3.1230 \(\int (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)} \, dx\)
Optimal. Leaf size=157 \[ \frac{4 a b \sqrt{c+d \tan (e+f x)}}{f}-\frac{i (a-i b)^2 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{i (a+i b)^2 \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f} \]
[Out]
((-I)*(a - I*b)^2*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + (I*(a + I*b)^2*Sqrt[c + I
*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (4*a*b*Sqrt[c + d*Tan[e + f*x]])/f + (2*b^2*(c + d*Ta
n[e + f*x])^(3/2))/(3*d*f)
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Rubi [A] time = 0.349869, antiderivative size = 157, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used =
{3543, 3528, 3539, 3537, 63, 208} \[ \frac{4 a b \sqrt{c+d \tan (e+f x)}}{f}-\frac{i (a-i b)^2 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{i (a+i b)^2 \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]],x]
[Out]
((-I)*(a - I*b)^2*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + (I*(a + I*b)^2*Sqrt[c + I
*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (4*a*b*Sqrt[c + d*Tan[e + f*x]])/f + (2*b^2*(c + d*Ta
n[e + f*x])^(3/2))/(3*d*f)
Rule 3543
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !LeQ[m, -1] && !(EqQ[m, 2] && EqQ
[a, 0])
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 (e+f x))^2 \sqrt{c+d \tan (e+f x)} \, dx &=\frac{2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\int \left (a^2-b^2+2 a b \tan (e+f x)\right ) \sqrt{c+d \tan (e+f x)} \, dx\\ &=\frac{4 a b \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\int \frac{a^2 c-b^2 c-2 a b d+\left (2 a b c+a^2 d-b^2 d\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{4 a b \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac{1}{2} \left ((a-i b)^2 (c-i d)\right ) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{1}{2} \left ((a+i b)^2 (c+i d)\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{4 a b \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}-\frac{\left ((a+i b)^2 (i c-d)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 f}+\frac{\left ((a-i b)^2 (i c+d)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 f}\\ &=\frac{4 a b \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}-\frac{\left ((a-i b)^2 (c-i d)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{d f}-\frac{\left ((a+i b)^2 (c+i d)\right ) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac{i (a-i b)^2 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{i (a+i b)^2 \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{4 a b \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\\ \end{align*}
Mathematica [A] time = 0.53634, size = 149, normalized size = 0.95 \[ \frac{2 b \sqrt{c+d \tan (e+f x)} (6 a d+b c+b d \tan (e+f x))-3 i d (a-i b)^2 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )+3 i d (a+i b)^2 \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{3 d f} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]],x]
[Out]
((-3*I)*(a - I*b)^2*Sqrt[c - I*d]*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + (3*I)*(a + I*b)^2*Sqrt[c
+ I*d]*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + 2*b*Sqrt[c + d*Tan[e + f*x]]*(b*c + 6*a*d + b*d*Ta
n[e + f*x]))/(3*d*f)
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Maple [B] time = 0.048, size = 1507, 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((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,x)
[Out]
2/3*b^2*(c+d*tan(f*x+e))^(3/2)/d/f+4*a*b*(c+d*tan(f*x+e))^(1/2)/f+d/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2
*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b^2+d/f/(2*(c^2+d^2)^(1/2
)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a^
2-d/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^
2)^(1/2)-2*c)^(1/2))*b^2+1/4/d/f*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d
^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^2-1/4/d/f*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1
/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b^2-1/4/d/f*ln((c
+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/
2)*a^2*c+1/2/f*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2
+d^2)^(1/2)+2*c)^(1/2)*a*b+1/4/d/f*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2
+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2*c-2/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+
2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a*b*c+2/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*ar
ctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*a
*b-d/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d
^2)^(1/2)-2*c)^(1/2))*a^2-1/4/d/f*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+
d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^2+1/4/d/f*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)
*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b^2+1/4/d/f*ln(d
*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1
/2)*a^2*c-1/2/f*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^
2+d^2)^(1/2)+2*c)^(1/2)*a*b-1/4/d/f*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^
2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^2*c-2/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))
^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*a*b+2/f/(2*(c^2+d^2)^(1/2
)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a*
b*c
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right ) + a\right )}^{2} \sqrt{d \tan \left (f x + e\right ) + c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,x, algorithm="maxima")
[Out]
integrate((b*tan(f*x + e) + a)^2*sqrt(d*tan(f*x + e) + 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((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,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 (e + f x \right )}\right )^{2} \sqrt{c + d \tan{\left (e + f x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))**(1/2)*(a+b*tan(f*x+e))**2,x)
[Out]
Integral((a + b*tan(e + f*x))**2*sqrt(c + d*tan(e + f*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right ) + a\right )}^{2} \sqrt{d \tan \left (f x + e\right ) + c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,x, algorithm="giac")
[Out]
integrate((b*tan(f*x + e) + a)^2*sqrt(d*tan(f*x + e) + c), x)