3.2.8 \(\int \frac {\tan ^2(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx\) [108]

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

[Out]

1/32*cos(f*x+e)/f/(a+a*sin(f*x+e))^(3/2)-1/4*sec(f*x+e)/f/(a+a*sin(f*x+e))^(3/2)+1/64*arctanh(1/2*cos(f*x+e)*a
^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(3/2)/f*2^(1/2)+5/8*sec(f*x+e)/a/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 134, normalized size of antiderivative = 1.00, 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 = {2791, 2934, 2729, 2728, 212} \begin {gather*} \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}} \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

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 2729

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 2791

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 2934

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]

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 {\text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.29, size = 128, normalized size = 0.96 \begin {gather*} -\frac {\sec (e+f x) \left (-25-\cos (2 (e+f x))+(2+2 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-40 \sin (e+f x)\right )}{64 f (a (1+\sin (e+f x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*(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]))
/(f*(a*(1 + Sin[e + f*x]))^(3/2))

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Maple [A]
time = 2.05, size = 202, normalized size = 1.51

method	result	size
default	\(\frac {\sin \left (f x +e \right ) \left (2 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+40 a^{\frac {5}{2}}\right )+\left (-\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+2 a^{\frac {5}{2}}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+24 a^{\frac {5}{2}}}{64 a^{\frac {7}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(202\)

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,method=_RETURNVERBOSE)

[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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (119) = 238\).
time = 0.36, size = 259, normalized size = 1.93 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 [A]
time = 12.39, size = 127, normalized size = 0.95 \begin {gather*} \frac {\frac {8 \, \sqrt {2}}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )} - \frac {\sqrt {2} {\left (9 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 7 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{64 \, f} \end {gather*}

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]

1/64*(8*sqrt(2)/(a^(3/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(3/4*pi + 1/2*f*x + 1/2*e)) - sqrt(2)*(9*sqrt(
a)*sin(3/4*pi + 1/2*f*x + 1/2*e)^3 - 7*sqrt(a)*sin(3/4*pi + 1/2*f*x + 1/2*e))/((sin(3/4*pi + 1/2*f*x + 1/2*e)^
2 - 1)^2*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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