3.533 \(\int \frac{\tan ^3(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=153 \[ \frac{2 a-3 b}{2 f (a+b)^3 \sqrt{a+b \sin ^2(e+f x)}}+\frac{2 a-3 b}{6 f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{2 f (a+b)^{7/2}}+\frac{\sec ^2(e+f x)}{2 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]
[Out]
-((2*a - 3*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(2*(a + b)^(7/2)*f) + (2*a - 3*b)/(6*(a + b)^2*
f*(a + b*Sin[e + f*x]^2)^(3/2)) + Sec[e + f*x]^2/(2*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*a - 3*b)/(2*(
a + b)^3*f*Sqrt[a + b*Sin[e + f*x]^2])
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Rubi [A] time = 0.154389, antiderivative size = 153, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3194, 78, 51, 63, 208} \[ \frac{2 a-3 b}{2 f (a+b)^3 \sqrt{a+b \sin ^2(e+f x)}}+\frac{2 a-3 b}{6 f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{2 f (a+b)^{7/2}}+\frac{\sec ^2(e+f x)}{2 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(5/2),x]
[Out]
-((2*a - 3*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(2*(a + b)^(7/2)*f) + (2*a - 3*b)/(6*(a + b)^2*
f*(a + b*Sin[e + f*x]^2)^(3/2)) + Sec[e + f*x]^2/(2*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*a - 3*b)/(2*(
a + b)^3*f*Sqrt[a + b*Sin[e + f*x]^2])
Rule 3194
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
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 \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{(1-x)^2 (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(2 a-3 b) \operatorname{Subst}\left (\int \frac{1}{(1-x) (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f}\\ &=\frac{2 a-3 b}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(2 a-3 b) \operatorname{Subst}\left (\int \frac{1}{(1-x) (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b)^2 f}\\ &=\frac{2 a-3 b}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 a-3 b}{2 (a+b)^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a-3 b) \operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b)^3 f}\\ &=\frac{2 a-3 b}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 a-3 b}{2 (a+b)^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a-3 b) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}-\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sin ^2(e+f x)}\right )}{2 b (a+b)^3 f}\\ &=-\frac{(2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{2 (a+b)^{7/2} f}+\frac{2 a-3 b}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 a-3 b}{2 (a+b)^3 f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}
Mathematica [C] time = 0.118391, size = 76, normalized size = 0.5 \[ \frac{(2 a-3 b) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b \sin ^2(e+f x)+a}{a+b}\right )+3 (a+b) \sec ^2(e+f x)}{6 f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(5/2),x]
[Out]
((2*a - 3*b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sin[e + f*x]^2)/(a + b)] + 3*(a + b)*Sec[e + f*x]^2)/(6*(
a + b)^2*f*(a + b*Sin[e + f*x]^2)^(3/2))
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Maple [B] time = 6.123, size = 1256, normalized size = 8.2 \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*sin(f*x+e)^2)^(5/2),x)
[Out]
1/2/f*b^3/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*
sin(f*x+e)+2*a)/(-1+sin(f*x+e)))*a-1/2/f*b^4/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(a+b)^(1/2)*ln((2*(a+b)^(1
/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(-1+sin(f*x+e)))+1/2/f*b^3/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1
/2))^3/(-a*b)^(1/2)/(sin(f*x+e)+(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)*a-1/2/f*b^4/(b+(-a*b)^(1/2
))^3/(-b+(-a*b)^(1/2))^3/(-a*b)^(1/2)/(sin(f*x+e)+(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/2/f*b^
3/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(-a*b)^(1/2)/(sin(f*x+e)-(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b
)^(1/2)*a+1/2/f*b^4/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(-a*b)^(1/2)/(sin(f*x+e)-(-a*b)^(1/2)/b)*(-b*cos(f*
x+e)^2+(a*b+b^2)/b)^(1/2)-1/12/f*b/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(sin(f*x+e)-(-a*b)^(1/2)/b)^2*(-b*co
s(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/12/f*b*(-a*b)^(1/2)/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/a/(sin(f*x+e)-(-a*b
)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/12/f*b/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(sin(f*x+e)+(-a
*b)^(1/2)/b)^2*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)+1/12/f*b*(-a*b)^(1/2)/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^
2/a/(sin(f*x+e)+(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)+1/4/f*b^2/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1
/2))^2/(a+b)/(1+sin(f*x+e))*(a+b-b*cos(f*x+e)^2)^(1/2)+1/4/f*b^3/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(a+b)^
(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e)))+1/2/f*b^3/(b+(-a*b)^(1/
2))^3/(-b+(-a*b)^(1/2))^3/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(
f*x+e)))*a-1/2/f*b^4/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)
^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e)))-1/4/f*b^2/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(a+b)/(-1+sin(f*x+
e))*(a+b-b*cos(f*x+e)^2)^(1/2)+1/4/f*b^3/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*
(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(-1+sin(f*x+e)))
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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*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 3.42309, size = 1746, normalized size = 11.41 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")
[Out]
[-1/12*(3*((2*a*b^2 - 3*b^3)*cos(f*x + e)^6 - 2*(2*a^2*b - a*b^2 - 3*b^3)*cos(f*x + e)^4 + (2*a^3 + a^2*b - 4*
a*b^2 - 3*b^3)*cos(f*x + e)^2)*sqrt(a + b)*log((b*cos(f*x + e)^2 - 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a +
b) - 2*a - 2*b)/cos(f*x + e)^2) + 2*(3*(2*a^2*b - a*b^2 - 3*b^3)*cos(f*x + e)^4 - 3*a^3 - 9*a^2*b - 9*a*b^2 -
3*b^3 - 4*(2*a^3 + a^2*b - 4*a*b^2 - 3*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b^2 + 4*a^3
*b^3 + 6*a^2*b^4 + 4*a*b^5 + b^6)*f*cos(f*x + e)^6 - 2*(a^5*b + 5*a^4*b^2 + 10*a^3*b^3 + 10*a^2*b^4 + 5*a*b^5
+ b^6)*f*cos(f*x + e)^4 + (a^6 + 6*a^5*b + 15*a^4*b^2 + 20*a^3*b^3 + 15*a^2*b^4 + 6*a*b^5 + b^6)*f*cos(f*x + e
)^2), 1/6*(3*((2*a*b^2 - 3*b^3)*cos(f*x + e)^6 - 2*(2*a^2*b - a*b^2 - 3*b^3)*cos(f*x + e)^4 + (2*a^3 + a^2*b -
4*a*b^2 - 3*b^3)*cos(f*x + e)^2)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b)) -
(3*(2*a^2*b - a*b^2 - 3*b^3)*cos(f*x + e)^4 - 3*a^3 - 9*a^2*b - 9*a*b^2 - 3*b^3 - 4*(2*a^3 + a^2*b - 4*a*b^2 -
3*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b^2 + 4*a^3*b^3 + 6*a^2*b^4 + 4*a*b^5 + b^6)*f*
cos(f*x + e)^6 - 2*(a^5*b + 5*a^4*b^2 + 10*a^3*b^3 + 10*a^2*b^4 + 5*a*b^5 + b^6)*f*cos(f*x + e)^4 + (a^6 + 6*a
^5*b + 15*a^4*b^2 + 20*a^3*b^3 + 15*a^2*b^4 + 6*a*b^5 + b^6)*f*cos(f*x + e)^2)]
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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(tan(f*x+e)**3/(a+b*sin(f*x+e)**2)**(5/2),x)
[Out]
Timed out
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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 \sin \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*sin(f*x+e)^2)^(5/2),x, algorithm="giac")
[Out]
integrate(tan(f*x + e)^3/(b*sin(f*x + e)^2 + a)^(5/2), x)