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

Optimal. Leaf size=81 \[ \frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 f (a+b)}-\frac {(2 a+b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f (a+b)^{3/2}} \]

[Out]

-1/2*(2*a+b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)/f+1/2*sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(
1/2)/(a+b)/f

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Rubi [A]  time = 0.10, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3194, 78, 63, 208} \[ \frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 f (a+b)}-\frac {(2 a+b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a + b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(2*(a + b)^(3/2)*f) + (Sec[e + f*x]^2*Sqrt[a + b*
Sin[e + f*x]^2])/(2*(a + b)*f)

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 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 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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 77, normalized size = 0.95 \[ -\frac {\frac {(2 a+b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a+b}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(((2*a + b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(a + b)^(3/2) - (Sec[e + f*x]^2*Sqrt[a + b*S
in[e + f*x]^2])/(a + b))/f

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fricas [A]  time = 0.59, size = 220, normalized size = 2.72 \[ \left [\frac {{\left (2 \, a + b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{2} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2}}, \frac {{\left (2 \, a + b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) \cos \left (f x + e\right )^{2} + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.29, size = 793, normalized size = 9.79 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

((2*a + b)*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2
*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) - sqrt(a))/sqrt(-a - b))/((a + b)*sqrt(-a - b)) - 2*(2*(sqrt(a)*tan(1/
2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a
))^3*a + (sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*ta
n(1/2*f*x + 1/2*e)^2 + a))^3*b + 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1
/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(3/2) + 5*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*ta
n(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*sqrt(a)*b - 2*(sqrt(a)*
tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)
^2 + a))*a^2 - (sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 +
4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a*b + 4*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a
*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*b^2 - 2*a^(5/2) - 5*a^(3/2)*b - 4*sqrt(a)*b^2)/(((s
qrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x +
1/2*e)^2 + a))^2 - 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*
e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*sqrt(a) - 3*a - 4*b)^2*(a + b)))/f

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maple [B]  time = 3.03, size = 353, normalized size = 4.36 \[ \frac {-\left (2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a b +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{2}+2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a b +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {3}{2}}}{4 \left (a +b \right )^{\frac {5}{2}} \cos \left (f x +e \right )^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x)

[Out]

1/4*(-(2*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2+3*ln(2/(sin(f*x+e)-1
)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b+ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x
+e)^2)^(1/2)+b*sin(f*x+e)+a))*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^2+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*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)*cos(f*x+e)^2+2*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^
(3/2))/(a+b)^(5/2)/cos(f*x+e)^2/f

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maxima [A]  time = 0.54, size = 124, normalized size = 1.53 \[ -\frac {\frac {2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )} {\left (a + b\right )} - a^{2} - 2 \, a b - b^{2}} - \frac {{\left (2 \, a b^{2} + b^{3}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}}}{4 \, b^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/4*(2*sqrt(b*sin(f*x + e)^2 + a)*b^3/((b*sin(f*x + e)^2 + a)*(a + b) - a^2 - 2*a*b - b^2) - (2*a*b^2 + b^3)*
log((sqrt(b*sin(f*x + e)^2 + a) - sqrt(a + b))/(sqrt(b*sin(f*x + e)^2 + a) + sqrt(a + b)))/(a + b)^(3/2))/(b^2
*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^3}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^3/(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(tan(e + f*x)^3/(a + b*sin(e + f*x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{3}{\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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