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

Optimal. Leaf size=167 \[ \frac{(4 a+b) (a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^3 d}+\frac{(2 a-b) (a-b) \sin (c+d x)}{2 a b^2 d \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac{(4 a-5 b) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}+\frac{\tan (c+d x) \sec (c+d x)}{2 b d \left (a-(a-b) \sin ^2(c+d x)\right )} \]

[Out]

-((4*a - 5*b)*ArcTanh[Sin[c + d*x]])/(2*b^3*d) + ((a - b)^(3/2)*(4*a + b)*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/S
qrt[a]])/(2*a^(3/2)*b^3*d) + ((a - b)*(2*a - b)*Sin[c + d*x])/(2*a*b^2*d*(a - (a - b)*Sin[c + d*x]^2)) + (Sec[
c + d*x]*Tan[c + d*x])/(2*b*d*(a - (a - b)*Sin[c + d*x]^2))

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Rubi [A]  time = 0.267281, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3676, 414, 527, 522, 206, 208} \[ \frac{(4 a+b) (a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^3 d}+\frac{(2 a-b) (a-b) \sin (c+d x)}{2 a b^2 d \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac{(4 a-5 b) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}+\frac{\tan (c+d x) \sec (c+d x)}{2 b d \left (a-(a-b) \sin ^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*a - 5*b)*ArcTanh[Sin[c + d*x]])/(2*b^3*d) + ((a - b)^(3/2)*(4*a + b)*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/S
qrt[a]])/(2*a^(3/2)*b^3*d) + ((a - b)*(2*a - b)*Sin[c + d*x])/(2*a*b^2*d*(a - (a - b)*Sin[c + d*x]^2)) + (Sec[
c + d*x]*Tan[c + d*x])/(2*b*d*(a - (a - b)*Sin[c + d*x]^2))

Rule 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 4.10419, size = 254, normalized size = 1.52 \[ \frac{-\frac{(4 a+b) (a-b)^{3/2} \log \left (\sqrt{a}-\sqrt{a-b} \sin (c+d x)\right )}{a^{3/2}}+\frac{(4 a+b) (a-b)^{3/2} \log \left (\sqrt{a-b} \sin (c+d x)+\sqrt{a}\right )}{a^{3/2}}+\frac{4 b (a-b)^2 \sin (c+d x)}{a ((a-b) \cos (2 (c+d x))+a+b)}+2 (4 a-5 b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 (5 b-4 a) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{b}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.109, size = 389, normalized size = 2.3 \begin{align*} -{\frac{\sin \left ( dx+c \right ) a}{2\,d{b}^{2} \left ( a \left ( \sin \left ( dx+c \right ) \right ) ^{2}-b \left ( \sin \left ( dx+c \right ) \right ) ^{2}-a \right ) }}+{\frac{\sin \left ( dx+c \right ) }{db \left ( a \left ( \sin \left ( dx+c \right ) \right ) ^{2}-b \left ( \sin \left ( dx+c \right ) \right ) ^{2}-a \right ) }}-{\frac{\sin \left ( dx+c \right ) }{2\,da \left ( a \left ( \sin \left ( dx+c \right ) \right ) ^{2}-b \left ( \sin \left ( dx+c \right ) \right ) ^{2}-a \right ) }}+2\,{\frac{{a}^{2}}{d{b}^{3}\sqrt{a \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \sin \left ( dx+c \right ) }{\sqrt{a \left ( a-b \right ) }}} \right ) }-{\frac{7\,a}{2\,d{b}^{2}}{\it Artanh} \left ({ \left ( a-b \right ) \sin \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}}+{\frac{1}{db}{\it Artanh} \left ({ \left ( a-b \right ) \sin \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}}+{\frac{1}{2\,da}{\it Artanh} \left ({ \left ( a-b \right ) \sin \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}}-{\frac{1}{4\,d{b}^{2} \left ( \sin \left ( dx+c \right ) +1 \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) +1 \right ) a}{d{b}^{3}}}+{\frac{5\,\ln \left ( \sin \left ( dx+c \right ) +1 \right ) }{4\,d{b}^{2}}}-{\frac{1}{4\,d{b}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) a}{d{b}^{3}}}-{\frac{5\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{4\,d{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d/b^2*a*sin(d*x+c)/(a*sin(d*x+c)^2-b*sin(d*x+c)^2-a)+1/d/b*sin(d*x+c)/(a*sin(d*x+c)^2-b*sin(d*x+c)^2-a)-1
/2/d/a*sin(d*x+c)/(a*sin(d*x+c)^2-b*sin(d*x+c)^2-a)+2/d/b^3/(a*(a-b))^(1/2)*arctanh((a-b)*sin(d*x+c)/(a*(a-b))
^(1/2))*a^2-7/2/d/b^2/(a*(a-b))^(1/2)*arctanh((a-b)*sin(d*x+c)/(a*(a-b))^(1/2))*a+1/d/b/(a*(a-b))^(1/2)*arctan
h((a-b)*sin(d*x+c)/(a*(a-b))^(1/2))+1/2/d/a/(a*(a-b))^(1/2)*arctanh((a-b)*sin(d*x+c)/(a*(a-b))^(1/2))-1/4/d/b^
2/(sin(d*x+c)+1)-1/d/b^3*ln(sin(d*x+c)+1)*a+5/4/d/b^2*ln(sin(d*x+c)+1)-1/4/d/b^2/(sin(d*x+c)-1)+1/d/b^3*ln(sin
(d*x+c)-1)*a-5/4/d/b^2*ln(sin(d*x+c)-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.42098, size = 1447, normalized size = 8.66 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(((4*a^3 - 7*a^2*b + 2*a*b^2 + b^3)*cos(d*x + c)^4 + (4*a^2*b - 3*a*b^2 - b^3)*cos(d*x + c)^2)*sqrt((a -
b)/a)*log(-((a - b)*cos(d*x + c)^2 + 2*a*sqrt((a - b)/a)*sin(d*x + c) - 2*a + b)/((a - b)*cos(d*x + c)^2 + b)
) + ((4*a^3 - 9*a^2*b + 5*a*b^2)*cos(d*x + c)^4 + (4*a^2*b - 5*a*b^2)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) -
((4*a^3 - 9*a^2*b + 5*a*b^2)*cos(d*x + c)^4 + (4*a^2*b - 5*a*b^2)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(
a*b^2 + (2*a^2*b - 3*a*b^2 + b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a*b^4*d*cos(d*x + c)^2 + (a^2*b^3 - a*b^4)*d*
cos(d*x + c)^4), -1/4*(2*((4*a^3 - 7*a^2*b + 2*a*b^2 + b^3)*cos(d*x + c)^4 + (4*a^2*b - 3*a*b^2 - b^3)*cos(d*x
+ c)^2)*sqrt(-(a - b)/a)*arctan(sqrt(-(a - b)/a)*sin(d*x + c)) + ((4*a^3 - 9*a^2*b + 5*a*b^2)*cos(d*x + c)^4
+ (4*a^2*b - 5*a*b^2)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - ((4*a^3 - 9*a^2*b + 5*a*b^2)*cos(d*x + c)^4 + (4
*a^2*b - 5*a*b^2)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(a*b^2 + (2*a^2*b - 3*a*b^2 + b^3)*cos(d*x + c)^2
)*sin(d*x + c))/(a*b^4*d*cos(d*x + c)^2 + (a^2*b^3 - a*b^4)*d*cos(d*x + c)^4)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.8084, size = 331, normalized size = 1.98 \begin{align*} -\frac{\frac{{\left (4 \, a - 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{b^{3}} - \frac{{\left (4 \, a - 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{b^{3}} - \frac{2 \,{\left (4 \, a^{3} - 7 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac{a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt{-a^{2} + a b}}\right )}{\sqrt{-a^{2} + a b} a b^{3}} + \frac{2 \,{\left (2 \, a^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{3} + b^{2} \sin \left (d x + c\right )^{3} - 2 \, a^{2} \sin \left (d x + c\right ) + 2 \, a b \sin \left (d x + c\right ) - b^{2} \sin \left (d x + c\right )\right )}}{{\left (a \sin \left (d x + c\right )^{4} - b \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{2} + b \sin \left (d x + c\right )^{2} + a\right )} a b^{2}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((4*a - 5*b)*log(abs(sin(d*x + c) + 1))/b^3 - (4*a - 5*b)*log(abs(sin(d*x + c) - 1))/b^3 - 2*(4*a^3 - 7*a
^2*b + 2*a*b^2 + b^3)*arctan(-(a*sin(d*x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/(sqrt(-a^2 + a*b)*a*b^3) + 2
*(2*a^2*sin(d*x + c)^3 - 3*a*b*sin(d*x + c)^3 + b^2*sin(d*x + c)^3 - 2*a^2*sin(d*x + c) + 2*a*b*sin(d*x + c) -
b^2*sin(d*x + c))/((a*sin(d*x + c)^4 - b*sin(d*x + c)^4 - 2*a*sin(d*x + c)^2 + b*sin(d*x + c)^2 + a)*a*b^2))/
d