3.450 \(\int \frac{\sec ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx\)

Optimal. Leaf size=90 \[ \frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} b^2 d}-\frac{(2 a-3 b) \tanh ^{-1}(\sin (c+d x))}{2 b^2 d}+\frac{\tan (c+d x) \sec (c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.14345, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3676, 414, 522, 206, 208} \[ \frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} b^2 d}-\frac{(2 a-3 b) \tanh ^{-1}(\sin (c+d x))}{2 b^2 d}+\frac{\tan (c+d x) \sec (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + b*Tan[c + d*x]^2),x]

[Out]

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

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

Mathematica [B]  time = 1.32839, size = 207, normalized size = 2.3 \[ \frac{-\frac{2 (a-b)^{3/2} \log \left (\sqrt{a}-\sqrt{a-b} \sin (c+d x)\right )}{\sqrt{a}}+\frac{2 (a-b)^{3/2} \log \left (\sqrt{a-b} \sin (c+d x)+\sqrt{a}\right )}{\sqrt{a}}+2 (2 a-3 b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 (3 b-2 a) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{b}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + b*Tan[c + d*x]^2),x]

[Out]

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

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Maple [B]  time = 0.087, size = 224, normalized size = 2.5 \begin{align*}{\frac{{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 ) }}}}-2\,{\frac{a}{db\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{1}{d}{\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\,db \left ( \sin \left ( dx+c \right ) +1 \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) +1 \right ) a}{2\,d{b}^{2}}}+{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) +1 \right ) }{4\,db}}-{\frac{1}{4\,db \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) a}{2\,d{b}^{2}}}-{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{4\,db}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+b*tan(d*x+c)^2),x)

[Out]

1/d/b^2/(a*(a-b))^(1/2)*arctanh((a-b)*sin(d*x+c)/(a*(a-b))^(1/2))*a^2-2/d/b/(a*(a-b))^(1/2)*arctanh((a-b)*sin(
d*x+c)/(a*(a-b))^(1/2))*a+1/d/(a*(a-b))^(1/2)*arctanh((a-b)*sin(d*x+c)/(a*(a-b))^(1/2))-1/4/d/b/(sin(d*x+c)+1)
-1/2/d/b^2*ln(sin(d*x+c)+1)*a+3/4/d/b*ln(sin(d*x+c)+1)-1/4/d/b/(sin(d*x+c)-1)+1/2/d/b^2*ln(sin(d*x+c)-1)*a-3/4
/d/b*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)^5/(a+b*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.03724, size = 724, normalized size = 8.04 \begin{align*} \left [-\frac{2 \,{\left (a - b\right )} \sqrt{\frac{a - b}{a}} \cos \left (d x + c\right )^{2} \log \left (-\frac{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + 2 \, a \sqrt{\frac{a - b}{a}} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) +{\left (2 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, b \sin \left (d x + c\right )}{4 \, b^{2} d \cos \left (d x + c\right )^{2}}, -\frac{4 \,{\left (a - b\right )} \sqrt{-\frac{a - b}{a}} \arctan \left (\sqrt{-\frac{a - b}{a}} \sin \left (d x + c\right )\right ) \cos \left (d x + c\right )^{2} +{\left (2 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, b \sin \left (d x + c\right )}{4 \, b^{2} d \cos \left (d x + c\right )^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\left (c + d x \right )}}{a + b \tan ^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+b*tan(d*x+c)**2),x)

[Out]

Integral(sec(c + d*x)**5/(a + b*tan(c + d*x)**2), x)

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Giac [A]  time = 1.75967, size = 177, normalized size = 1.97 \begin{align*} -\frac{\frac{{\left (2 \, a - 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{b^{2}} - \frac{{\left (2 \, a - 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{b^{2}} - \frac{4 \,{\left (a^{2} - 2 \, a b + b^{2}\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} b^{2}} + \frac{2 \, \sin \left (d x + c\right )}{{\left (\sin \left (d x + c\right )^{2} - 1\right )} b}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*tan(d*x+c)^2),x, algorithm="giac")

[Out]

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