3.528 \(\int \frac {\tan ^5(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\)

Optimal. Leaf size=229 \[ -\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 b^4 d}+\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{105 b^3 d}-\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{35 b^2 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}} \]

[Out]

-arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d/(a-I*b)^(1/2)-arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d
/(a+I*b)^(1/2)-4/105*a*(24*a^2-35*b^2)*(a+b*tan(d*x+c))^(1/2)/b^4/d+2/105*(24*a^2-35*b^2)*(a+b*tan(d*x+c))^(1/
2)*tan(d*x+c)/b^3/d-12/35*a*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^2/b^2/d+2/7*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^3/
b/d

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Rubi [A]  time = 0.56, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3566, 3647, 3648, 3630, 12, 3539, 3537, 63, 208} \[ \frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{105 b^3 d}-\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 b^4 d}-\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{35 b^2 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/Sqrt[a + b*Tan[c + d*x]],x]

[Out]

-(ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]]/(Sqrt[a - I*b]*d)) - ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a
+ I*b]]/(Sqrt[a + I*b]*d) - (4*a*(24*a^2 - 35*b^2)*Sqrt[a + b*Tan[c + d*x]])/(105*b^4*d) + (2*(24*a^2 - 35*b^
2)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(105*b^3*d) - (12*a*Tan[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]])/(35*b^2
*d) + (2*Tan[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]])/(7*b*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3648

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m
+ n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m
+ n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b - b*C)*(m + n + 1)*Tan[e + f*x] - C*m*(b*c - a*d)*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \frac {\tan ^5(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx &=\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}+\frac {2 \int \frac {\tan ^2(c+d x) \left (-3 a-\frac {7}{2} b \tan (c+d x)-3 a \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx}{7 b}\\ &=-\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{35 b^2 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}+\frac {4 \int \frac {\tan (c+d x) \left (6 a^2+\frac {1}{4} \left (24 a^2-35 b^2\right ) \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx}{35 b^2}\\ &=\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{105 b^3 d}-\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{35 b^2 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}+\frac {8 \int \frac {-\frac {1}{4} a \left (24 a^2-35 b^2\right )+\frac {105}{8} b^3 \tan (c+d x)-\frac {1}{4} a \left (24 a^2-35 b^2\right ) \tan ^2(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx}{105 b^3}\\ &=-\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 b^4 d}+\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{105 b^3 d}-\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{35 b^2 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}+\frac {8 \int \frac {105 b^3 \tan (c+d x)}{8 \sqrt {a+b \tan (c+d x)}} \, dx}{105 b^3}\\ &=-\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 b^4 d}+\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{105 b^3 d}-\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{35 b^2 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}+\int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 b^4 d}+\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{105 b^3 d}-\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{35 b^2 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}+\frac {1}{2} i \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} i \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 b^4 d}+\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{105 b^3 d}-\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{35 b^2 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}+\frac {\operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}+\frac {\operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 b^4 d}+\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{105 b^3 d}-\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{35 b^2 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {i \operatorname {Subst}\left (\int \frac {1}{-1+\frac {i a}{b}-\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}+\frac {i \operatorname {Subst}\left (\int \frac {1}{-1-\frac {i a}{b}+\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d}-\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 b^4 d}+\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{105 b^3 d}-\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{35 b^2 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}\\ \end {align*}

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Mathematica [A]  time = 3.47, size = 185, normalized size = 0.81 \[ \frac {\frac {2 \sqrt {a+b \tan (c+d x)} \left (-48 a^3+b \left (24 a^2-35 b^2\right ) \tan (c+d x)-18 a b^2 \tan ^2(c+d x)+70 a b^2+15 b^3 \tan ^3(c+d x)\right )}{b^4}-\frac {105 \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{\sqrt {a-\sqrt {-b^2}}}-\frac {105 \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\sqrt {a+\sqrt {-b^2}}}}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/Sqrt[a + b*Tan[c + d*x]],x]

[Out]

((-105*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - Sqrt[-b^2]]])/Sqrt[a - Sqrt[-b^2]] - (105*ArcTanh[Sqrt[a + b*
Tan[c + d*x]]/Sqrt[a + Sqrt[-b^2]]])/Sqrt[a + Sqrt[-b^2]] + (2*Sqrt[a + b*Tan[c + d*x]]*(-48*a^3 + 70*a*b^2 +
b*(24*a^2 - 35*b^2)*Tan[c + d*x] - 18*a*b^2*Tan[c + d*x]^2 + 15*b^3*Tan[c + d*x]^3))/b^4)/(105*d)

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fricas [B]  time = 0.96, size = 2309, normalized size = 10.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-1/420*(420*sqrt(2)*(a^2*b^4 + b^6)*d^5*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*s
qrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*(1/((a^2 + b^2)*d^4))^(3/4)*arctan(-((a^4 + 2*a^2*b^2 + b^4)*d^4*sqrt(b
^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^3 + a*b^2)*d^2*sqrt(b^2/((a^4 + 2*a^2*b^2 + b
^4)*d^4)) - sqrt(2)*((a^5 + 2*a^3*b^2 + a*b^4)*d^7*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)
*d^4)) + (a^4 + 2*a^2*b^2 + b^4)*d^5*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(((a^2 + b^2)*d^2*sqrt(1/((a
^2 + b^2)*d^4))*cos(d*x + c) + sqrt(2)*((a^2 + b^2)*d^3*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x +
c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) -
a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4) + a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*
b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(3/4) + sqrt(2)*((a^5 + 2*a^3*b^2 +
a*b^4)*d^7*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^4 + 2*a^2*b^2 + b^4)*d^5*sq
rt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^
2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(3/4))/b^2)*cos(d*x + c)^3 + 420*sqrt
(2)*(a^2*b^4 + b^6)*d^5*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*sqrt(b^2/((a^4 +
2*a^2*b^2 + b^4)*d^4))*(1/((a^2 + b^2)*d^4))^(3/4)*arctan(((a^4 + 2*a^2*b^2 + b^4)*d^4*sqrt(b^2/((a^4 + 2*a^2*
b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^3 + a*b^2)*d^2*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + sqrt(
2)*((a^5 + 2*a^3*b^2 + a*b^4)*d^7*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^4 + 2
*a^2*b^2 + b^4)*d^5*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(((a^2 + b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4))*c
os(d*x + c) - sqrt(2)*((a^2 + b^2)*d^3*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x + c))*sqrt((a*cos(
d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*
(1/((a^2 + b^2)*d^4))^(1/4) + a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/(
(a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(3/4) - sqrt(2)*((a^5 + 2*a^3*b^2 + a*b^4)*d^7*sqrt(
b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^4 + 2*a^2*b^2 + b^4)*d^5*sqrt(b^2/((a^4 + 2*
a^2*b^2 + b^4)*d^4)))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a
^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(3/4))/b^2)*cos(d*x + c)^3 + 105*sqrt(2)*(a*b^4*d^3*sq
rt(1/((a^2 + b^2)*d^4))*cos(d*x + c)^3 + b^4*d*cos(d*x + c)^3)*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^
4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4)*log(((a^2 + b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c)
+ sqrt(2)*((a^2 + b^2)*d^3*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x + c))*sqrt((a*cos(d*x + c) + b
*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b
^2)*d^4))^(1/4) + a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c)) - 105*sqrt(2)*(a*b^4*d^3*sqrt(1/((a^2 + b^2)*
d^4))*cos(d*x + c)^3 + b^4*d*cos(d*x + c)^3)*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b
^2)*(1/((a^2 + b^2)*d^4))^(1/4)*log(((a^2 + b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) - sqrt(2)*((a^2 +
b^2)*d^3*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x + c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos
(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4) +
a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c)) + 8*(18*a*b^2*cos(d*x + c) + 8*(6*a^3 - 11*a*b^2)*cos(d*x + c)^
3 - (15*b^3 + 2*(12*a^2*b - 25*b^3)*cos(d*x + c)^2)*sin(d*x + c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d
*x + c)))/(b^4*d*cos(d*x + c)^3)

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giac [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(tan(d*x+c)^5/(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.32, size = 607, normalized size = 2.65 \[ \frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d \,b^{4}}-\frac {6 \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}} a}{5 d \,b^{4}}+\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}} a^{2}}{d \,b^{4}}-\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d \,b^{2}}-\frac {2 a^{3} \sqrt {a +b \tan \left (d x +c \right )}}{d \,b^{4}}+\frac {2 a \sqrt {a +b \tan \left (d x +c \right )}}{b^{2} d}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d \sqrt {a^{2}+b^{2}}}-\frac {\arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {a^{2}+b^{2}}\, \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 d \sqrt {a^{2}+b^{2}}}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {a^{2}+b^{2}}\, \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+b*tan(d*x+c))^(1/2),x)

[Out]

2/7/d/b^4*(a+b*tan(d*x+c))^(7/2)-6/5/d/b^4*(a+b*tan(d*x+c))^(5/2)*a+2/d/b^4*(a+b*tan(d*x+c))^(3/2)*a^2-2/3/d/b
^2*(a+b*tan(d*x+c))^(3/2)-2/d/b^4*a^3*(a+b*tan(d*x+c))^(1/2)+2*a*(a+b*tan(d*x+c))^(1/2)/b^2/d-1/4/d/(a^2+b^2)^
(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^
2+b^2)^(1/2))-1/d/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^
(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c)
)^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+1/4/d/(a^2+b^2)^(1/2)*(2*(a^2+b^2)^(1/2)
+2*a)^(1/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))+1/d/(a^2+b
^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^
2+b^2)^(1/2)-2*a)^(1/2))*a-1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(
d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(tan(d*x + c)^5/sqrt(b*tan(d*x + c) + a), x)

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mupad [B]  time = 11.48, size = 938, normalized size = 4.10 \[ \sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (2\,a\,\left (\frac {8\,a^2}{b^4\,d}-\frac {2\,\left (a^2+b^2\right )}{b^4\,d}\right )-\frac {20\,a^3}{b^4\,d}+\frac {6\,a\,\left (a^2+b^2\right )}{b^4\,d}\right )+\left (\frac {8\,a^2}{3\,b^4\,d}-\frac {2\,\left (a^2+b^2\right )}{3\,b^4\,d}\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}+\frac {2\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{7/2}}{7\,b^4\,d}-\frac {6\,a\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,b^4\,d}-\mathrm {atan}\left (-\frac {b^2\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {16\,b^2}{d}-\frac {64\,a^2\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^3\,d^2\,64{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}+\frac {a^2\,b^2\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,128{}\mathrm {i}}{\frac {64\,b^4}{d}+\frac {64\,a^2\,b^2}{d}-\frac {256\,a^2\,b^4\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a^3\,b^3\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}-\frac {256\,a^4\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^5\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}+\frac {128\,a\,b^3\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {64\,b^4}{d}+\frac {64\,a^2\,b^2}{d}-\frac {256\,a^2\,b^4\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a^3\,b^3\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}-\frac {256\,a^4\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^5\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}\right )\,\sqrt {\frac {a-b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {b^2\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,16{}\mathrm {i}}{\frac {16\,b^2}{d}-\frac {16\,a\,b^2\,d^2}{a\,d^3-b\,d^3\,1{}\mathrm {i}}}+\frac {a\,b^2\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,16{}\mathrm {i}}{\frac {b^3\,16{}\mathrm {i}}{d}-\frac {16\,a\,b^2}{d}-\frac {a\,b^3\,d^2\,16{}\mathrm {i}}{a\,d^3-b\,d^3\,1{}\mathrm {i}}+\frac {16\,a^2\,b^2\,d^2}{a\,d^3-b\,d^3\,1{}\mathrm {i}}}\right )\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}}\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^5/(a + b*tan(c + d*x))^(1/2),x)

[Out]

(a + b*tan(c + d*x))^(1/2)*(2*a*((8*a^2)/(b^4*d) - (2*(a^2 + b^2))/(b^4*d)) - (20*a^3)/(b^4*d) + (6*a*(a^2 + b
^2))/(b^4*d)) - atan((a^2*b^2*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan(c
+ d*x))^(1/2)*128i)/((64*b^4)/d + (64*a^2*b^2)/d - (256*a^2*b^4*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a^3*b^3*d^2*25
6i)/(4*a^2*d^3 + 4*b^2*d^3) - (256*a^4*b^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^5*d^2*256i)/(4*a^2*d^3 + 4*b^2*
d^3)) - (b^2*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*32i
)/((16*b^2)/d - (64*a^2*b^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^3*d^2*64i)/(4*a^2*d^3 + 4*b^2*d^3)) + (128*a*b
^3*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((64*b^4)/d
+ (64*a^2*b^2)/d - (256*a^2*b^4*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a^3*b^3*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) - (2
56*a^4*b^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^5*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3)))*((a - b*1i)/(4*a^2*d^2 +
4*b^2*d^2))^(1/2)*2i + ((8*a^2)/(3*b^4*d) - (2*(a^2 + b^2))/(3*b^4*d))*(a + b*tan(c + d*x))^(3/2) + atan((b^2*
(1/(a*d^2 - b*d^2*1i))^(1/2)*(a + b*tan(c + d*x))^(1/2)*16i)/((16*b^2)/d - (16*a*b^2*d^2)/(a*d^3 - b*d^3*1i))
+ (a*b^2*(1/(a*d^2 - b*d^2*1i))^(1/2)*(a + b*tan(c + d*x))^(1/2)*16i)/((b^3*16i)/d - (16*a*b^2)/d - (a*b^3*d^2
*16i)/(a*d^3 - b*d^3*1i) + (16*a^2*b^2*d^2)/(a*d^3 - b*d^3*1i)))*(1/(a*d^2 - b*d^2*1i))^(1/2)*1i + (2*(a + b*t
an(c + d*x))^(7/2))/(7*b^4*d) - (6*a*(a + b*tan(c + d*x))^(5/2))/(5*b^4*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5/(a+b*tan(d*x+c))**(1/2),x)

[Out]

Integral(tan(c + d*x)**5/sqrt(a + b*tan(c + d*x)), x)

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