\(\int (a+b \tan ^2(c+d x))^{5/2} \, dx\) [319]

Optimal result

Integrand size = 16, antiderivative size = 170 \[ \int \left (a+b \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {(a-b)^{5/2} \arctan \left (\frac {\sqrt {a-b} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)}}\right )}{d}+\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)}}\right )}{8 d}+\frac {(7 a-4 b) b \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)}}{8 d}+\frac {b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d} \]

[Out]

(a-b)^(5/2)*arctan((a-b)^(1/2)*tan(d*x+c)/(a+b*tan(d*x+c)^2)^(1/2))/d+1/8*(15*a^2-20*a*b+8*b^2)*arctanh(b^(1/2
)*tan(d*x+c)/(a+b*tan(d*x+c)^2)^(1/2))*b^(1/2)/d+1/8*(7*a-4*b)*b*(a+b*tan(d*x+c)^2)^(1/2)*tan(d*x+c)/d+1/4*b*t
an(d*x+c)*(a+b*tan(d*x+c)^2)^(3/2)/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3742, 427, 542, 537, 223, 212, 385, 209} \[ \int \left (a+b \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)}}\right )}{8 d}+\frac {(a-b)^{5/2} \arctan \left (\frac {\sqrt {a-b} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)}}\right )}{d}+\frac {b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac {b (7 a-4 b) \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)}}{8 d} \]

[In]

Int[(a + b*Tan[c + d*x]^2)^(5/2),x]

[Out]

((a - b)^(5/2)*ArcTan[(Sqrt[a - b]*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]^2]])/d + (Sqrt[b]*(15*a^2 - 20*a*b +
8*b^2)*ArcTanh[(Sqrt[b]*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]^2]])/(8*d) + ((7*a - 4*b)*b*Tan[c + d*x]*Sqrt[a
+ b*Tan[c + d*x]^2])/(8*d) + (b*Tan[c + d*x]*(a + b*Tan[c + d*x]^2)^(3/2))/(4*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 385

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

Rule 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 542

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

Rule 3742

Int[((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[(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, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^{5/2}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac {\text {Subst}\left (\int \frac {\sqrt {a+b x^2} \left (a (4 a-b)+(7 a-4 b) b x^2\right )}{1+x^2} \, dx,x,\tan (c+d x)\right )}{4 d} \\ & = \frac {(7 a-4 b) b \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)}}{8 d}+\frac {b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac {\text {Subst}\left (\int \frac {a \left (8 a^2-9 a b+4 b^2\right )+b \left (15 a^2-20 a b+8 b^2\right ) x^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (c+d x)\right )}{8 d} \\ & = \frac {(7 a-4 b) b \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)}}{8 d}+\frac {b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac {(a-b)^3 \text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (b \left (15 a^2-20 a b+8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (c+d x)\right )}{8 d} \\ & = \frac {(7 a-4 b) b \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)}}{8 d}+\frac {b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac {(a-b)^3 \text {Subst}\left (\int \frac {1}{1-(-a+b) x^2} \, dx,x,\frac {\tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)}}\right )}{d}+\frac {\left (b \left (15 a^2-20 a b+8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)}}\right )}{8 d} \\ & = \frac {(a-b)^{5/2} \arctan \left (\frac {\sqrt {a-b} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)}}\right )}{d}+\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)}}\right )}{8 d}+\frac {(7 a-4 b) b \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)}}{8 d}+\frac {b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.23 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.99 \[ \int \left (a+b \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {-8 (a-b)^{5/2} \arctan \left (\frac {\sqrt {b}+\sqrt {b} \tan ^2(c+d x)-\tan (c+d x) \sqrt {a+b \tan ^2(c+d x)}}{\sqrt {a-b}}\right )-\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \log \left (-\sqrt {b} \tan (c+d x)+\sqrt {a+b \tan ^2(c+d x)}\right )+b \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)} \left (9 a-4 b+2 b \tan ^2(c+d x)\right )}{8 d} \]

[In]

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

[Out]

(-8*(a - b)^(5/2)*ArcTan[(Sqrt[b] + Sqrt[b]*Tan[c + d*x]^2 - Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]^2])/Sqrt[a -
b]] - Sqrt[b]*(15*a^2 - 20*a*b + 8*b^2)*Log[-(Sqrt[b]*Tan[c + d*x]) + Sqrt[a + b*Tan[c + d*x]^2]] + b*Tan[c +
d*x]*Sqrt[a + b*Tan[c + d*x]^2]*(9*a - 4*b + 2*b*Tan[c + d*x]^2))/(8*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(460\) vs. \(2(148)=296\).

Time = 0.16 (sec) , antiderivative size = 461, normalized size of antiderivative = 2.71

method	result	size
derivativedivides	\(\frac {b^{\frac {5}{2}} \ln \left (\sqrt {b}\, \tan \left (d x +c \right )+\sqrt {a +b \tan \left (d x +c \right )^{2}}\right )}{d}+\frac {b^{2} \tan \left (d x +c \right )^{3} \sqrt {a +b \tan \left (d x +c \right )^{2}}}{4 d}+\frac {9 b a \tan \left (d x +c \right ) \sqrt {a +b \tan \left (d x +c \right )^{2}}}{8 d}+\frac {15 \sqrt {b}\, a^{2} \ln \left (\sqrt {b}\, \tan \left (d x +c \right )+\sqrt {a +b \tan \left (d x +c \right )^{2}}\right )}{8 d}-\frac {b^{2} \tan \left (d x +c \right ) \sqrt {a +b \tan \left (d x +c \right )^{2}}}{2 d}-\frac {5 b^{\frac {3}{2}} a \ln \left (\sqrt {b}\, \tan \left (d x +c \right )+\sqrt {a +b \tan \left (d x +c \right )^{2}}\right )}{2 d}-\frac {b \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (d x +c \right )^{2}}}\right )}{d \left (a -b \right )}+\frac {3 a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (d x +c \right )^{2}}}\right )}{d \left (a -b \right )}-\frac {3 a^{2} \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (d x +c \right )^{2}}}\right )}{d b \left (a -b \right )}+\frac {a^{3} \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (d x +c \right )^{2}}}\right )}{d \,b^{2} \left (a -b \right )}\)	\(461\)
default	\(\frac {b^{\frac {5}{2}} \ln \left (\sqrt {b}\, \tan \left (d x +c \right )+\sqrt {a +b \tan \left (d x +c \right )^{2}}\right )}{d}+\frac {b^{2} \tan \left (d x +c \right )^{3} \sqrt {a +b \tan \left (d x +c \right )^{2}}}{4 d}+\frac {9 b a \tan \left (d x +c \right ) \sqrt {a +b \tan \left (d x +c \right )^{2}}}{8 d}+\frac {15 \sqrt {b}\, a^{2} \ln \left (\sqrt {b}\, \tan \left (d x +c \right )+\sqrt {a +b \tan \left (d x +c \right )^{2}}\right )}{8 d}-\frac {b^{2} \tan \left (d x +c \right ) \sqrt {a +b \tan \left (d x +c \right )^{2}}}{2 d}-\frac {5 b^{\frac {3}{2}} a \ln \left (\sqrt {b}\, \tan \left (d x +c \right )+\sqrt {a +b \tan \left (d x +c \right )^{2}}\right )}{2 d}-\frac {b \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (d x +c \right )^{2}}}\right )}{d \left (a -b \right )}+\frac {3 a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (d x +c \right )^{2}}}\right )}{d \left (a -b \right )}-\frac {3 a^{2} \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (d x +c \right )^{2}}}\right )}{d b \left (a -b \right )}+\frac {a^{3} \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (d x +c \right )^{2}}}\right )}{d \,b^{2} \left (a -b \right )}\)	\(461\)

[In]

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

[Out]

1/d*b^(5/2)*ln(b^(1/2)*tan(d*x+c)+(a+b*tan(d*x+c)^2)^(1/2))+1/4/d*b^2*tan(d*x+c)^3*(a+b*tan(d*x+c)^2)^(1/2)+9/
8/d*b*a*tan(d*x+c)*(a+b*tan(d*x+c)^2)^(1/2)+15/8/d*b^(1/2)*a^2*ln(b^(1/2)*tan(d*x+c)+(a+b*tan(d*x+c)^2)^(1/2))
-1/2/d*b^2*tan(d*x+c)*(a+b*tan(d*x+c)^2)^(1/2)-5/2/d*b^(3/2)*a*ln(b^(1/2)*tan(d*x+c)+(a+b*tan(d*x+c)^2)^(1/2))
-1/d*b*(b^4*(a-b))^(1/2)/(a-b)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(d*x+c)^2)^(1/2)*tan(d*x+c))+3/d*a*(
b^4*(a-b))^(1/2)/(a-b)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(d*x+c)^2)^(1/2)*tan(d*x+c))-3/d*a^2/b*(b^4*
(a-b))^(1/2)/(a-b)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(d*x+c)^2)^(1/2)*tan(d*x+c))+1/d*a^3*(b^4*(a-b))
^(1/2)/b^2/(a-b)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(d*x+c)^2)^(1/2)*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 1.30 (sec) , antiderivative size = 703, normalized size of antiderivative = 4.14 \[ \int \left (a+b \tan ^2(c+d x)\right )^{5/2} \, dx=\left [\frac {{\left (15 \, a^{2} - 20 \, a b + 8 \, b^{2}\right )} \sqrt {b} \log \left (2 \, b \tan \left (d x + c\right )^{2} + 2 \, \sqrt {b \tan \left (d x + c\right )^{2} + a} \sqrt {b} \tan \left (d x + c\right ) + a\right ) + 8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (d x + c\right )^{2} + 2 \, \sqrt {b \tan \left (d x + c\right )^{2} + a} \sqrt {-a + b} \tan \left (d x + c\right ) - a}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, {\left (2 \, b^{2} \tan \left (d x + c\right )^{3} + {\left (9 \, a b - 4 \, b^{2}\right )} \tan \left (d x + c\right )\right )} \sqrt {b \tan \left (d x + c\right )^{2} + a}}{16 \, d}, \frac {16 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a - b} \arctan \left (-\frac {\sqrt {b \tan \left (d x + c\right )^{2} + a}}{\sqrt {a - b} \tan \left (d x + c\right )}\right ) + {\left (15 \, a^{2} - 20 \, a b + 8 \, b^{2}\right )} \sqrt {b} \log \left (2 \, b \tan \left (d x + c\right )^{2} + 2 \, \sqrt {b \tan \left (d x + c\right )^{2} + a} \sqrt {b} \tan \left (d x + c\right ) + a\right ) + 2 \, {\left (2 \, b^{2} \tan \left (d x + c\right )^{3} + {\left (9 \, a b - 4 \, b^{2}\right )} \tan \left (d x + c\right )\right )} \sqrt {b \tan \left (d x + c\right )^{2} + a}}{16 \, d}, -\frac {{\left (15 \, a^{2} - 20 \, a b + 8 \, b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (d x + c\right )^{2} + a} \sqrt {-b}}{b \tan \left (d x + c\right )}\right ) - 4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (d x + c\right )^{2} + 2 \, \sqrt {b \tan \left (d x + c\right )^{2} + a} \sqrt {-a + b} \tan \left (d x + c\right ) - a}{\tan \left (d x + c\right )^{2} + 1}\right ) - {\left (2 \, b^{2} \tan \left (d x + c\right )^{3} + {\left (9 \, a b - 4 \, b^{2}\right )} \tan \left (d x + c\right )\right )} \sqrt {b \tan \left (d x + c\right )^{2} + a}}{8 \, d}, \frac {8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a - b} \arctan \left (-\frac {\sqrt {b \tan \left (d x + c\right )^{2} + a}}{\sqrt {a - b} \tan \left (d x + c\right )}\right ) - {\left (15 \, a^{2} - 20 \, a b + 8 \, b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (d x + c\right )^{2} + a} \sqrt {-b}}{b \tan \left (d x + c\right )}\right ) + {\left (2 \, b^{2} \tan \left (d x + c\right )^{3} + {\left (9 \, a b - 4 \, b^{2}\right )} \tan \left (d x + c\right )\right )} \sqrt {b \tan \left (d x + c\right )^{2} + a}}{8 \, d}\right ] \]

[In]

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

[Out]

[1/16*((15*a^2 - 20*a*b + 8*b^2)*sqrt(b)*log(2*b*tan(d*x + c)^2 + 2*sqrt(b*tan(d*x + c)^2 + a)*sqrt(b)*tan(d*x
+ c) + a) + 8*(a^2 - 2*a*b + b^2)*sqrt(-a + b)*log(-((a - 2*b)*tan(d*x + c)^2 + 2*sqrt(b*tan(d*x + c)^2 + a)*
sqrt(-a + b)*tan(d*x + c) - a)/(tan(d*x + c)^2 + 1)) + 2*(2*b^2*tan(d*x + c)^3 + (9*a*b - 4*b^2)*tan(d*x + c))
*sqrt(b*tan(d*x + c)^2 + a))/d, 1/16*(16*(a^2 - 2*a*b + b^2)*sqrt(a - b)*arctan(-sqrt(b*tan(d*x + c)^2 + a)/(s
qrt(a - b)*tan(d*x + c))) + (15*a^2 - 20*a*b + 8*b^2)*sqrt(b)*log(2*b*tan(d*x + c)^2 + 2*sqrt(b*tan(d*x + c)^2
+ a)*sqrt(b)*tan(d*x + c) + a) + 2*(2*b^2*tan(d*x + c)^3 + (9*a*b - 4*b^2)*tan(d*x + c))*sqrt(b*tan(d*x + c)^
2 + a))/d, -1/8*((15*a^2 - 20*a*b + 8*b^2)*sqrt(-b)*arctan(sqrt(b*tan(d*x + c)^2 + a)*sqrt(-b)/(b*tan(d*x + c)
)) - 4*(a^2 - 2*a*b + b^2)*sqrt(-a + b)*log(-((a - 2*b)*tan(d*x + c)^2 + 2*sqrt(b*tan(d*x + c)^2 + a)*sqrt(-a
+ b)*tan(d*x + c) - a)/(tan(d*x + c)^2 + 1)) - (2*b^2*tan(d*x + c)^3 + (9*a*b - 4*b^2)*tan(d*x + c))*sqrt(b*ta
n(d*x + c)^2 + a))/d, 1/8*(8*(a^2 - 2*a*b + b^2)*sqrt(a - b)*arctan(-sqrt(b*tan(d*x + c)^2 + a)/(sqrt(a - b)*t
an(d*x + c))) - (15*a^2 - 20*a*b + 8*b^2)*sqrt(-b)*arctan(sqrt(b*tan(d*x + c)^2 + a)*sqrt(-b)/(b*tan(d*x + c))
) + (2*b^2*tan(d*x + c)^3 + (9*a*b - 4*b^2)*tan(d*x + c))*sqrt(b*tan(d*x + c)^2 + a))/d]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

int((a + b*tan(c + d*x)^2)^(5/2),x)

[Out]

int((a + b*tan(c + d*x)^2)^(5/2), x)