3.1260 \(\int \frac{(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=219 \[ -\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{(-b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}}+\frac{(b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.649914, antiderivative size = 219, 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 = {3565, 3628, 3539, 3537, 63, 208} \[ -\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{(-b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}}+\frac{(b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^3/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

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

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3628

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^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 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 \frac{(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \int \frac{\frac{1}{2} \left (2 b^3 c^2+3 a^3 c d-7 a b^2 c d+8 a^2 b d^2\right )+\frac{3}{2} d \left (3 a^2 b c-b^3 c-a^3 d+3 a b^2 d\right ) \tan (e+f x)+\frac{1}{2} b \left (2 a b c d-a^2 d^2+b^2 \left (2 c^2+3 d^2\right )\right ) \tan ^2(e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx}{3 d \left (c^2+d^2\right )}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 \int \frac{\frac{3}{2} d \left (6 a^2 b c d-2 b^3 c d+a^3 \left (c^2-d^2\right )-3 a b^2 \left (c^2-d^2\right )\right )-\frac{3}{2} d \left (2 a^3 c d-6 a b^2 c d-3 a^2 b \left (c^2-d^2\right )+b^3 \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d \left (c^2+d^2\right )^2}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{(a-i b)^3 \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)^2}+\frac{(a+i b)^3 \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)^2}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{(i a+b)^3 \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d)^2 f}+\frac{(i a-b)^3 \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d)^2 f}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{(a-i b)^3 \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 )}{(c-i d)^2 d f}-\frac{(a+i b)^3 \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 )}{(c+i d)^2 d f}\\ &=\frac{(i a+b)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(c-i d)^{5/2} f}-\frac{(i a-b)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(c+i d)^{5/2} f}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.44385, size = 284, normalized size = 1.3 \[ -\frac{-3 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x)) \left (i (c+i d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c-i d}\right )-(d+i c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c+i d}\right )\right )-d \left (-3 a^2 b c+a^3 d-3 a b^2 d+b^3 c\right ) \left (i (c+i d) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c+d \tan (e+f x)}{c-i d}\right )-(d+i c) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c+d \tan (e+f x)}{c+i d}\right )\right )+6 b^2 d (c-i d) (c+i d) (a+b \tan (e+f x))+4 b^3 c \left (c^2+d^2\right )}{3 d^2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^3/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

-(4*b^3*c*(c^2 + d^2) - d*(-3*a^2*b*c + b^3*c + a^3*d - 3*a*b^2*d)*(I*(c + I*d)*Hypergeometric2F1[-3/2, 1, -1/
2, (c + d*Tan[e + f*x])/(c - I*d)] - (I*c + d)*Hypergeometric2F1[-3/2, 1, -1/2, (c + d*Tan[e + f*x])/(c + I*d)
]) + 6*b^2*(c - I*d)*(c + I*d)*d*(a + b*Tan[e + f*x]) - 3*b*(3*a^2 - b^2)*d*(I*(c + I*d)*Hypergeometric2F1[-1/
2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)] - (I*c + d)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c
+ I*d)])*(c + d*Tan[e + f*x]))/(3*d^2*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2))

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Maple [B]  time = 0.111, size = 26303, normalized size = 120.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^3/(c+d*tan(f*x+e))^(5/2),x)

[Out]

result too large to display

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Maxima [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))^3/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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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))^3/(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 \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{3}}{\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))**3/(c+d*tan(f*x+e))**(5/2),x)

[Out]

Integral((a + b*tan(e + f*x))**3/(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 \frac{{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{{\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))^3/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^3/(d*tan(f*x + e) + c)^(5/2), x)