3.325 \(\int \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=214 \[ \frac{2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}-\frac{2 (a B+A b) \sqrt{a+b \tan (c+d x)}}{d}+\frac{(a-i b)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}-\frac{(a+i b)^{3/2} (-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}+\frac{2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac{2 B (a+b \tan (c+d x))^{3/2}}{3 d} \]

[Out]

((a - I*b)^(3/2)*(I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d - ((a + I*b)^(3/2)*(I*A - B)*Arc
Tanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/d - (2*(A*b + a*B)*Sqrt[a + b*Tan[c + d*x]])/d - (2*B*(a + b*Tan
[c + d*x])^(3/2))/(3*d) + (2*(7*A*b - 2*a*B)*(a + b*Tan[c + d*x])^(5/2))/(35*b^2*d) + (2*B*Tan[c + d*x]*(a + b
*Tan[c + d*x])^(5/2))/(7*b*d)

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Rubi [A]  time = 0.623399, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {3607, 3630, 3528, 3539, 3537, 63, 208} \[ \frac{2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}-\frac{2 (a B+A b) \sqrt{a+b \tan (c+d x)}}{d}+\frac{(a-i b)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}-\frac{(a+i b)^{3/2} (-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}+\frac{2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac{2 B (a+b \tan (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - I*b)^(3/2)*(I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d - ((a + I*b)^(3/2)*(I*A - B)*Arc
Tanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/d - (2*(A*b + a*B)*Sqrt[a + b*Tan[c + d*x]])/d - (2*B*(a + b*Tan
[c + d*x])^(3/2))/(3*d) + (2*(7*A*b - 2*a*B)*(a + b*Tan[c + d*x])^(5/2))/(35*b^2*d) + (2*B*Tan[c + d*x]*(a + b
*Tan[c + d*x])^(5/2))/(7*b*d)

Rule 3607

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*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*
f*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[a^2*A*d*(m +
n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m
- 1) - b*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&
!(IGtQ[n, 1] && ( !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 3528

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 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 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac{2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac{2 \int (a+b \tan (c+d x))^{3/2} \left (-a B-\frac{7}{2} b B \tan (c+d x)+\frac{1}{2} (7 A b-2 a B) \tan ^2(c+d x)\right ) \, dx}{7 b}\\ &=\frac{2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac{2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac{2 \int (a+b \tan (c+d x))^{3/2} \left (-\frac{7 A b}{2}-\frac{7}{2} b B \tan (c+d x)\right ) \, dx}{7 b}\\ &=-\frac{2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac{2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac{2 \int \sqrt{a+b \tan (c+d x)} \left (-\frac{7}{2} b (a A-b B)-\frac{7}{2} b (A b+a B) \tan (c+d x)\right ) \, dx}{7 b}\\ &=-\frac{2 (A b+a B) \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac{2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac{2 \int \frac{-\frac{7}{2} b \left (a^2 A-A b^2-2 a b B\right )-\frac{7}{2} b \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{7 b}\\ &=-\frac{2 (A b+a B) \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac{2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac{1}{2} \left ((a-i b)^2 (A-i B)\right ) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx-\frac{1}{2} \left ((a+i b)^2 (A+i B)\right ) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 (A b+a B) \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac{2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac{\left ((a+i b)^2 (i A-B)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}-\frac{\left ((a-i b)^2 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}\\ &=-\frac{2 (A b+a B) \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac{2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac{\left ((a-i b)^2 (A-i B)\right ) \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{\left ((a+i b)^2 (A+i B)\right ) \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{(a-i b)^{3/2} (i A+B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}-\frac{(a+i b)^{3/2} (i A-B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}-\frac{2 (A b+a B) \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac{2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}\\ \end{align*}

Mathematica [A]  time = 2.38791, size = 252, normalized size = 1.18 \[ \frac{\frac{2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{b}+\frac{35}{3} b (A-i B) \left (3 \sqrt{a-i b} (b+i a) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )-i \sqrt{a+b \tan (c+d x)} (4 a+b \tan (c+d x)-3 i b)\right )+\frac{35}{3} b (A+i B) \left (i \sqrt{a+b \tan (c+d x)} (4 a+b \tan (c+d x)+3 i b)+3 \sqrt{a+i b} (b-i a) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )\right )+10 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{35 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(7*A*b - 2*a*B)*(a + b*Tan[c + d*x])^(5/2))/b + 10*B*Tan[c + d*x]*(a + b*Tan[c + d*x])^(5/2) + (35*b*(A -
I*B)*(3*Sqrt[a - I*b]*(I*a + b)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] - I*Sqrt[a + b*Tan[c + d*x]]*(
4*a - (3*I)*b + b*Tan[c + d*x])))/3 + (35*b*(A + I*B)*(3*Sqrt[a + I*b]*((-I)*a + b)*ArcTanh[Sqrt[a + b*Tan[c +
d*x]]/Sqrt[a + I*b]] + I*Sqrt[a + b*Tan[c + d*x]]*(4*a + (3*I)*b + b*Tan[c + d*x])))/3)/(35*b*d)

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Maple [B]  time = 0.111, size = 1729, normalized size = 8.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d/b*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))*A*(2*(a^2+b^2
)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a-1/4/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))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)+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))*B*(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))*B*a^
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))*B*a^2+1/4/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))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)-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))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-2/5/d/b^2*B*(a+b*tan(d*x+c))^(5/2
)*a-1/4/d*b*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))*A*(2*(a^2+
b^2)^(1/2)+2*a)^(1/2)+1/4/d*b*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))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+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))*B-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))*B+2/d*b/(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*a-1/d*b/(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*(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))*B*(a^2+b^2)^(1/2)*a+1/4/d/b*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))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^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))*B*(a^2+b^2)^(1/2)*a-1/4/d/b*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))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2+1/d*b/(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*(a^2+b^2)^(1/2)-2/d*b/(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*a+2/7/d/b^2*B*(a+b*tan(d*x+c))^(7/2)-2/d*B*a*(a+b*tan(d*x+c))^(1/2)+2/5/d/b*A*(a+b*tan(d*x+c))^(5/2)-2
/d*b*A*(a+b*tan(d*x+c))^(1/2)+1/4/d/b*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))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a-2/3*B*(a+b*tan(d*x+c))^(3/2)/d

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \tan{\left (c + d x \right )}\right ) \left (a + b \tan{\left (c + d x \right )}\right )^{\frac{3}{2}} \tan ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [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)^2*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out