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

Optimal. Leaf size=175 \[ \frac {2 (a A-b B) \sqrt {a+b \tan (c+d x)}}{d}-\frac {(a-i b)^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (A+i B) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 A (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3592, 3528, 3539, 3537, 63, 208} \[ \frac {2 (a A-b B) \sqrt {a+b \tan (c+d x)}}{d}-\frac {(a-i b)^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (A+i B) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 A (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - I*b)^(3/2)*(A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d) - ((a + I*b)^(3/2)*(A + I*B)*
ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/d + (2*(a*A - b*B)*Sqrt[a + b*Tan[c + d*x]])/d + (2*A*(a + b*
Tan[c + d*x])^(3/2))/(3*d) + (2*B*(a + b*Tan[c + d*x])^(5/2))/(5*b*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 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 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 3592

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

Rubi steps

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

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Mathematica [A]  time = 1.39, size = 192, normalized size = 1.10 \[ \frac {5 (A-i B) \left (\sqrt {a+b \tan (c+d x)} (4 a+b \tan (c+d x)-3 i b)-3 (a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )\right )+5 (A+i B) \left (\sqrt {a+b \tan (c+d x)} (4 a+b \tan (c+d x)+3 i b)-3 (a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )\right )+\frac {6 B (a+b \tan (c+d x))^{5/2}}{b}}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*B*(a + b*Tan[c + d*x])^(5/2))/b + 5*(A - I*B)*(-3*(a - I*b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a
- I*b]] + Sqrt[a + b*Tan[c + d*x]]*(4*a - (3*I)*b + b*Tan[c + d*x])) + 5*(A + I*B)*(-3*(a + I*b)^(3/2)*ArcTanh
[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + Sqrt[a + b*Tan[c + d*x]]*(4*a + (3*I)*b + b*Tan[c + d*x])))/(15*d)

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

[Out]

Timed out

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

[Out]

Timed out

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maple [B]  time = 0.32, size = 1686, normalized size = 9.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 29.60, size = 2868, normalized size = 16.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((16*B^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((16*b^2*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*
a*b^2*d^2)/d^4)^(1/2)*(B*b^3 + B*a^2*b + a*d*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^
2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d + (16*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b
^2))/d^2)*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d^4)^(1/2))/2)*((6*B^4*a^2*b
^4*d^4 - B^4*b^6*d^4 - 9*B^4*a^4*b^2*d^4)^(1/2)/(4*d^4) - (B^2*a^3)/(4*d^2) + (3*B^2*a*b^2)/(4*d^2))^(1/2) - l
og((16*B^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((16*b^2*(-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + B^2*a^3*d^2 - 3*B^2*
a*b^2*d^2)/d^4)^(1/2)*(B*b^3 + B*a^2*b - a*d*(-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b
^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d - (16*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*
b^2))/d^2)*(-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/d^4)^(1/2))/2)*(-((6*B^4*a
^2*b^4*d^4 - B^4*b^6*d^4 - 9*B^4*a^4*b^2*d^4)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/(4*d^4))^(1/2) - log((16*
B^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((16*b^2*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^
2)/d^4)^(1/2)*(B*b^3 + B*a^2*b - a*d*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d
^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d - (16*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2
)*(((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/d^4)^(1/2))/2)*(((6*B^4*a^2*b^4*d^4
- B^4*b^6*d^4 - 9*B^4*a^4*b^2*d^4)^(1/2) - B^2*a^3*d^2 + 3*B^2*a*b^2*d^2)/(4*d^4))^(1/2) - ((2*B*(a^2 + b^2))/
(b*d) - (2*B*a^2)/(b*d))*(a + b*tan(c + d*x))^(1/2) + log((16*B^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((16*b^2*(-((-B^
4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(B*b^3 + B*a^2*b + a*d*(-((-B^4*b
^2*d^4*(3*a^2 - b^2)^2)^(1/2) + B^2*a^3*d^2 - 3*B^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d + (16
*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2)*(-((-B^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + B^
2*a^3*d^2 - 3*B^2*a*b^2*d^2)/d^4)^(1/2))/2)*((3*B^2*a*b^2)/(4*d^2) - (B^2*a^3)/(4*d^2) - (6*B^4*a^2*b^4*d^4 -
B^4*b^6*d^4 - 9*B^4*a^4*b^2*d^4)^(1/2)/(4*d^4))^(1/2) - log(- ((((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^
3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*((16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2 - (16
*a*b^2*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(A*a^2 + A*b^2 + d*(
((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))
/d))/2 - (8*A^3*b^2*(a^2 - b^2)*(a^2 + b^2)^2)/d^3)*(((6*A^4*a^2*b^4*d^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^(1
/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/(4*d^4))^(1/2) - log(- ((-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3
*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*((16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2 - (16*
a*b^2*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(A*a^2 + A*b^2 + d*(
-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2))
)/d))/2 - (8*A^3*b^2*(a^2 - b^2)*(a^2 + b^2)^2)/d^3)*(-((6*A^4*a^2*b^4*d^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^
(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/(4*d^4))^(1/2) + log(((((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*
d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*((16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2 + (16*a
*b^2*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(A*a^2 + A*b^2 - d*(((
-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d
))/2 - (8*A^3*b^2*(a^2 - b^2)*(a^2 + b^2)^2)/d^3)*((6*A^4*a^2*b^4*d^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^(1/2)
/(4*d^4) + (A^2*a^3)/(4*d^2) - (3*A^2*a*b^2)/(4*d^2))^(1/2) + log(((-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A
^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*((16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2
+ (16*a*b^2*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(A*a^2 + A*b^2
- d*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^
(1/2)))/d))/2 - (8*A^3*b^2*(a^2 - b^2)*(a^2 + b^2)^2)/d^3)*((A^2*a^3)/(4*d^2) - (6*A^4*a^2*b^4*d^4 - A^4*b^6*d
^4 - 9*A^4*a^4*b^2*d^4)^(1/2)/(4*d^4) - (3*A^2*a*b^2)/(4*d^2))^(1/2) + (2*A*(a + b*tan(c + d*x))^(3/2))/(3*d)
+ (2*A*a*(a + b*tan(c + d*x))^(1/2))/d + (2*B*(a + b*tan(c + d*x))^(5/2))/(5*b*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(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), x)

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