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

Optimal. Leaf size=80 \[ \frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b f}-\frac {\sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 80, 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 = {4139, 446, 80, 50, 63, 208} \[ \frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b f}-\frac {\sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/f - Sqrt[a + b*Sec[e + f*x]^2]/f + (a + b*Sec[e + f*x]^2
)^(3/2)/(3*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 80

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

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4139

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p)/x
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])

Rubi steps

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

________________________________________________________________________________________

Mathematica [F]  time = 1.01, size = 0, normalized size = 0.00 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^3(e+f x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.24, size = 386, normalized size = 4.82 \[ \left [\frac {3 \, \sqrt {a} b \cos \left (f x + e\right )^{2} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, {\left ({\left (a - 3 \, b\right )} \cos \left (f x + e\right )^{2} + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, b f \cos \left (f x + e\right )^{2}}, -\frac {3 \, \sqrt {-a} b \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a - 3 \, b\right )} \cos \left (f x + e\right )^{2} + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, b f \cos \left (f x + e\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(3*sqrt(a)*b*cos(f*x + e)^2*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x
+ e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 + 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x
+ e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) + 8*((a - 3*b)*cos(f*x + e)
^2 + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b*f*cos(f*x + e)^2), -1/12*(3*sqrt(-a)*b*arctan(1/4*(8*a
^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(2*a^3*co
s(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2))*cos(f*x + e)^2 - 4*((a - 3*b)*cos(f*x + e)^2 + b)*sqrt((a*cos(
f*x + e)^2 + b)/cos(f*x + e)^2))/(b*f*cos(f*x + e)^2)]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)16*(1/12*(3*a*(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4
-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^5+(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*ta
n(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))*
(-9*a^3-48*b^3+15*a*b^2+6*a^2*b)+sqrt(a+b)*(9*a-12*b)*(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+
exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^4+(-sqrt(a+
b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^
2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^3*(6*a^2+16*b^2-18*a*b)+sqrt(a+b)*(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt
(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a
+b))^2*(-6*a^2+24*b^2-6*a*b)+sqrt(a+b)*(-3*a^3+20*b^3-19*a*b^2+6*a^2*b))/(-2*sqrt(a+b)*(-sqrt(a+b)*tan(1/2*(f*
x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*
(f*x+exp(1)))^2+a+b))-(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)
))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^2-a+3*b)^3-1/8*a*atan(1/2*(-sqrt(a+b)*tan(1
/2*(f*x+exp(1)))^2-sqrt(a+b)+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1))
)^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))/sqrt(-a))/sqrt(-a))*sign(cos(f*x+exp(1)))/f

________________________________________________________________________________________

maple [B]  time = 1.56, size = 648, normalized size = 8.10 \[ -\frac {\sqrt {4}\, \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}}\, \left (-1+\cos \left (f x +e \right )\right ) \left (3 \sqrt {a +b}\, \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {a}\, \ln \left (4 \cos \left (f x +e \right ) \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}+4 a \cos \left (f x +e \right )+4 \sqrt {a}\, \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right ) b +\sqrt {a +b}\, \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a -3 \sqrt {a +b}\, \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b -3 \left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (-\frac {4 \left (-1+\cos \left (f x +e \right )\right ) \left (\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right ) \sqrt {a +b}+\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a +b}-a \cos \left (f x +e \right )+b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a +b}}\right ) a b +3 \left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (-\frac {2 \left (-1+\cos \left (f x +e \right )\right ) \left (\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right ) \sqrt {a +b}+\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a +b}-a \cos \left (f x +e \right )+b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a +b}}\right ) a b +\sqrt {a +b}\, \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a -3 \sqrt {a +b}\, \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b +\sqrt {a +b}\, \cos \left (f x +e \right ) \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b +\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b \sqrt {a +b}\right )}{6 f \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{2} \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a +b}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sec(f*x + e)^2 + a)*tan(f*x + e)^3, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________