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

Optimal. Leaf size=148 \[ \frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 f (a+b)}+\frac {(2 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}-\frac {\sqrt {a+b} (2 a+5 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 f (a+b)} \]

[Out]

1/6*(2*a+5*b)*(a+b*sin(f*x+e)^2)^(3/2)/(a+b)/f+1/2*sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(5/2)/(a+b)/f-1/2*(2*a+5*b)
*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))*(a+b)^(1/2)/f+1/2*(2*a+5*b)*(a+b*sin(f*x+e)^2)^(1/2)/f

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Rubi [A]  time = 0.14, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 f (a+b)}+\frac {(2 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}-\frac {\sqrt {a+b} (2 a+5 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 f (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a + b]*(2*a + 5*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(2*f) + ((2*a + 5*b)*Sqrt[a + b*Sin
[e + f*x]^2])/(2*f) + ((2*a + 5*b)*(a + b*Sin[e + f*x]^2)^(3/2))/(6*(a + b)*f) + (Sec[e + f*x]^2*(a + b*Sin[e
+ f*x]^2)^(5/2))/(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 \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x (a+b x)^{3/2}}{(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 )^{5/2}}{2 (a+b) f}-\frac {(2 a+5 b) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f}\\ &=\frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}-\frac {(2 a+5 b) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{4 f}\\ &=\frac {(2 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}-\frac {((a+b) (2 a+5 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+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}-\frac {((a+b) (2 a+5 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 {\sqrt {a+b} (2 a+5 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f}+\frac {(2 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}\\ \end {align*}

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Mathematica [A]  time = 0.51, size = 116, normalized size = 0.78 \[ \frac {(2 a+5 b) \left (\sqrt {a+b \sin ^2(e+f x)} \left (4 a+b \sin ^2(e+f x)+3 b\right )-3 (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )\right )+3 \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 f (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2) + (2*a + 5*b)*(-3*(a + b)^(3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]
^2]/Sqrt[a + b]] + Sqrt[a + b*Sin[e + f*x]^2]*(4*a + 3*b + b*Sin[e + f*x]^2)))/(6*(a + b)*f)

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fricas [A]  time = 0.82, size = 265, normalized size = 1.79 \[ \left [\frac {3 \, {\left (2 \, a + 5 \, b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{2} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, a - 3 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, f \cos \left (f x + e\right )^{2}}, \frac {3 \, {\left (2 \, a + 5 \, b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) \cos \left (f x + e\right )^{2} - {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, a - 3 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, f \cos \left (f x + e\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*(2*a + 5*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*b*cos(f*x + e)^4 - 2*(4*a + 7*b)*cos(f*x + e)^2 - 3*a - 3*b)*sqrt(
-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2), 1/6*(3*(2*a + 5*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*b*cos(f*x + e)^4 - 2*(4*a + 7*b)*cos(f*x + e)^2 - 3*a - 3*b
)*sqrt(-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2)]

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giac [B]  time = 2.38, size = 2185, normalized size = 14.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(2*a^2 + 7*a*b + 5*b^2)*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) - 6*(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^2 + 3*(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*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*tan(1/2*f*x + 1/2*e)^2 + a))^3*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))^2*a^(5/2) + 7*(
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)*b + 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 - 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^3 - 3*(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*b + 3*(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^2 + 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^3 - 2*a^(7/2) - 7*a^(5/2)*b - 9*a^(3/2)*b^2 - 4*s
qrt(a)*b^3)/((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 - 8*(3*(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))^5*a*b +
3*(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))^5*b^2 + 9*(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))^4*a^(3/2)*b + 15*(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))^4*sqrt(a)*b^2 + 6*(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^2*b + 30*(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*b^2 + 32*(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^3 - 6*(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^(5/2)*
b + 6*(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)*b^2 + 24*(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^3 - 9*(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^3*b - 33*(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))*a^2*b^2 + 48*(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^4 - 3*a^(7/2)*b - 21*a^(5/2)*b^2 - 56*a^(3/2)*b^3 - 48*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))^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)^3)/f

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maple [B]  time = 3.89, size = 567, normalized size = 3.83 \[ \frac {-4 \left (a +b \right )^{\frac {3}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, b \left (\cos ^{4}\left (f x +e \right )\right )-\left (-16 \left (a +b \right )^{\frac {3}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, a -28 \left (a +b \right )^{\frac {3}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, b +6 \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^{3}+27 a^{2} 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 )+36 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 )+15 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 )+6 \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^{3}+27 a^{2} 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 )+36 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}{1+\sin \left (f x +e \right )}\right )+15 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 )\right ) \left (\cos ^{2}\left (f x +e \right )\right )+6 \left (a +b \right )^{\frac {3}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, a +6 \left (a +b \right )^{\frac {3}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, b}{12 \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)^(3/2)*tan(f*x+e)^3,x)

[Out]

1/12*(-4*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*b*cos(f*x+e)^4-(-16*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a-2
8*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*b+6*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^3+27*a^2*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))+36*a*b^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))+15*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))+6*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+27*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))
+36*a*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))+15*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)))*cos(f*x+e)^2+6*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^
(1/2)*a+6*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*b)/(a+b)^(3/2)/cos(f*x+e)^2/f

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maxima [A]  time = 0.42, size = 169, normalized size = 1.14 \[ \frac {4 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{2} + 12 \, {\left (a b^{2} + 2 \, b^{3}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a} + \frac {3 \, {\left (2 \, a^{2} b^{2} + 7 \, a b^{3} + 5 \, b^{4}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{\sqrt {a + b}} - \frac {6 \, {\left (a b^{3} + b^{4}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}}{b \sin \left (f x + e\right )^{2} - b}}{12 \, b^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(b*sin(f*x + e)^2 + a)^(3/2)*b^2 + 12*(a*b^2 + 2*b^3)*sqrt(b*sin(f*x + e)^2 + a) + 3*(2*a^2*b^2 + 7*a*
b^3 + 5*b^4)*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) - 6*(a*b^3 + b^4)*sqrt(b*sin(f*x + e)^2 + a)/(b*sin(f*x + e)^2 - b))/(b^2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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