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

Optimal. Leaf size=134 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{32 \sqrt{2} a^{3/2} f}+\frac{\cos (e+f x)}{32 f (a \sin (e+f x)+a)^{3/2}}+\frac{5 \sec (e+f x)}{8 a f \sqrt{a \sin (e+f x)+a}}-\frac{\sec (e+f x)}{4 f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])]/(32*Sqrt[2]*a^(3/2)*f) + Cos[e + f*x]/(32*f
*(a + a*Sin[e + f*x])^(3/2)) - Sec[e + f*x]/(4*f*(a + a*Sin[e + f*x])^(3/2)) + (5*Sec[e + f*x])/(8*a*f*Sqrt[a
+ a*Sin[e + f*x]])

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Rubi [A]  time = 0.223266, antiderivative size = 134, 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 = {2712, 2855, 2650, 2649, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{32 \sqrt{2} a^{3/2} f}+\frac{\cos (e+f x)}{32 f (a \sin (e+f x)+a)^{3/2}}+\frac{5 \sec (e+f x)}{8 a f \sqrt{a \sin (e+f x)+a}}-\frac{\sec (e+f x)}{4 f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])]/(32*Sqrt[2]*a^(3/2)*f) + Cos[e + f*x]/(32*f
*(a + a*Sin[e + f*x])^(3/2)) - Sec[e + f*x]/(4*f*(a + a*Sin[e + f*x])^(3/2)) + (5*Sec[e + f*x])/(8*a*f*Sqrt[a
+ a*Sin[e + f*x]])

Rule 2712

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[(b*(a + b*Sin[e +
f*x])^m)/(a*f*(2*m - 1)*Cos[e + f*x]), x] - Dist[1/(a^2*(2*m - 1)), Int[((a + b*Sin[e + f*x])^(m + 1)*(a*m -
b*(2*m - 1)*Sin[e + f*x]))/Cos[e + f*x]^2, x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ
[m] && LtQ[m, 0]

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 2650

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

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

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])

Rubi steps

\begin{align*} \int \frac{\tan ^2(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac{\sec (e+f x)}{4 f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{\sec ^2(e+f x) \left (-\frac{3 a}{2}+4 a \sin (e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx}{4 a^2}\\ &=-\frac{\sec (e+f x)}{4 f (a+a \sin (e+f x))^{3/2}}+\frac{5 \sec (e+f x)}{8 a f \sqrt{a+a \sin (e+f x)}}-\frac{1}{16} \int \frac{1}{(a+a \sin (e+f x))^{3/2}} \, dx\\ &=\frac{\cos (e+f x)}{32 f (a+a \sin (e+f x))^{3/2}}-\frac{\sec (e+f x)}{4 f (a+a \sin (e+f x))^{3/2}}+\frac{5 \sec (e+f x)}{8 a f \sqrt{a+a \sin (e+f x)}}-\frac{\int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{64 a}\\ &=\frac{\cos (e+f x)}{32 f (a+a \sin (e+f x))^{3/2}}-\frac{\sec (e+f x)}{4 f (a+a \sin (e+f x))^{3/2}}+\frac{5 \sec (e+f x)}{8 a f \sqrt{a+a \sin (e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{32 a f}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{32 \sqrt{2} a^{3/2} f}+\frac{\cos (e+f x)}{32 f (a+a \sin (e+f x))^{3/2}}-\frac{\sec (e+f x)}{4 f (a+a \sin (e+f x))^{3/2}}+\frac{5 \sec (e+f x)}{8 a f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.443707, size = 128, normalized size = 0.96 \[ -\frac{\sec (e+f x) \left (-40 \sin (e+f x)-\cos (2 (e+f x))+(2+2 i) (-1)^{3/4} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )-25\right )}{64 f (a (\sin (e+f x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[e + f*x]*(-25 - Cos[2*(e + f*x)] + (2 + 2*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*
x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 40*Sin[e + f*x]))/(64*
f*(a*(1 + Sin[e + f*x]))^(3/2))

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Maple [A]  time = 0.607, size = 202, normalized size = 1.5 \begin{align*}{\frac{1}{ \left ( 64+64\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( \sin \left ( fx+e \right ) \left ( 2\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}+40\,{a}^{5/2} \right ) + \left ( -\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{2}+2\,{a}^{5/2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}+24\,{a}^{5/2} \right ){a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64/a^(7/2)*(sin(f*x+e)*(2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))
*a^2+40*a^(5/2))+(-(a-a*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2+2*a^
(5/2))*cos(f*x+e)^2+2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2+2
4*a^(5/2))/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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Maxima [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)^2/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.63547, size = 644, normalized size = 4.81 \begin{align*} \frac{\sqrt{2}{\left (\cos \left (f x + e\right )^{3} - 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (f x + e\right )^{2} + 2 \, \sqrt{2} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \,{\left (\cos \left (f x + e\right )^{2} + 20 \, \sin \left (f x + e\right ) + 12\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{128 \,{\left (a^{2} f \cos \left (f x + e\right )^{3} - 2 \, a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{2} f \cos \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(sqrt(2)*(cos(f*x + e)^3 - 2*cos(f*x + e)*sin(f*x + e) - 2*cos(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2
+ 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x
+ e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(c
os(f*x + e)^2 + 20*sin(f*x + e) + 12)*sqrt(a*sin(f*x + e) + a))/(a^2*f*cos(f*x + e)^3 - 2*a^2*f*cos(f*x + e)*s
in(f*x + e) - 2*a^2*f*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (e + f x \right )}}{\left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(e + f*x)**2/(a*(sin(e + f*x) + 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2