\(\int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\) [519]

Optimal result

Integrand size = 23, antiderivative size = 211 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {(a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}-\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d} \]

[Out]

(a-I*b)^(5/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+(a+I*b)^(5/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+
I*b)^(1/2))/d-2*(a^2-b^2)*(a+b*tan(d*x+c))^(1/2)/d-2/3*a*(a+b*tan(d*x+c))^(3/2)/d-2/5*(a+b*tan(d*x+c))^(5/2)/d
-4/63*a*(a+b*tan(d*x+c))^(7/2)/b^2/d+2/9*tan(d*x+c)*(a+b*tan(d*x+c))^(7/2)/b/d

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3647, 3711, 12, 3609, 3620, 3618, 65, 214} \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=-\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {(a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}-\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d} \]

[In]

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

[Out]

((a - I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + ((a + I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[
c + d*x]]/Sqrt[a + I*b]])/d - (2*(a^2 - b^2)*Sqrt[a + b*Tan[c + d*x]])/d - (2*a*(a + b*Tan[c + d*x])^(3/2))/(3
*d) - (2*(a + b*Tan[c + d*x])^(5/2))/(5*d) - (4*a*(a + b*Tan[c + d*x])^(7/2))/(63*b^2*d) + (2*Tan[c + d*x]*(a
+ b*Tan[c + d*x])^(7/2))/(9*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 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 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 3609

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

Rule 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^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 3620

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 3647

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 3711

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}+\frac {2 \int (a+b \tan (c+d x))^{5/2} \left (-a-\frac {9}{2} b \tan (c+d x)-a \tan ^2(c+d x)\right ) \, dx}{9 b} \\ & = -\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}+\frac {2 \int -\frac {9}{2} b \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx}{9 b} \\ & = -\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\int \tan (c+d x) (a+b \tan (c+d x))^{5/2} \, dx \\ & = -\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}-\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\int (-b+a \tan (c+d x)) (a+b \tan (c+d x))^{3/2} \, dx \\ & = -\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}-\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\int \sqrt {a+b \tan (c+d x)} \left (-2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx \\ & = -\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}-\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\int \frac {-b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}-\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}+\frac {1}{2} (i a-b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} (i a+b)^3 \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}-\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {(a-i b)^3 \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac {(a+i b)^3 \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d} \\ & = -\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}-\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {(i a-b)^3 \text {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 a+b)^3 \text {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 {(a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}-\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.92 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.94 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {(a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-10 a^4-483 a^2 b^2+315 b^4+a b \left (5 a^2-231 b^2\right ) \tan (c+d x)+\left (75 a^2 b^2-63 b^4\right ) \tan ^2(c+d x)+95 a b^3 \tan ^3(c+d x)+35 b^4 \tan ^4(c+d x)\right )}{315 b^2 d} \]

[In]

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

[Out]

((a - I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + ((a + I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[
c + d*x]]/Sqrt[a + I*b]])/d + (2*Sqrt[a + b*Tan[c + d*x]]*(-10*a^4 - 483*a^2*b^2 + 315*b^4 + a*b*(5*a^2 - 231*
b^2)*Tan[c + d*x] + (75*a^2*b^2 - 63*b^4)*Tan[c + d*x]^2 + 95*a*b^3*Tan[c + d*x]^3 + 35*b^4*Tan[c + d*x]^4))/(
315*b^2*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1239\) vs. \(2(177)=354\).

Time = 0.66 (sec) , antiderivative size = 1240, normalized size of antiderivative = 5.88

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1240\)
default	\(\text {Expression too large to display}\)	\(1240\)

[In]

int(tan(d*x+c)^3*(a+b*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/9/d/b^2*(a+b*tan(d*x+c))^(9/2)-2/7*a*(a+b*tan(d*x+c))^(7/2)/b^2/d-2/5*(a+b*tan(d*x+c))^(5/2)/d-2/3*a*(a+b*ta
n(d*x+c))^(3/2)/d-2/d*a^2*(a+b*tan(d*x+c))^(1/2)+2*b^2*(a+b*tan(d*x+c))^(1/2)/d+1/2/d*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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*
a-3/4/d*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))*(2*(a^2+b^2)^(
1/2)+2*a)^(1/2)*a^2+1/4/d*b^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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-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))*(a^2+b^2)^(1/2)*a^2+1/d*b^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^2+b^2)^
(1/2)+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))*a^3-3/d*b^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-1/2/d*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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a+3/4/d*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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-1/4/d*
b^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))*(2*(a^2+b^2)^(1/2)
+2*a)^(1/2)+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))*(a^2+b^2)^(1/2)*a^2-1/d*b^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^2+b^2)^(1/2)-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))*a
^3+3/d*b^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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1193 vs. \(2 (173) = 346\).

Time = 0.27 (sec) , antiderivative size = 1193, normalized size of antiderivative = 5.65 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/630*(315*b^2*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^
2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a) + (2*a*d^3*sqrt(-
(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) - (5*a^6 - 15*a^4*b^2 + 11*a^2*b^4 - b^6)*d)
*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4
))/d^2)) - 315*b^2*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20
*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a) - (2*a*d^3*sqr
t(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) - (5*a^6 - 15*a^4*b^2 + 11*a^2*b^4 - b^6)
*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/
d^4))/d^2)) - 315*b^2*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 -
20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a) + (2*a*d^3*
sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) + (5*a^6 - 15*a^4*b^2 + 11*a^2*b^4 - b
^6)*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^1
0)/d^4))/d^2)) + 315*b^2*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^
6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a) - (2*a*d
^3*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) + (5*a^6 - 15*a^4*b^2 + 11*a^2*b^4
- b^6)*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 +
b^10)/d^4))/d^2)) - 4*(35*b^4*tan(d*x + c)^4 + 95*a*b^3*tan(d*x + c)^3 - 10*a^4 - 483*a^2*b^2 + 315*b^4 + 3*(2
5*a^2*b^2 - 21*b^4)*tan(d*x + c)^2 + (5*a^3*b - 231*a*b^3)*tan(d*x + c))*sqrt(b*tan(d*x + c) + a))/(b^2*d)

Sympy [F]

\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(tan(d*x+c)**3*(a+b*tan(d*x+c))**(5/2),x)

[Out]

Integral((a + b*tan(c + d*x))**(5/2)*tan(c + d*x)**3, x)

Maxima [F]

\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

integrate((b*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^3, x)

Giac [F(-1)]

Timed out. \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 54.57 (sec) , antiderivative size = 2384, normalized size of antiderivative = 11.30 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]

[In]

int(tan(c + d*x)^3*(a + b*tan(c + d*x))^(5/2),x)

[Out]

(a + b*tan(c + d*x))^(3/2)*((2*a*((2*a^2)/(b^2*d) - (2*(a^2 + b^2))/(b^2*d)))/3 - (2*a^3)/(3*b^2*d) + (2*a*(a^
2 + b^2))/(3*b^2*d)) - atan(((((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a
*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i
- a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 -
a^6*b^2))/d^2)*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i - (((8*(4*b^6
*d^2 - 4*a^4*b^2*d^2))/d^3 + 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3
*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1
/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*((5*a*b^4 + a^4*b*5i + a^
5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i)/((((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 - 64*a*b^2*(a
+ b*tan(c + d*x))^(1/2)*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*((5*a
*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(a + b*tan(c + d*x))^(1/2)*(b^
8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(
4*d^2))^(1/2) + (((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 + 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a*b^4 + a^4*b*
5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10
i - 10*a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2
)*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(3*a*b^10 + 8*a^3*b^8 +
6*a^5*b^6 - a^9*b^2))/d^3))*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*2i
- atan(((((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a*b^4 - a^4*b*5i + a^
5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*
a^3*b^2)/(4*d^2))^(1/2) - (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*((5*a
*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i - (((8*(4*b^6*d^2 - 4*a^4*b^2*d^2
))/d^3 + 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(
4*d^2))^(1/2))*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan
(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3
*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i)/((((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x))^(
1/2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 - a^4*b*5i + a^
5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*
a^4*b^4 - a^6*b^2))/d^2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) + (((8
*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 + 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i +
a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d
^2))^(1/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*((5*a*b^4 - a^4*b*
5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(3*a*b^10 + 8*a^3*b^8 + 6*a^5*b^6 - a^9*b^2
))/d^3))*((5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*2i + ((2*a^2)/(5*b^2*d
) - (2*(a^2 + b^2))/(5*b^2*d))*(a + b*tan(c + d*x))^(5/2) - (((2*a^2)/(b^2*d) - (2*(a^2 + b^2))/(b^2*d))*(a^2
+ b^2) - 2*a*(2*a*((2*a^2)/(b^2*d) - (2*(a^2 + b^2))/(b^2*d)) - (2*a^3)/(b^2*d) + (2*a*(a^2 + b^2))/(b^2*d)))*
(a + b*tan(c + d*x))^(1/2) + (2*(a + b*tan(c + d*x))^(9/2))/(9*b^2*d) - (2*a*(a + b*tan(c + d*x))^(7/2))/(7*b^
2*d)