3.306 \(\int \tan ^5(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=145 \[ \frac {\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}+\frac {(a-b) \sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f} \]

[Out]

-(a-b)^(3/2)*arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))/f+(a-b)*(a+b*tan(f*x+e)^2)^(1/2)/f+1/3*(a+b*tan(f*x
+e)^2)^(3/2)/f-1/5*(a+b)*(a+b*tan(f*x+e)^2)^(5/2)/b^2/f+1/7*(a+b*tan(f*x+e)^2)^(7/2)/b^2/f

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Rubi [A]  time = 0.17, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3670, 446, 88, 50, 63, 208} \[ \frac {\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}+\frac {(a-b) \sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]])/f) + ((a - b)*Sqrt[a + b*Tan[e + f*x]^2])/f
+ (a + b*Tan[e + f*x]^2)^(3/2)/(3*f) - ((a + b)*(a + b*Tan[e + f*x]^2)^(5/2))/(5*b^2*f) + (a + b*Tan[e + f*x]^
2)^(7/2)/(7*b^2*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 3670

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

Rubi steps

\begin {align*} \int \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5 \left (a+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2 (a+b x)^{3/2}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {(-a-b) (a+b x)^{3/2}}{b}+\frac {(a+b x)^{3/2}}{1+x}+\frac {(a+b x)^{5/2}}{b}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac {(a-b) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {(a-b) \sqrt {a+b \tan ^2(e+f x)}}{f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac {(a-b)^2 \operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {(a-b) \sqrt {a+b \tan ^2(e+f x)}}{f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac {(a-b)^2 \operatorname {Subst}\left (\int \frac {1}{1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \tan ^2(e+f x)}\right )}{b f}\\ &=-\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {(a-b) \sqrt {a+b \tan ^2(e+f x)}}{f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}\\ \end {align*}

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Mathematica [A]  time = 1.39, size = 139, normalized size = 0.96 \[ \frac {\frac {2 \left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2}-\frac {2 (a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2}+\frac {2}{3} \left (a+b \tan ^2(e+f x)\right )^{3/2}+2 (a-b) \left (\sqrt {a+b \tan ^2(e+f x)}-\sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(a + b*Tan[e + f*x]^2)^(3/2))/3 - (2*(a + b)*(a + b*Tan[e + f*x]^2)^(5/2))/(5*b^2) + (2*(a + b*Tan[e + f*x
]^2)^(7/2))/(7*b^2) + 2*(a - b)*(-(Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]]) + Sqrt[a + b*T
an[e + f*x]^2]))/(2*f)

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fricas [A]  time = 0.55, size = 414, normalized size = 2.86 \[ \left [-\frac {105 \, {\left (a b^{2} - b^{3}\right )} \sqrt {a - b} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left (15 \, b^{3} \tan \left (f x + e\right )^{6} + 3 \, {\left (8 \, a b^{2} - 7 \, b^{3}\right )} \tan \left (f x + e\right )^{4} - 6 \, a^{3} - 21 \, a^{2} b + 140 \, a b^{2} - 105 \, b^{3} + {\left (3 \, a^{2} b - 42 \, a b^{2} + 35 \, b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{420 \, b^{2} f}, \frac {105 \, {\left (a b^{2} - b^{3}\right )} \sqrt {-a + b} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, {\left (15 \, b^{3} \tan \left (f x + e\right )^{6} + 3 \, {\left (8 \, a b^{2} - 7 \, b^{3}\right )} \tan \left (f x + e\right )^{4} - 6 \, a^{3} - 21 \, a^{2} b + 140 \, a b^{2} - 105 \, b^{3} + {\left (3 \, a^{2} b - 42 \, a b^{2} + 35 \, b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{210 \, b^{2} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/420*(105*(a*b^2 - b^3)*sqrt(a - b)*log(-(b^2*tan(f*x + e)^4 + 2*(4*a*b - 3*b^2)*tan(f*x + e)^2 + 4*(b*tan(
f*x + e)^2 + 2*a - b)*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 8*a^2 - 8*a*b + b^2)/(tan(f*x + e)^4 + 2*tan(f*
x + e)^2 + 1)) - 4*(15*b^3*tan(f*x + e)^6 + 3*(8*a*b^2 - 7*b^3)*tan(f*x + e)^4 - 6*a^3 - 21*a^2*b + 140*a*b^2
- 105*b^3 + (3*a^2*b - 42*a*b^2 + 35*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f), 1/210*(105*(a*b
^2 - b^3)*sqrt(-a + b)*arctan(2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(b*tan(f*x + e)^2 + 2*a - b)) + 2*(15*
b^3*tan(f*x + e)^6 + 3*(8*a*b^2 - 7*b^3)*tan(f*x + e)^4 - 6*a^3 - 21*a^2*b + 140*a*b^2 - 105*b^3 + (3*a^2*b -
42*a*b^2 + 35*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f)]

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giac [A]  time = 0.89, size = 196, normalized size = 1.35 \[ \frac {{\left (a^{2} - 2 \, a b + b^{2}\right )} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {-a + b}}\right )}{\sqrt {-a + b} f} + \frac {15 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {7}{2}} b^{12} f^{6} - 21 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} a b^{12} f^{6} - 21 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} b^{13} f^{6} + 35 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{14} f^{6} + 105 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} a b^{14} f^{6} - 105 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b^{15} f^{6}}{105 \, b^{14} f^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

(a^2 - 2*a*b + b^2)*arctan(sqrt(b*tan(f*x + e)^2 + a)/sqrt(-a + b))/(sqrt(-a + b)*f) + 1/105*(15*(b*tan(f*x +
e)^2 + a)^(7/2)*b^12*f^6 - 21*(b*tan(f*x + e)^2 + a)^(5/2)*a*b^12*f^6 - 21*(b*tan(f*x + e)^2 + a)^(5/2)*b^13*f
^6 + 35*(b*tan(f*x + e)^2 + a)^(3/2)*b^14*f^6 + 105*sqrt(b*tan(f*x + e)^2 + a)*a*b^14*f^6 - 105*sqrt(b*tan(f*x
+ e)^2 + a)*b^15*f^6)/(b^14*f^7)

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maple [B]  time = 0.34, size = 256, normalized size = 1.77 \[ \frac {\left (\tan ^{2}\left (f x +e \right )\right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {5}{2}}}{7 f b}-\frac {2 a \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {5}{2}}}{35 f \,b^{2}}-\frac {\left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {5}{2}}}{5 b f}+\frac {b \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{3 f}+\frac {4 a \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{3 f}-\frac {b \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{f}+\frac {b^{2} \arctan \left (\frac {\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}-\frac {2 a b \arctan \left (\frac {\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}+\frac {a^{2} \arctan \left (\frac {\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7/f*tan(f*x+e)^2*(a+b*tan(f*x+e)^2)^(5/2)/b-2/35/f*a/b^2*(a+b*tan(f*x+e)^2)^(5/2)-1/5*(a+b*tan(f*x+e)^2)^(5/
2)/b/f+1/3/f*b*tan(f*x+e)^2*(a+b*tan(f*x+e)^2)^(1/2)+4/3/f*a*(a+b*tan(f*x+e)^2)^(1/2)-b*(a+b*tan(f*x+e)^2)^(1/
2)/f+1/f*b^2/(-a+b)^(1/2)*arctan((a+b*tan(f*x+e)^2)^(1/2)/(-a+b)^(1/2))-2/f*a*b/(-a+b)^(1/2)*arctan((a+b*tan(f
*x+e)^2)^(1/2)/(-a+b)^(1/2))+1/f*a^2/(-a+b)^(1/2)*arctan((a+b*tan(f*x+e)^2)^(1/2)/(-a+b)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 40.84, size = 233, normalized size = 1.61 \[ \frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{7/2}}{7\,b^2\,f}-\left (\frac {2\,a}{5\,b^2\,f}-\frac {a-b}{5\,b^2\,f}\right )\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}-\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\left (a-b\right )\,\left (\left (\frac {2\,a}{b^2\,f}-\frac {a-b}{b^2\,f}\right )\,\left (a-b\right )-\frac {a^2}{b^2\,f}\right )-{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}\,\left (\frac {\left (\frac {2\,a}{b^2\,f}-\frac {a-b}{b^2\,f}\right )\,\left (a-b\right )}{3}-\frac {a^2}{3\,b^2\,f}\right )+\frac {\mathrm {atan}\left (\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,{\left (a-b\right )}^{3/2}\,1{}\mathrm {i}}{a^2-2\,a\,b+b^2}\right )\,{\left (a-b\right )}^{3/2}\,1{}\mathrm {i}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a + b*tan(e + f*x)^2)^(7/2)/(7*b^2*f) - ((2*a)/(5*b^2*f) - (a - b)/(5*b^2*f))*(a + b*tan(e + f*x)^2)^(5/2) -
(a + b*tan(e + f*x)^2)^(1/2)*(a - b)*(((2*a)/(b^2*f) - (a - b)/(b^2*f))*(a - b) - a^2/(b^2*f)) - (a + b*tan(e
+ f*x)^2)^(3/2)*((((2*a)/(b^2*f) - (a - b)/(b^2*f))*(a - b))/3 - a^2/(3*b^2*f)) + (atan(((a + b*tan(e + f*x)^2
)^(1/2)*(a - b)^(3/2)*1i)/(a^2 - 2*a*b + b^2))*(a - b)^(3/2)*1i)/f

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{5}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**5*(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Integral((a + b*tan(e + f*x)**2)**(3/2)*tan(e + f*x)**5, x)

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