3.416 \(\int \frac{\tan ^3(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=63 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{a^{3/2} f}-\frac{a+b}{a b f \sqrt{a+b \sec ^2(e+f x)}} \]
[Out]
ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]]/(a^(3/2)*f) - (a + b)/(a*b*f*Sqrt[a + b*Sec[e + f*x]^2])
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Rubi [A] time = 0.116453, antiderivative size = 63, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{4139, 446, 78, 63, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{a^{3/2} f}-\frac{a+b}{a b f \sqrt{a+b \sec ^2(e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]^3/(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]]/(a^(3/2)*f) - (a + b)/(a*b*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 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 )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-1+x^2}{x \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{-1+x}{x (a+b x)^{3/2}} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=-\frac{a+b}{a b 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 f}\\ &=-\frac{a+b}{a b 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 b f}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{a^{3/2} f}-\frac{a+b}{a b f \sqrt{a+b \sec ^2(e+f x)}}\\ \end{align*}
Mathematica [C] time = 4.26875, size = 187, normalized size = 2.97 \[ \frac{3 (a+b) \tan ^4(e+f x) F_1\left (2;\frac{1}{2},\frac{3}{2};3;\sin ^2(e+f x),\frac{a \sin ^2(e+f x)}{a+b}\right )}{2 f \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (\sin ^2(e+f x) \left (3 a F_1\left (3;\frac{1}{2},\frac{5}{2};4;\sin ^2(e+f x),\frac{a \sin ^2(e+f x)}{a+b}\right )+(a+b) F_1\left (3;\frac{3}{2},\frac{3}{2};4;\sin ^2(e+f x),\frac{a \sin ^2(e+f x)}{a+b}\right )\right )+6 (a+b) F_1\left (2;\frac{1}{2},\frac{3}{2};3;\sin ^2(e+f x),\frac{a \sin ^2(e+f x)}{a+b}\right )\right )} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Tan[e + f*x]^3/(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
(3*(a + b)*AppellF1[2, 1/2, 3/2, 3, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Tan[e + f*x]^4)/(2*f*(a + b*Se
c[e + f*x]^2)^(3/2)*(6*(a + b)*AppellF1[2, 1/2, 3/2, 3, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (3*a*App
ellF1[3, 1/2, 5/2, 4, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a + b)*AppellF1[3, 3/2, 3/2, 4, Sin[e + f
*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2))
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Maple [B] time = 0.372, size = 2371, normalized size = 37.6 \begin{align*} \text{result 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)^(3/2),x)
[Out]
1/f/a^(3/2)/((-a*b)^(1/2)-a)/((-a*b)^(1/2)+a)/b*(-1+cos(f*x+e))^2*(1+cos(f*x+e))^2*(3*cos(f*x+e)^5*a^(11/2)*b-
4*cos(f*x+e)^3*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b^3-2*cos(f*x+e)^3*
ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^
2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^4-2*cos(f*x+e)^2*ln(4*cos(f*x+e)
*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e
))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^4*b^2-4*cos(f*x+e)^2*ln(4*cos(f*x+e)*((b+a*cos(f*x+
e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b^3-2*cos(f*x+e)^2*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*
x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^
2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^4-cos(f*x+e)*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1
/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2
)^(1/2)*a^3*b^3-2*cos(f*x+e)*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e
)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^4-c
os(f*x+e)*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*c
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^5-cos(f*x+e)^5*ln(4*cos(
f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos
(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^5*b-2*cos(f*x+e)^5*ln(4*cos(f*x+e)*((b+a*cos(
f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)
)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^4*b^2-cos(f*x+e)^5*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(
f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e
)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b^3-cos(f*x+e)^4*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*
a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e
))^2)^(1/2)*a^5*b-2*cos(f*x+e)^4*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f
*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^4*b
^2-cos(f*x+e)^4*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*(
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b^3-2*cos(f*x+e)^3
*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)
^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^4*b^2+6*cos(f*x+e)^3*a^(9/2)*b^2+6*
cos(f*x+e)^3*a^(7/2)*b^3+2*cos(f*x+e)^3*a^(5/2)*b^4+cos(f*x+e)*a^(9/2)*b^2+3*cos(f*x+e)*a^(7/2)*b^3+3*cos(f*x+
e)*a^(5/2)*b^4+cos(f*x+e)*a^(3/2)*b^5+3*cos(f*x+e)^5*a^(9/2)*b^2+cos(f*x+e)^5*a^(7/2)*b^3+2*cos(f*x+e)^3*a^(11
/2)*b-ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b^3-2*ln(4*cos(f*x+e)*((b+a*
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(
1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^4-ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2
)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+c
os(f*x+e))^2)^(1/2)*a*b^5+cos(f*x+e)^5*a^(13/2))*cos(f*x+e)^3*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)/(b+a*cos
(f*x+e)^2)^4/(a+b)/sin(f*x+e)^4
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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)^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 1.03414, size = 1022, normalized size = 16.22 \begin{align*} \left [-\frac{8 \,{\left (a^{2} + a b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )^{2} -{\left (a b \cos \left (f x + e\right )^{2} + b^{2}\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 (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} f\right )}}, -\frac{4 \,{\left (a^{2} + a b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )^{2} +{\left (a b \cos \left (f x + e\right )^{2} + b^{2}\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 (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} 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)^(3/2),x, algorithm="fricas")
[Out]
[-1/8*(8*(a^2 + a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)^2 - (a*b*cos(f*x + e)^2 + b^2)*s
qrt(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)))/(a^3*b*f*cos(f*x + e)^2 + a^2*b^2*f), -1/4*(4*(a^2 +
a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)^2 + (a*b*cos(f*x + e)^2 + b^2)*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)*sqrt((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)))/(a^3*b*f*cos(f*x + e)^2 + a^2*b^2*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{3}{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)**(3/2),x)
[Out]
Integral(tan(e + f*x)**3/(a + b*sec(e + f*x)**2)**(3/2), 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{3}{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)^(3/2),x, algorithm="giac")
[Out]
integrate(tan(f*x + e)^3/(b*sec(f*x + e)^2 + a)^(3/2), x)