3.429 \(\int \frac{\tan ^3(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=89 \[ -\frac{1}{a^2 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{a^{5/2} f}-\frac{a+b}{3 a b f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[Out]

ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]]/(a^(5/2)*f) - (a + b)/(3*a*b*f*(a + b*Sec[e + f*x]^2)^(3/2)) - 1/(
a^2*f*Sqrt[a + b*Sec[e + f*x]^2])

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Rubi [A]  time = 0.128431, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4139, 446, 78, 51, 63, 208} \[ -\frac{1}{a^2 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{a^{5/2} f}-\frac{a+b}{3 a b f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^3/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]]/(a^(5/2)*f) - (a + b)/(3*a*b*f*(a + b*Sec[e + f*x]^2)^(3/2)) - 1/(
a^2*f*Sqrt[a + b*Sec[e + f*x]^2])

Rule 4139

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p)/x
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])

Rule 446

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 51

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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

Mathematica [C]  time = 10.4458, size = 613, normalized size = 6.89 \[ -\frac{e^{i (e+f x)} \sec ^5(e+f x) \sqrt{4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \left (\frac{-12 \log \left (\sqrt{a} \sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}+a e^{2 i (e+f x)}+a+2 b\right )-12 \log \left (\sqrt{a} \sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}+a e^{2 i (e+f x)}+a+2 b e^{2 i (e+f x)}\right )+24 i f x}{\sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}}-\frac{\sqrt{a} \left (1+e^{2 i (e+f x)}\right ) \left (-6 a^2 b \left (e^{2 i (e+f x)}+e^{4 i (e+f x)}+1\right )+a^3 \left (1+e^{2 i (e+f x)}\right )^2-32 a b^2 \left (1+e^{2 i (e+f x)}\right )^2-96 b^3 e^{2 i (e+f x)}\right )}{b^2 \left (a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}\right )^2}\right ) (a \cos (2 e+2 f x)+a+2 b)^{5/2}}{96 \sqrt{2} a^{5/2} f \left (a+b \sec ^2(e+f x)\right )^{5/2}}-\frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+3 b) (a \cos (2 e+2 f x)+a+2 b)^{5/2}}{48 b^2 f (a \cos (2 (e+f x))+a+2 b)^{3/2} \left (a+b \sec ^2(e+f x)\right )^{5/2}}+\frac{\sec ^4(e+f x) ((a-2 b) \cos (2 (e+f x))+a+b) (a \cos (2 e+2 f x)+a+2 b)^{5/2}}{96 b^2 f (a \cos (2 (e+f x))+a+2 b)^{3/2} \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^3/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-((a + 3*b + a*Cos[2*(e + f*x)])*(a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^4)/(48*b^2*f*(a + 2*b + a*C
os[2*(e + f*x)])^(3/2)*(a + b*Sec[e + f*x]^2)^(5/2)) + ((a + b + (a - 2*b)*Cos[2*(e + f*x)])*(a + 2*b + a*Cos[
2*e + 2*f*x])^(5/2)*Sec[e + f*x]^4)/(96*b^2*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2)*(a + b*Sec[e + f*x]^2)^(5/2
)) - (E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*(a + 2*b + a*Cos[2*e + 2
*f*x])^(5/2)*(-((Sqrt[a]*(1 + E^((2*I)*(e + f*x)))*(-96*b^3*E^((2*I)*(e + f*x)) + a^3*(1 + E^((2*I)*(e + f*x))
)^2 - 32*a*b^2*(1 + E^((2*I)*(e + f*x)))^2 - 6*a^2*b*(1 + E^((2*I)*(e + f*x)) + E^((4*I)*(e + f*x)))))/(b^2*(4
*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2)^2)) + ((24*I)*f*x - 12*Log[a + 2*b + a*E^((2*I)*(e + f
*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]] - 12*Log[a + a*E^((2*I)*(e + f*x
)) + 2*b*E^((2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]])/Sqrt[4*
b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*Sec[e + f*x]^5)/(96*Sqrt[2]*a^(5/2)*f*(a + b*Sec[e + f
*x]^2)^(5/2))

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Maple [B]  time = 2.165, size = 10839, normalized size = 121.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x)

[Out]

result too large to display

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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(tan(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.8122, size = 1264, normalized size = 14.2 \begin{align*} \left [\frac{3 \,{\left (a^{2} b \cos \left (f x + e\right )^{4} + 2 \, a b^{2} \cos \left (f x + e\right )^{2} + b^{3}\right )} \sqrt{a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} + 8 \,{\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt{a} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) - 8 \,{\left (3 \, a b^{2} \cos \left (f x + e\right )^{2} +{\left (a^{3} + 4 \, a^{2} b\right )} \cos \left (f x + e\right )^{4}\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \,{\left (a^{5} b f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b^{2} f \cos \left (f x + e\right )^{2} + a^{3} b^{3} f\right )}}, -\frac{3 \,{\left (a^{2} b \cos \left (f x + e\right )^{4} + 2 \, a b^{2} \cos \left (f x + e\right )^{2} + b^{3}\right )} \sqrt{-a} \arctan \left (\frac{{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \,{\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) + 4 \,{\left (3 \, a b^{2} \cos \left (f x + e\right )^{2} +{\left (a^{3} + 4 \, a^{2} b\right )} \cos \left (f x + e\right )^{4}\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \,{\left (a^{5} b f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b^{2} f \cos \left (f x + e\right )^{2} + a^{3} b^{3} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/24*(3*(a^2*b*cos(f*x + e)^4 + 2*a*b^2*cos(f*x + e)^2 + b^3)*sqrt(a)*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*
cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 + 8*(16*a^3*cos(f*x + e)^8 + 24*a^
2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x
+ e)^2)) - 8*(3*a*b^2*cos(f*x + e)^2 + (a^3 + 4*a^2*b)*cos(f*x + e)^4)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x +
e)^2))/(a^5*b*f*cos(f*x + e)^4 + 2*a^4*b^2*f*cos(f*x + e)^2 + a^3*b^3*f), -1/12*(3*(a^2*b*cos(f*x + e)^4 + 2*a
*b^2*cos(f*x + e)^2 + b^3)*sqrt(-a)*arctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sq
rt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2)) + 4*(3*a*b^
2*cos(f*x + e)^2 + (a^3 + 4*a^2*b)*cos(f*x + e)^4)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^5*b*f*cos(f
*x + e)^4 + 2*a^4*b^2*f*cos(f*x + e)^2 + a^3*b^3*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{3}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**3/(a+b*sec(f*x+e)**2)**(5/2),x)

[Out]

Integral(tan(e + f*x)**3/(a + b*sec(e + f*x)**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{3}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(tan(f*x + e)^3/(b*sec(f*x + e)^2 + a)^(5/2), x)