*x])^2,x]

[Out]

((-a + b)^3*(a + b)^3*(b + a*Cos[c + d*x])*Sec[c + d*x]^2)/(a^2*b^5*d*(a + b*Sec[c + d*x])^2) + ((-5*a^4 + 9*a
^2*b^2 - 3*b^4)*(b + a*Cos[c + d*x])^2*Log[Cos[c + d*x]]*Sec[c + d*x]^2)/(b^6*d*(a + b*Sec[c + d*x])^2) + ((5*
a^6 - 9*a^4*b^2 + 3*a^2*b^4 + b^6)*(b + a*Cos[c + d*x])^2*Log[b + a*Cos[c + d*x]]*Sec[c + d*x]^2)/(a^2*b^6*d*(
a + b*Sec[c + d*x])^2) + (2*a*(-2*a^2 + 3*b^2)*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^3)/(b^5*d*(a + b*Sec[c + d*
x])^2) - (3*(-a + b)*(a + b)*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^4)/(2*b^4*d*(a + b*Sec[c + d*x])^2) - (2*a*(b
+ a*Cos[c + d*x])^2*Sec[c + d*x]^5)/(3*b^3*d*(a + b*Sec[c + d*x])^2) + ((b + a*Cos[c + d*x])^2*Sec[c + d*x]^6
)/(4*b^2*d*(a + b*Sec[c + d*x])^2)

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Maple [A]  time = 0.069, size = 324, normalized size = 1.8 \begin{align*} -{\frac{{a}^{4}}{d{b}^{5} \left ( b+a\cos \left ( dx+c \right ) \right ) }}+3\,{\frac{{a}^{2}}{d{b}^{3} \left ( b+a\cos \left ( dx+c \right ) \right ) }}-3\,{\frac{1}{db \left ( b+a\cos \left ( dx+c \right ) \right ) }}+{\frac{b}{d{a}^{2} \left ( b+a\cos \left ( dx+c \right ) \right ) }}+5\,{\frac{{a}^{4}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{b}^{6}}}-9\,{\frac{{a}^{2}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{b}^{4}}}+3\,{\frac{\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{b}^{2}}}+{\frac{\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d{a}^{2}}}+{\frac{3\,{a}^{2}}{2\,d{b}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3}{2\,d{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-5\,{\frac{\ln \left ( \cos \left ( dx+c \right ) \right ){a}^{4}}{d{b}^{6}}}+9\,{\frac{\ln \left ( \cos \left ( dx+c \right ) \right ){a}^{2}}{d{b}^{4}}}-3\,{\frac{\ln \left ( \cos \left ( dx+c \right ) \right ) }{d{b}^{2}}}+{\frac{1}{4\,d{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{2\,a}{3\,d{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-4\,{\frac{{a}^{3}}{d{b}^{5}\cos \left ( dx+c \right ) }}+6\,{\frac{a}{d{b}^{3}\cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^7/(a+b*sec(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 0.984654, size = 306, normalized size = 1.71 \begin{align*} -\frac{\frac{5 \, a^{3} b^{3} \cos \left (d x + c\right ) - 3 \, a^{2} b^{4} + 12 \,{\left (5 \, a^{6} - 9 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{4} + 6 \,{\left (5 \, a^{5} b - 9 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} - 2 \,{\left (5 \, a^{4} b^{2} - 9 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2}}{a^{3} b^{5} \cos \left (d x + c\right )^{5} + a^{2} b^{6} \cos \left (d x + c\right )^{4}} + \frac{12 \,{\left (5 \, a^{4} - 9 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left (\cos \left (d x + c\right )\right )}{b^{6}} - \frac{12 \,{\left (5 \, a^{6} - 9 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{2} b^{6}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.39557, size = 690, normalized size = 3.85 \begin{align*} -\frac{5 \, a^{3} b^{4} \cos \left (d x + c\right ) - 3 \, a^{2} b^{5} + 12 \,{\left (5 \, a^{6} b - 9 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} + 6 \,{\left (5 \, a^{5} b^{2} - 9 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \,{\left (5 \, a^{4} b^{3} - 9 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} - 12 \,{\left ({\left (5 \, a^{7} - 9 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{5} +{\left (5 \, a^{6} b - 9 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + 12 \,{\left ({\left (5 \, a^{7} - 9 \, a^{5} b^{2} + 3 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{5} +{\left (5 \, a^{6} b - 9 \, a^{4} b^{3} + 3 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\cos \left (d x + c\right )\right )}{12 \,{\left (a^{3} b^{6} d \cos \left (d x + c\right )^{5} + a^{2} b^{7} d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/12*(5*a^3*b^4*cos(d*x + c) - 3*a^2*b^5 + 12*(5*a^6*b - 9*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)^4 + 6*(5*a
^5*b^2 - 9*a^3*b^4)*cos(d*x + c)^3 - 2*(5*a^4*b^3 - 9*a^2*b^5)*cos(d*x + c)^2 - 12*((5*a^7 - 9*a^5*b^2 + 3*a^3
*b^4 + a*b^6)*cos(d*x + c)^5 + (5*a^6*b - 9*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(d*x + c)^4)*log(a*cos(d*x + c) + b)
+ 12*((5*a^7 - 9*a^5*b^2 + 3*a^3*b^4)*cos(d*x + c)^5 + (5*a^6*b - 9*a^4*b^3 + 3*a^2*b^5)*cos(d*x + c)^4)*log(
-cos(d*x + c)))/(a^3*b^6*d*cos(d*x + c)^5 + a^2*b^7*d*cos(d*x + c)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{7}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**7/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**7/(a + b*sec(c + d*x))**2, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

Exception raised: TypeError