3.1243 \(\int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=188 \[ \frac{2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{(b+i a) (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{(-b+i a) (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{2 b (c+d \tan (e+f x))^{5/2}}{5 f} \]

[Out]

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

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Rubi [A]  time = 0.427378, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ \frac{2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{(b+i a) (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{(-b+i a) (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{2 b (c+d \tan (e+f x))^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2),x]

[Out]

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

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

Mathematica [A]  time = 1.07993, size = 233, normalized size = 1.24 \[ \frac{i \left ((a-i b) \left (\frac{2}{5} (c+d \tan (e+f x))^{5/2}+\frac{2}{3} (c-i d) \left (\sqrt{c+d \tan (e+f x)} (4 c+d \tan (e+f x)-3 i d)-3 (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )\right )\right )-(a+i b) \left (\frac{2}{5} (c+d \tan (e+f x))^{5/2}+\frac{2}{3} (c+i d) \left (\sqrt{c+d \tan (e+f x)} (4 c+d \tan (e+f x)+3 i d)-3 (c+i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )\right )\right )\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2),x]

[Out]

((I/2)*((a - I*b)*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (2*(c - I*d)*(-3*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[
e + f*x]]/Sqrt[c - I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c - (3*I)*d + d*Tan[e + f*x])))/3) - (a + I*b)*((2*(c +
d*Tan[e + f*x])^(5/2))/5 + (2*(c + I*d)*(-3*(c + I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] +
Sqrt[c + d*Tan[e + f*x]]*(4*c + (3*I)*d + d*Tan[e + f*x])))/3)))/f

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Maple [B]  time = 0.05, size = 2405, normalized size = 12.8 \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(f*x+e))*(c+d*tan(f*x+e))^(5/2),x)

[Out]

2/3/f*(c+d*tan(f*x+e))^(3/2)*b*c+2/f*b*c^2*(c+d*tan(f*x+e))^(1/2)-2/f*b*d^2*(c+d*tan(f*x+e))^(1/2)+3/f*a/(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+2/3/f*a*(c+d*tan(f*x+e))^(3/2)*d+1/f*a/(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))*d^3+1/4/f*d^2*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+2/5*b*(c+d*tan(f
*x+e))^(5/2)/f+4/f*a*(c+d*tan(f*x+e))^(1/2)*c*d-1/f*a/(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))*d^3+1/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arc
tan(((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)*b*
c^2+1/f*d^2/(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)*b-3/f*d^2/(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*c+1/4/f*d*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+1/4/
f/d*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*c^3-3/4/f*d*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*c-1/f*d^2/(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))*(c^2+d^2)^(1/2)*b+3/f*d^2/(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*c-1/4/
f*d*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-1/4/f/d*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*c^3+3/4/f*d*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*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)*(c^2+d^2)^(1/2)*b*c-1/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)*b*c^2-1/2/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)*(c^2+d^2)^(1/2)*b*c-1/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arcta
n(((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*c^3+3/4/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*c^2-1/4/f*d^2*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+1/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*c^3-3/4/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*c^2-3/f*a/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arc
tan(((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/f*d/(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*c-1/4/f/d*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*c^2+1/4/f/d*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*c^2+2/f*d
/(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))*(c^2+d^2)^(1/2)*a*c

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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(f*x+e))*(c+d*tan(f*x+e))^(5/2),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(f*x+e))*(c+d*tan(f*x+e))^(5/2),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 ) \left (c + d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))*(c+d*tan(f*x+e))**(5/2),x)

[Out]

Integral((a + b*tan(e + f*x))*(c + d*tan(e + f*x))**(5/2), 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 )}{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))*(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)*(d*tan(f*x + e) + c)^(5/2), x)