3.5.99 \(\int (a+b \sin ^2(e+f x))^{3/2} \tan ^5(e+f x) \, dx\) [499]

Optimal. Leaf size=220 \[ \frac {\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 \sqrt {a+b} f}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b) f}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 (a+b)^2 f}-\frac {(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f} \]

[Out]

-1/24*(8*a^2+40*a*b+35*b^2)*(a+b*sin(f*x+e)^2)^(3/2)/(a+b)^2/f-1/8*(8*a+9*b)*sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(
5/2)/(a+b)^2/f+1/4*sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(5/2)/(a+b)/f+1/8*(8*a^2+40*a*b+35*b^2)*arctanh((a+b*sin(f*
x+e)^2)^(1/2)/(a+b)^(1/2))/f/(a+b)^(1/2)-1/8*(8*a^2+40*a*b+35*b^2)*(a+b*sin(f*x+e)^2)^(1/2)/(a+b)/f

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Rubi [A]
time = 0.17, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3273, 91, 79, 52, 65, 214} \begin {gather*} -\frac {\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 f (a+b)^2}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 f (a+b)}+\frac {\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 f \sqrt {a+b}}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 f (a+b)}-\frac {(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 f (a+b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x]^5,x]

[Out]

((8*a^2 + 40*a*b + 35*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*Sqrt[a + b]*f) - ((8*a^2 + 40*a
*b + 35*b^2)*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)*f) - ((8*a^2 + 40*a*b + 35*b^2)*(a + b*Sin[e + f*x]^2)^(3/
2))/(24*(a + b)^2*f) - ((8*a + 9*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2))/(8*(a + b)^2*f) + (Sec[e + f*
x]^4*(a + b*Sin[e + f*x]^2)^(5/2))/(4*(a + b)*f)

Rule 52

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 65

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 79

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] || L
tQ[p, n]))))

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 214

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

Rule 3273

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 ^5(e+f x) \, dx &=\frac {\text {Subst}\left (\int \frac {x^2 (a+b x)^{3/2}}{(1-x)^3} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}-\frac {\text {Subst}\left (\int \frac {(a+b x)^{3/2} \left (\frac {1}{2} (4 a+5 b)+2 (a+b) x\right )}{(1-x)^2} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f}\\ &=-\frac {(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}+\frac {\left (8 a^2+40 a b+35 b^2\right ) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b)^2 f}\\ &=-\frac {\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 (a+b)^2 f}-\frac {(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}+\frac {\left (8 a^2+40 a b+35 b^2\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b) f}\\ &=-\frac {\left (8 a^2+40 a b+35 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b) f}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 (a+b)^2 f}-\frac {(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}+\frac {\left (8 a^2+40 a b+35 b^2\right ) \text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 f}\\ &=-\frac {\left (8 a^2+40 a b+35 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b) f}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 (a+b)^2 f}-\frac {(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}+\frac {\left (8 a^2+40 a b+35 b^2\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^2(e+f x)}\right )}{8 b f}\\ &=\frac {\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 \sqrt {a+b} f}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b) f}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 (a+b)^2 f}-\frac {(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}\\ \end {align*}

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Mathematica [A]
time = 1.34, size = 160, normalized size = 0.73 \begin {gather*} -\frac {3 (8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}-6 (a+b) \sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}+\left (8 a^2+40 a b+35 b^2\right ) \left (-3 (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )+\sqrt {a+b \sin ^2(e+f x)} \left (4 a+3 b+b \sin ^2(e+f x)\right )\right )}{24 (a+b)^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x]^5,x]

[Out]

-1/24*(3*(8*a + 9*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2) - 6*(a + b)*Sec[e + f*x]^4*(a + b*Sin[e + f*x
]^2)^(5/2) + (8*a^2 + 40*a*b + 35*b^2)*(-3*(a + b)^(3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] + Sqr
t[a + b*Sin[e + f*x]^2]*(4*a + 3*b + b*Sin[e + f*x]^2)))/((a + b)^2*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(710\) vs. \(2(196)=392\).
time = 25.31, size = 711, normalized size = 3.23

method	result	size
default	\(\frac {16 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}} b \left (\cos ^{6}\left (f x +e \right )\right )+\left (-64 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}} a -160 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}} b +24 \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^{4}+168 \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} b +369 \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} b^{2}+330 \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}+105 \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^{4}+24 \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^{4}+168 \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} b +369 \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} b^{2}+330 \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}+105 \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^{4}\right ) \left (\cos ^{4}\left (f x +e \right )\right )-6 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}} \left (8 a +13 b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+12 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}} a +12 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}} b}{48 \left (a +b \right )^{\frac {5}{2}} \cos \left (f x +e \right )^{4} f}\)	\(711\)

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)^5,x,method=_RETURNVERBOSE)

[Out]

1/48*(16*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*b*cos(f*x+e)^6+(-64*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*a-1
60*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*b+24*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*s
in(f*x+e)+a))*a^4+168*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*b+369*l
n(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*b^2+330*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^3+105*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^4+24*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^4+168*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*b+369*ln(2/(1+s
in(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*b^2+330*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+105*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^4)*cos(f*x+e)^4-6*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*(8*a+13*b)*cos(f*x+e)^2+12*
(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*a+12*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*b)/(a+b)^(5/2)/cos(f*x+e)^4
/f

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Maxima [A]
time = 0.51, size = 246, normalized size = 1.12 \begin {gather*} -\frac {16 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{3} + 48 \, {\left (a b^{3} + 3 \, b^{4}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a} + \frac {3 \, {\left (8 \, a^{2} b^{3} + 40 \, a b^{4} + 35 \, b^{5}\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 ({\left (8 \, a b^{4} + 13 \, b^{5}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - {\left (8 \, a^{2} b^{4} + 19 \, a b^{5} + 11 \, b^{6}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}\right )}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{2} - 2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )} {\left (a + b\right )} + a^{2} + 2 \, a b + b^{2}}}{48 \, b^{3} f} \end {gather*}

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)^5,x, algorithm="maxima")

[Out]

-1/48*(16*(b*sin(f*x + e)^2 + a)^(3/2)*b^3 + 48*(a*b^3 + 3*b^4)*sqrt(b*sin(f*x + e)^2 + a) + 3*(8*a^2*b^3 + 40
*a*b^4 + 35*b^5)*log((sqrt(b*sin(f*x + e)^2 + a) - sqrt(a + b))/(sqrt(b*sin(f*x + e)^2 + a) + sqrt(a + b)))/sq
rt(a + b) - 6*((8*a*b^4 + 13*b^5)*(b*sin(f*x + e)^2 + a)^(3/2) - (8*a^2*b^4 + 19*a*b^5 + 11*b^6)*sqrt(b*sin(f*
x + e)^2 + a))/((b*sin(f*x + e)^2 + a)^2 - 2*(b*sin(f*x + e)^2 + a)*(a + b) + a^2 + 2*a*b + b^2))/(b^3*f)

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Fricas [A]
time = 0.79, size = 385, normalized size = 1.75 \begin {gather*} \left [\frac {3 \, {\left (8 \, a^{2} + 40 \, a b + 35 \, b^{2}\right )} \sqrt {a + b} \cos \left (f x + e\right )^{4} \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 (8 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{6} - 16 \, {\left (2 \, a^{2} + 7 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 3 \, {\left (8 \, a^{2} + 21 \, a b + 13 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, a^{2} + 12 \, a b + 6 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{48 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{4}}, -\frac {3 \, {\left (8 \, a^{2} + 40 \, a b + 35 \, b^{2}\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 )^{4} - {\left (8 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{6} - 16 \, {\left (2 \, a^{2} + 7 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 3 \, {\left (8 \, a^{2} + 21 \, a b + 13 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, a^{2} + 12 \, a b + 6 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{24 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{4}}\right ] \end {gather*}

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)^5,x, algorithm="fricas")

[Out]

[1/48*(3*(8*a^2 + 40*a*b + 35*b^2)*sqrt(a + b)*cos(f*x + e)^4*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*(8*(a*b + b^2)*cos(f*x + e)^6 - 16*(2*a^2 + 7*a*b + 5*b
^2)*cos(f*x + e)^4 - 3*(8*a^2 + 21*a*b + 13*b^2)*cos(f*x + e)^2 + 6*a^2 + 12*a*b + 6*b^2)*sqrt(-b*cos(f*x + e)
^2 + a + b))/((a + b)*f*cos(f*x + e)^4), -1/24*(3*(8*a^2 + 40*a*b + 35*b^2)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*
x + e)^2 + a + b)*sqrt(-a - b)/(a + b))*cos(f*x + e)^4 - (8*(a*b + b^2)*cos(f*x + e)^6 - 16*(2*a^2 + 7*a*b + 5
*b^2)*cos(f*x + e)^4 - 3*(8*a^2 + 21*a*b + 13*b^2)*cos(f*x + e)^2 + 6*a^2 + 12*a*b + 6*b^2)*sqrt(-b*cos(f*x +
e)^2 + a + b))/((a + b)*f*cos(f*x + e)^4)]

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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)**5,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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)^5,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 1.54Error: Bad Argument Type

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {tan}\left (e+f\,x\right )}^5\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^5*(a + b*sin(e + f*x)^2)^(3/2),x)

[Out]

int(tan(e + f*x)^5*(a + b*sin(e + f*x)^2)^(3/2), x)

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