∫ tan4 (c+dx)(A+B tan(c+dx))____________________

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

[Out]

-(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(5/2)/d+(I*A-B)*arctanh((a+b*tan(d*x+c))^(1/2)/

(a+I*b)^(1/2))/(a+I*b)^(5/2)/d+2/3*(8*A*a^4*b+17*A*a^2*b^3+3*A*b^5-16*B*a^5-30*B*a^3*b^2-8*B*a*b^4)*(a+b*tan(d

*x+c))^(1/2)/b^4/(a^2+b^2)^2/d-2/3*(4*A*a^3*b+10*A*a*b^3-8*B*a^4-15*B*a^2*b^2-B*b^4)*(a+b*tan(d*x+c))^(1/2)*ta

n(d*x+c)/b^3/(a^2+b^2)^2/d+2*a*(A*a^2*b+3*A*b^3-2*B*a^3-4*B*a*b^2)*tan(d*x+c)^2/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x

+c))^(1/2)+2/3*a*(A*b-B*a)*tan(d*x+c)^3/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)

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Rubi [A] time = 1.05, antiderivative size = 371, normalized size of antiderivative = 1.00, 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 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 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 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 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])))

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*}

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Mathematica [C] time = 6.35, size = 450, normalized size = 1.21 \[ \frac {2 B \tan ^3(c+d x)}{3 b d (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (\frac {3 (A b-2 a B) \tan ^2(c+d x)}{b d (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (\frac {3 \left (-8 a^2 B+4 a A b+b^2 B\right ) \tan (c+d x)}{2 b d (a+b \tan (c+d x))^{3/2}}-\frac {3 \left (-\frac {2 \left (-16 a^3 B+8 a^2 A b+2 a b^2 B+A b^3\right )}{3 b (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (\frac {\left (\frac {3}{2} a b^4 B-\frac {3 A b^5}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {a+b \tan (c+d x)}{a+i b}\right )}{3 (-b+i a) (a+b \tan (c+d x))^{3/2}}-\frac {\, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {a+b \tan (c+d x)}{a-i b}\right )}{3 (b+i a) (a+b \tan (c+d x))^{3/2}}\right )}{b}-\frac {3}{2} b^3 B \left (\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \tan (c+d x)}{a+i b}\right )}{(-b+i a) \sqrt {a+b \tan (c+d x)}}-\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \tan (c+d x)}{a-i b}\right )}{(b+i a) \sqrt {a+b \tan (c+d x)}}\right )\right )}{3 b}\right )}{4 b d}\right )}{b}\right )}{3 b} \]

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)*(-1/3*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a - I*b)]/((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

*Tan[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)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab