3.5.88 \(\int \sqrt {a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx\) [488]

Optimal. Leaf size=177 \[ \frac {\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{3/2} f}-\frac {\left (8 a^2+24 a b+15 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}-\frac {(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) f} \]

[Out]

1/8*(8*a^2+24*a*b+15*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)/f-1/8*(8*a+7*b)*sec(f*x+e)
^2*(a+b*sin(f*x+e)^2)^(3/2)/(a+b)^2/f+1/4*sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(3/2)/(a+b)/f-1/8*(8*a^2+24*a*b+15*b
^2)*(a+b*sin(f*x+e)^2)^(1/2)/(a+b)^2/f

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Rubi [A]
time = 0.14, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, 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+24 a b+15 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 f (a+b)^2}+\frac {\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 f (a+b)^{3/2}}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f (a+b)}-\frac {(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 f (a+b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x]^5,x]

[Out]

((8*a^2 + 24*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(3/2)*f) - ((8*a^2 + 24
*a*b + 15*b^2)*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)^2*f) - ((8*a + 7*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2
)^(3/2))/(8*(a + b)^2*f) + (Sec[e + f*x]^4*(a + b*Sin[e + f*x]^2)^(3/2))/(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 \sqrt {a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx &=\frac {\text {Subst}\left (\int \frac {x^2 \sqrt {a+b x}}{(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 )^{3/2}}{4 (a+b) f}-\frac {\text {Subst}\left (\int \frac {\sqrt {a+b x} \left (\frac {1}{2} (4 a+3 b)+2 (a+b) x\right )}{(1-x)^2} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f}\\ &=-\frac {(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) f}+\frac {\left (8 a^2+24 a b+15 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)^2 f}\\ &=-\frac {\left (8 a^2+24 a b+15 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}-\frac {(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) f}+\frac {\left (8 a^2+24 a b+15 b^2\right ) \text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b) f}\\ &=-\frac {\left (8 a^2+24 a b+15 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}-\frac {(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) f}+\frac {\left (8 a^2+24 a b+15 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 (a+b) f}\\ &=\frac {\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{3/2} f}-\frac {\left (8 a^2+24 a b+15 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}-\frac {(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) f}\\ \end {align*}

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Mathematica [A]
time = 0.39, size = 143, normalized size = 0.81 \begin {gather*} \frac {-\left ((8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )+2 (a+b) \sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}+\left (8 a^2+24 a b+15 b^2\right ) \left (\sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )-\sqrt {a+b \sin ^2(e+f x)}\right )}{8 (a+b)^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x]^5,x]

[Out]

(-((8*a + 7*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(3/2)) + 2*(a + b)*Sec[e + f*x]^4*(a + b*Sin[e + f*x]^2)^
(3/2) + (8*a^2 + 24*a*b + 15*b^2)*(Sqrt[a + b]*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] - Sqrt[a + b*Si
n[e + f*x]^2]))/(8*(a + b)^2*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(720\) vs. \(2(157)=314\).
time = 35.18, size = 721, normalized size = 4.07

method	result	size
default	\(\frac {\left (-16 \left (a +b \right )^{\frac {3}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, a^{2}-48 \left (a +b \right )^{\frac {3}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, a b -30 b^{2} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {3}{2}}+8 \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}+40 \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 +71 \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}+54 \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}+15 \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}+8 \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}+40 \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 +71 \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}+54 \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}+15 \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}\right ) \left (\cos ^{4}\left (f x +e \right )\right )-2 \left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (a +b \right )^{\frac {3}{2}} \left (8 a +7 b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+4 \left (a +b \right )^{\frac {3}{2}} \left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} a +4 b \left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (a +b \right )^{\frac {3}{2}}}{16 \left (a +b \right )^{\frac {3}{2}} \cos \left (f x +e \right )^{4} \left (a^{2}+2 a b +b^{2}\right ) f}\)	\(721\)

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)^5,x,method=_RETURNVERBOSE)

[Out]

1/16*((-16*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a^2-48*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a*b-30*b^2*(a+
b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)+8*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+40*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+71*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*b^2+54*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+15*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+8*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^4+40*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^3*b+71*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*b^2+54*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+15*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)*cos(f*x+e)^4-2*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(3/2)*(8*a+7*b)*cos(f*x+e)^2+4*(a+b)^(3/2)*(a+b-b*c
os(f*x+e)^2)^(3/2)*a+4*b*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(3/2))/(a+b)^(3/2)/cos(f*x+e)^4/(a^2+2*a*b+b^2)/f

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Maxima [A]
time = 0.51, size = 237, normalized size = 1.34 \begin {gather*} -\frac {16 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{3} + \frac {{\left (8 \, a^{2} b^{3} + 24 \, a b^{4} + 15 \, 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 )}{{\left (a + b\right )}^{\frac {3}{2}}} - \frac {2 \, {\left ({\left (8 \, a b^{4} + 9 \, b^{5}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - {\left (8 \, a^{2} b^{4} + 15 \, a b^{5} + 7 \, 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} {\left (a + b\right )} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} - 2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )} {\left (a^{2} + 2 \, a b + b^{2}\right )}}}{16 \, b^{3} f} \end {gather*}

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)^5,x, algorithm="maxima")

[Out]

-1/16*(16*sqrt(b*sin(f*x + e)^2 + a)*b^3 + (8*a^2*b^3 + 24*a*b^4 + 15*b^5)*log((sqrt(b*sin(f*x + e)^2 + a) - s
qrt(a + b))/(sqrt(b*sin(f*x + e)^2 + a) + sqrt(a + b)))/(a + b)^(3/2) - 2*((8*a*b^4 + 9*b^5)*(b*sin(f*x + e)^2
+ a)^(3/2) - (8*a^2*b^4 + 15*a*b^5 + 7*b^6)*sqrt(b*sin(f*x + e)^2 + a))/((b*sin(f*x + e)^2 + a)^2*(a + b) + a
^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(b*sin(f*x + e)^2 + a)*(a^2 + 2*a*b + b^2)))/(b^3*f)

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Fricas [A]
time = 0.93, size = 354, normalized size = 2.00 \begin {gather*} \left [\frac {{\left (8 \, a^{2} + 24 \, a b + 15 \, 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^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + {\left (8 \, a^{2} + 17 \, a b + 9 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4}}, -\frac {{\left (8 \, a^{2} + 24 \, a b + 15 \, 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^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + {\left (8 \, a^{2} + 17 \, a b + 9 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{2} + 2 \, a b + b^{2}\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)^(1/2)*tan(f*x+e)^5,x, algorithm="fricas")

[Out]

[1/16*((8*a^2 + 24*a*b + 15*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^2 + 2*a*b + b^2)*cos(f*x + e)^4 + (8*a^2 + 17*a*b +
9*b^2)*cos(f*x + e)^2 - 2*a^2 - 4*a*b - 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^2 + 2*a*b + b^2)*f*cos(f*
x + e)^4), -1/8*((8*a^2 + 24*a*b + 15*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^2 + 2*a*b + b^2)*cos(f*x + e)^4 + (8*a^2 + 17*a*b + 9*b^2)*cos(f*x + e)^2 - 2*a^
2 - 4*a*b - 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \tan ^{5}{\left (e + f x \right )}\, dx \end {gather*}

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)**5,x)

[Out]

Integral(sqrt(a + b*sin(e + f*x)**2)*tan(e + f*x)**5, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 2790 vs. \(2 (163) = 326\).
time = 2.05, size = 2790, normalized size = 15.76 \begin {gather*} \text {Too large to display} \end {gather*}

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)^5,x, algorithm="giac")

[Out]

-1/4*((8*a^2 + 24*a*b + 15*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))/((a + b)*sqrt(-a - b)) -
16*((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 - sqrt(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*t
an(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) - 2*(8*(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))^7*a^2 + 16*(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))^7*a*b + 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))^7*b^2 - 56*(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))^6*a^(5/2)
- 80*(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))^6*a^(3/2)*b - 17*(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))^6*sqrt(a)*b^2 - 120*(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^3 - 464*(
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^2*b - 425*(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^2 - 60*(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^3 + 136*(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^(7/2) + 144*(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^(5/2)*b - 425*(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^2 - 468*(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^3 + 344*(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^4 + 1520*(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^3*b + 2093*(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^2
+ 712*(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^3 - 240*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*ta
n(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*b^4 + 24*(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*a^(9/2) + 592*(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^(7/2)*b + 2165*(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^2 + 2808*(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*a^(3/2)*b^3 + 1232*(sqr
t(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^4 - 232*(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^5 - 1072*(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^4*b - 1675*(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^2 - 652*(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 + 624*(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 + 448*(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^5 - 104*a^(1
1/2) - 656*a^(9/2)*b - 1723*a^(7/2)*b^2 - 2340*...

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {tan}\left (e+f\,x\right )}^5\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,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)^(1/2),x)

[Out]

int(tan(e + f*x)^5*(a + b*sin(e + f*x)^2)^(1/2), x)

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