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

Optimal. Leaf size=118 \[ \frac{2 a-b}{2 f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a-b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{2 f (a+b)^{5/2}}+\frac{\sec ^2(e+f x)}{2 f (a+b) \sqrt{a+b \sin ^2(e+f x)}} \]

[Out]

-((2*a - b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(2*(a + b)^(5/2)*f) + (2*a - b)/(2*(a + b)^2*f*Sq
rt[a + b*Sin[e + f*x]^2]) + Sec[e + f*x]^2/(2*(a + b)*f*Sqrt[a + b*Sin[e + f*x]^2])

________________________________________________________________________________________

Rubi [A]  time = 0.122448, antiderivative size = 118, 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 = {3194, 78, 51, 63, 208} \[ \frac{2 a-b}{2 f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a-b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{2 f (a+b)^{5/2}}+\frac{\sec ^2(e+f x)}{2 f (a+b) \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a - b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(2*(a + b)^(5/2)*f) + (2*a - b)/(2*(a + b)^2*f*Sq
rt[a + b*Sin[e + f*x]^2]) + Sec[e + f*x]^2/(2*(a + b)*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 )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{(1-x)^2 (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{\sec ^2(e+f x)}{2 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a-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) f}\\ &=\frac{2 a-b}{2 (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sec ^2(e+f x)}{2 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a-b) \operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b)^2 f}\\ &=\frac{2 a-b}{2 (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sec ^2(e+f x)}{2 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a-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)^2 f}\\ &=-\frac{(2 a-b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{2 (a+b)^{5/2} f}+\frac{2 a-b}{2 (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sec ^2(e+f x)}{2 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.1166, size = 75, normalized size = 0.64 \[ \frac{(2 a-b) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \sin ^2(e+f x)+a}{a+b}\right )+(a+b) \sec ^2(e+f x)}{2 f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a - b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[e + f*x]^2)/(a + b)] + (a + b)*Sec[e + f*x]^2)/(2*(a + b
)^2*f*Sqrt[a + b*Sin[e + f*x]^2])

________________________________________________________________________________________

Maple [B]  time = 9.714, size = 2199, normalized size = 18.6 \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*sin(f*x+e)^2)^(3/2),x)

[Out]

1/4/(a^3*b^2*cos(f*x+e)^4+3*a^2*b^3*cos(f*x+e)^4+3*a*b^4*cos(f*x+e)^4+b^5*cos(f*x+e)^4-2*a^4*b*cos(f*x+e)^2-8*
a^3*b^2*cos(f*x+e)^2-12*a^2*b^3*cos(f*x+e)^2-8*a*b^4*cos(f*x+e)^2-2*b^5*cos(f*x+e)^2+a^5+5*a^4*b+10*a^3*b^2+10
*a^2*b^3+5*a*b^4+b^5)/cos(f*x+e)^2/(a+b)^(1/2)*(2*b^3/(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(3/2)+2/(a+b)^(1/2)*(a+
b-b*cos(f*x+e)^2)^(3/2)*a^3+6*b/(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a^2+6*b^2/(a+b)^(1/2)*(a+b-b*cos(f*x+e)
^2)^(3/2)*a-cos(f*x+e)^6*(2*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2+l
n(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b-ln(2/(1+sin(f*x+e))*((a+b)^(1/
2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^2-4*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a+2*b*(a+b-b*cos(f
*x+e)^2)^(1/2)*(a+b)^(1/2)+2*(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a-6*(a+b)^(1/2)*(-b*cos(f*x+e
)^2+(a*b^2+b^3)/b^2)^(1/2)*b+2*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a
^2+ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b-ln(2/(-1+sin(f*x+e))*((a+
b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^2)*b^2+2*cos(f*x+e)^4*((a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)
^(1/2)*b+(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)*a+(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(
3/2)*b+2*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3+3*a^2*b*ln(2/(1+sin(
f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))-b^3*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*c
os(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))-4*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(1/2)*a^2-2*(a+b-b*cos(f*x+e)^2)^(1/2)*
(a+b)^(1/2)*a*b+2*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(1/2)*b^2+2*(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(
1/2)*a^2-4*(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a*b-6*(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/
b^2)^(1/2)*b^2+2*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^3+3*a^2*b*ln(
2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))-b^3*ln(2/(-1+sin(f*x+e))*((a+b)^(1/
2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a)))*b-cos(f*x+e)^2*(4*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(1/2)*a*b+4
*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(1/2)*b^2-2*(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)*a^2+2*(a+b)^
(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)*b^2+2*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2
)-b*sin(f*x+e)+a))*a^4+5*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3*b+3*
ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^2-ln(2/(1+sin(f*x+e))*((a+b
)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b^3-ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2
)^(1/2)-b*sin(f*x+e)+a))*b^4-4*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(1/2)*a^3-6*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(
1/2)*a^2*b+2*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(1/2)*b^3+2*(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*
a^3-2*(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^2*b-10*(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^
2)^(1/2)*a*b^2-6*(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^3+2*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(
a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^4+5*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+
b*sin(f*x+e)+a))*a^3*b+3*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2*b^2
-ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b^3-ln(2/(-1+sin(f*x+e))*((a+
b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^4))/f

________________________________________________________________________________________

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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 3.27331, size = 1026, normalized size = 8.69 \begin{align*} \left [-\frac{{\left ({\left (2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} -{\left (2 \, a^{2} + a b - b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{a + b} \log \left (\frac{b \cos \left (f x + e\right )^{2} - 2 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \,{\left ({\left (2 \, a^{2} + a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{4 \,{\left ({\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{4} -{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2}\right )}}, \frac{{\left ({\left (2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} -{\left (2 \, a^{2} + a b - b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{-a - b} \arctan \left (\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a - b}}{a + b}\right ) -{\left ({\left (2 \, a^{2} + a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{2 \,{\left ({\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{4} -{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2}\right )}}\right ] \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)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(((2*a*b - b^2)*cos(f*x + e)^4 - (2*a^2 + a*b - b^2)*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*((2*a^2 + a*b - b^2)*cos(f*x +
e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*f*cos(f*x + e
)^4 - (a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*f*cos(f*x + e)^2), 1/2*(((2*a*b - b^2)*cos(f*x + e)^4 - (2*a
^2 + a*b - b^2)*cos(f*x + e)^2)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b)) - ((
2*a^2 + a*b - b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3*b + 3*a^2*b^2 +
3*a*b^3 + b^4)*f*cos(f*x + e)^4 - (a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*f*cos(f*x + e)^2)]

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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{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*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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