3.327 \(\int (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\)
Optimal. Leaf size=150 \[ \frac{2 (a B+A b) \sqrt{a+b \tan (c+d x)}}{d}-\frac{(a-i b)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{(a+i b)^{3/2} (-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}+\frac{2 B (a+b \tan (c+d x))^{3/2}}{3 d} \]
[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)*
ArcTanh[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)
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Rubi [A] time = 0.312391, antiderivative size = 150, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3528, 3539, 3537, 63, 208} \[ \frac{2 (a B+A b) \sqrt{a+b \tan (c+d x)}}{d}-\frac{(a-i b)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{(a+i b)^{3/2} (-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}+\frac{2 B (a+b \tan (c+d x))^{3/2}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[(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)*
ArcTanh[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)
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 (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac{2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\int \sqrt{a+b \tan (c+d x)} (a A-b B+(A b+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}+\int \frac{a^2 A-A b^2-2 a b 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\\ &=\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{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{\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{\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}\\ \end{align*}
Mathematica [A] time = 0.482307, size = 140, normalized size = 0.93 \[ \frac{2 \sqrt{a+b \tan (c+d x)} (4 a B+3 A b+b B \tan (c+d x))-3 i (a-i b)^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )+3 i (a+i b)^{3/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]
[Out]
((-3*I)*(a - I*b)^(3/2)*(A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + (3*I)*(a + I*b)^(3/2)*(A +
I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + 2*Sqrt[a + b*Tan[c + d*x]]*(3*A*b + 4*a*B + b*B*Tan[c
+ d*x]))/(3*d)
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Maple [B] time = 0.088, size = 1665, normalized size = 11.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)
[Out]
2/d*b*A*(a+b*tan(d*x+c))^(1/2)+2/d*B*a*(a+b*tan(d*x+c))^(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+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)-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/(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*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)-1/4/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^2-1/4/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/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+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+1/4/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+1/4/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*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)+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/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(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-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+2/3*B*(a+b*tan(d*x+c))^(3/2)/d
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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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((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}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((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), 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((a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")
[Out]
Timed out