3.357 \(\int \frac{\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=371 \[ \frac{2 a (A b-a B) \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (a^2 A b-2 a^3 B-4 a b^2 B+3 A b^3\right ) \tan ^2(c+d x)}{b^2 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (4 a^3 A b-15 a^2 b^2 B-8 a^4 B+10 a A b^3-b^4 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^3 d \left (a^2+b^2\right )^2}+\frac{2 \left (17 a^2 A b^3+8 a^4 A b-30 a^3 b^2 B-16 a^5 B-8 a b^4 B+3 A b^5\right ) \sqrt{a+b \tan (c+d x)}}{3 b^4 d \left (a^2+b^2\right )^2}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{5/2}}+\frac{(-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{5/2}} \]

[Out]

-(((I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((a - I*b)^(5/2)*d)) + ((I*A - B)*ArcTanh[Sqrt[a
+ b*Tan[c + d*x]]/Sqrt[a + I*b]])/((a + I*b)^(5/2)*d) + (2*a*(A*b - a*B)*Tan[c + d*x]^3)/(3*b*(a^2 + b^2)*d*(
a + b*Tan[c + d*x])^(3/2)) + (2*a*(a^2*A*b + 3*A*b^3 - 2*a^3*B - 4*a*b^2*B)*Tan[c + d*x]^2)/(b^2*(a^2 + b^2)^2
*d*Sqrt[a + b*Tan[c + d*x]]) + (2*(8*a^4*A*b + 17*a^2*A*b^3 + 3*A*b^5 - 16*a^5*B - 30*a^3*b^2*B - 8*a*b^4*B)*S
qrt[a + b*Tan[c + d*x]])/(3*b^4*(a^2 + b^2)^2*d) - (2*(4*a^3*A*b + 10*a*A*b^3 - 8*a^4*B - 15*a^2*b^2*B - b^4*B
)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(3*b^3*(a^2 + b^2)^2*d)

________________________________________________________________________________________

Rubi [A]  time = 1.04536, antiderivative size = 371, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {3605, 3645, 3647, 3630, 3539, 3537, 63, 208} \[ \frac{2 a (A b-a B) \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (a^2 A b-2 a^3 B-4 a b^2 B+3 A b^3\right ) \tan ^2(c+d x)}{b^2 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (4 a^3 A b-15 a^2 b^2 B-8 a^4 B+10 a A b^3-b^4 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^3 d \left (a^2+b^2\right )^2}+\frac{2 \left (17 a^2 A b^3+8 a^4 A b-30 a^3 b^2 B-16 a^5 B-8 a b^4 B+3 A b^5\right ) \sqrt{a+b \tan (c+d x)}}{3 b^4 d \left (a^2+b^2\right )^2}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{5/2}}+\frac{(-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^4*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

-(((I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((a - I*b)^(5/2)*d)) + ((I*A - B)*ArcTanh[Sqrt[a
+ b*Tan[c + d*x]]/Sqrt[a + I*b]])/((a + I*b)^(5/2)*d) + (2*a*(A*b - a*B)*Tan[c + d*x]^3)/(3*b*(a^2 + b^2)*d*(
a + b*Tan[c + d*x])^(3/2)) + (2*a*(a^2*A*b + 3*A*b^3 - 2*a^3*B - 4*a*b^2*B)*Tan[c + d*x]^2)/(b^2*(a^2 + b^2)^2
*d*Sqrt[a + b*Tan[c + d*x]]) + (2*(8*a^4*A*b + 17*a^2*A*b^3 + 3*A*b^5 - 16*a^5*B - 30*a^3*b^2*B - 8*a*b^4*B)*S
qrt[a + b*Tan[c + d*x]])/(3*b^4*(a^2 + b^2)^2*d) - (2*(4*a^3*A*b + 10*a*A*b^3 - 8*a^4*B - 15*a^2*b^2*B - b^4*B
)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(3*b^3*(a^2 + b^2)^2*d)

Rule 3605

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

Rule 3645

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

Rubi steps

\begin{align*} \int \frac{\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx &=\frac{2 a (A b-a B) \tan ^3(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 \int \frac{\tan ^2(c+d x) \left (-3 a (A b-a B)+\frac{3}{2} b (A b-a B) \tan (c+d x)-\frac{3}{2} \left (a A b-2 a^2 B-b^2 B\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^{3/2}} \, dx}{3 b \left (a^2+b^2\right )}\\ &=\frac{2 a (A b-a B) \tan ^3(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (a^2 A b+3 A b^3-2 a^3 B-4 a b^2 B\right ) \tan ^2(c+d x)}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{4 \int \frac{\tan (c+d x) \left (-3 a \left (a^2 A b+3 A b^3-2 a^3 B-4 a b^2 B\right )-\frac{3}{4} b^2 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)-\frac{3}{4} \left (4 a^3 A b+10 a A b^3-8 a^4 B-15 a^2 b^2 B-b^4 B\right ) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx}{3 b^2 \left (a^2+b^2\right )^2}\\ &=\frac{2 a (A b-a B) \tan ^3(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (a^2 A b+3 A b^3-2 a^3 B-4 a b^2 B\right ) \tan ^2(c+d x)}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (4 a^3 A b+10 a A b^3-8 a^4 B-15 a^2 b^2 B-b^4 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right )^2 d}+\frac{8 \int \frac{\frac{3}{4} a \left (4 a^3 A b+10 a A b^3-8 a^4 B-15 a^2 b^2 B-b^4 B\right )-\frac{9}{8} b^3 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)+\frac{3}{8} \left (8 a^4 A b+17 a^2 A b^3+3 A b^5-16 a^5 B-30 a^3 b^2 B-8 a b^4 B\right ) \tan ^2(c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{9 b^3 \left (a^2+b^2\right )^2}\\ &=\frac{2 a (A b-a B) \tan ^3(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (a^2 A b+3 A b^3-2 a^3 B-4 a b^2 B\right ) \tan ^2(c+d x)}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (8 a^4 A b+17 a^2 A b^3+3 A b^5-16 a^5 B-30 a^3 b^2 B-8 a b^4 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^4 \left (a^2+b^2\right )^2 d}-\frac{2 \left (4 a^3 A b+10 a A b^3-8 a^4 B-15 a^2 b^2 B-b^4 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right )^2 d}+\frac{8 \int \frac{\frac{9}{8} b^3 \left (a^2 A-A b^2+2 a b B\right )-\frac{9}{8} b^3 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{9 b^3 \left (a^2+b^2\right )^2}\\ &=\frac{2 a (A b-a B) \tan ^3(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (a^2 A b+3 A b^3-2 a^3 B-4 a b^2 B\right ) \tan ^2(c+d x)}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (8 a^4 A b+17 a^2 A b^3+3 A b^5-16 a^5 B-30 a^3 b^2 B-8 a b^4 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^4 \left (a^2+b^2\right )^2 d}-\frac{2 \left (4 a^3 A b+10 a A b^3-8 a^4 B-15 a^2 b^2 B-b^4 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right )^2 d}+\frac{(A-i B) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}+\frac{(A+i B) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}\\ &=\frac{2 a (A b-a B) \tan ^3(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (a^2 A b+3 A b^3-2 a^3 B-4 a b^2 B\right ) \tan ^2(c+d x)}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (8 a^4 A b+17 a^2 A b^3+3 A b^5-16 a^5 B-30 a^3 b^2 B-8 a b^4 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^4 \left (a^2+b^2\right )^2 d}-\frac{2 \left (4 a^3 A b+10 a A b^3-8 a^4 B-15 a^2 b^2 B-b^4 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right )^2 d}-\frac{(i A-B) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 (a+i b)^2 d}+\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 (a-i b)^2 d}\\ &=\frac{2 a (A b-a B) \tan ^3(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (a^2 A b+3 A b^3-2 a^3 B-4 a b^2 B\right ) \tan ^2(c+d x)}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (8 a^4 A b+17 a^2 A b^3+3 A b^5-16 a^5 B-30 a^3 b^2 B-8 a b^4 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^4 \left (a^2+b^2\right )^2 d}-\frac{2 \left (4 a^3 A b+10 a A b^3-8 a^4 B-15 a^2 b^2 B-b^4 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right )^2 d}+\frac{(A-i B) \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 (i a+b)^2 d}+\frac{(A+i B) \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 )}{(i a-b)^2 b d}\\ &=-\frac{(i A+B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{(a-i b)^{5/2} d}+\frac{(i A-B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{(a+i b)^{5/2} d}+\frac{2 a (A b-a B) \tan ^3(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 a \left (a^2 A b+3 A b^3-2 a^3 B-4 a b^2 B\right ) \tan ^2(c+d x)}{b^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (8 a^4 A b+17 a^2 A b^3+3 A b^5-16 a^5 B-30 a^3 b^2 B-8 a b^4 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^4 \left (a^2+b^2\right )^2 d}-\frac{2 \left (4 a^3 A b+10 a A b^3-8 a^4 B-15 a^2 b^2 B-b^4 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right )^2 d}\\ \end{align*}

Mathematica [C]  time = 6.3233, size = 450, normalized size = 1.21 \[ \frac{2 B \tan ^3(c+d x)}{3 b d (a+b \tan (c+d x))^{3/2}}+\frac{2 \left (\frac{3 (A b-2 a B) \tan ^2(c+d x)}{b d (a+b \tan (c+d x))^{3/2}}+\frac{2 \left (\frac{3 \left (-8 a^2 B+4 a A b+b^2 B\right ) \tan (c+d x)}{2 b d (a+b \tan (c+d x))^{3/2}}-\frac{3 \left (-\frac{2 \left (8 a^2 A b-16 a^3 B+2 a b^2 B+A b^3\right )}{3 b (a+b \tan (c+d x))^{3/2}}+\frac{2 \left (\frac{\left (\frac{3}{2} a b^4 B-\frac{3 A b^5}{2}\right ) \left (\frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \tan (c+d x)}{a+i b}\right )}{3 (-b+i a) (a+b \tan (c+d x))^{3/2}}-\frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \tan (c+d x)}{a-i b}\right )}{3 (b+i a) (a+b \tan (c+d x))^{3/2}}\right )}{b}-\frac{3}{2} b^3 B \left (\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan (c+d x)}{a+i b}\right )}{(-b+i a) \sqrt{a+b \tan (c+d x)}}-\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan (c+d x)}{a-i b}\right )}{(b+i a) \sqrt{a+b \tan (c+d x)}}\right )\right )}{3 b}\right )}{4 b d}\right )}{b}\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^4*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x]

[Out]

(2*B*Tan[c + d*x]^3)/(3*b*d*(a + b*Tan[c + d*x])^(3/2)) + (2*((3*(A*b - 2*a*B)*Tan[c + d*x]^2)/(b*d*(a + b*Tan
[c + d*x])^(3/2)) + (2*((3*(4*a*A*b - 8*a^2*B + b^2*B)*Tan[c + d*x])/(2*b*d*(a + b*Tan[c + d*x])^(3/2)) - (3*(
(-2*(8*a^2*A*b + A*b^3 - 16*a^3*B + 2*a*b^2*B))/(3*b*(a + b*Tan[c + d*x])^(3/2)) + (2*((((-3*A*b^5)/2 + (3*a*b
^4*B)/2)*(-Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a - I*b)]/(3*(I*a + b)*(a + b*Tan[c + d*x])^
(3/2)) + Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a + I*b)]/(3*(I*a - b)*(a + b*Tan[c + d*x])^(3
/2))))/b - (3*b^3*B*(-(Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I*b)]/((I*a + b)*Sqrt[a + b*T
an[c + d*x]])) + Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)]/((I*a - b)*Sqrt[a + b*Tan[c +
d*x]])))/2))/(3*b)))/(4*b*d)))/b))/(3*b)

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Maple [B]  time = 0.137, size = 12953, normalized size = 34.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x)

[Out]

result too large to display

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{4}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^4/(b*tan(d*x + c) + a)^(5/2), x)