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

Optimal. Leaf size=118 \[ \frac {(2 a+3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f (a+b)}-\frac {(2 a+3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f \sqrt {a+b}}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 f (a+b)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3194, 78, 50, 63, 208} \[ \frac {(2 a+3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f (a+b)}-\frac {(2 a+3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f \sqrt {a+b}}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 f (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 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 \sqrt {a+b \sin ^2(e+f x)} \tan ^3(e+f x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x \sqrt {a+b x}}{(1-x)^2} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 (a+b) f}-\frac {(2 a+3 b) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f}\\ &=\frac {(2 a+3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 (a+b) f}-\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 f}\\ &=\frac {(2 a+3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 (a+b) f}-\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 f}\\ &=-\frac {(2 a+3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 \sqrt {a+b} f}+\frac {(2 a+3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 (a+b) f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.39, size = 84, normalized size = 0.71 \[ \frac {(\cos (2 (e+f x))+2) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}-\frac {(2 a+3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.86, size = 234, normalized size = 1.98 \[ \left [\frac {{\left (2 \, a + 3 \, b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{2} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (2 \, {\left (a + b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{4 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{2}}, \frac {{\left (2 \, a + 3 \, b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) \cos \left (f x + e\right )^{2} + {\left (2 \, {\left (a + b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{2 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.04, size = 1011, normalized size = 8.57 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*a + 3*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))/sqrt(-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 - sqr
t(a)*b)/((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))^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) + a + 4*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*tan(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*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*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 - (sq
rt(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)/((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 -
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)/f

________________________________________________________________________________________

maple [B]  time = 4.80, size = 403, normalized size = 3.42 \[ \frac {-\left (-4 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {a +b}\, a -6 b \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {a +b}+2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2}+5 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a b +3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{2}+2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2}+5 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a b +3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {a +b}}{4 \left (a +b \right )^{\frac {3}{2}} \cos \left (f x +e \right )^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.44, size = 127, normalized size = 1.08 \[ \frac {4 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2} - \frac {2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{3}}{b \sin \left (f x + e\right )^{2} - b} + \frac {{\left (2 \, a b^{2} + 3 \, b^{3}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{\sqrt {a + b}}}{4 \, b^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________