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[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)
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Rubi [A] time = 0.377947, antiderivative size = 175, normalized size of antiderivative = 1.,
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 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]
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 (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*}
Mathematica [A] time = 1.28854, size = 192, normalized size = 1.1 \[ \frac{5 (A-i B) \left (\sqrt{a+b \tan (c+d x)} (4 a+b \tan (c+d x)-3 i b)-3 (a-i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )\right )+5 (A+i B) \left (\sqrt{a+b \tan (c+d x)} (4 a+b \tan (c+d x)+3 i b)-3 (a+i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )\right )+\frac{6 B (a+b \tan (c+d x))^{5/2}}{b}}{15 d} \]
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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Maple [B] time = 0.107, size = 1686, normalized size = 9.6 \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)*(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)-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*tan(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*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/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)*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))*A*(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/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)*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+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^2-1/4/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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
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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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)*(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{\left (c + d x \right )}\, dx \end{align*}
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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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)*(a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")
[Out]
Timed out