3.1253 \(\int \frac{(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=317 \[ -\frac{2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 f \left (c^2+d^2\right )}-\frac{2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 f \left (c^2+d^2\right )}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{i (a-i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}}+\frac{i (a+i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}} \]

[Out]

((-I)*(a - I*b)^4*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(3/2)*f) + (I*(a + I*b)^4*ArcTan
h[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((c + I*d)^(3/2)*f) - (2*(b*c - a*d)^2*(a + b*Tan[e + f*x])^2)/(d*(
c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]) - (2*b*(15*a^2*b*c*d^2 - 6*a^3*d^3 - 12*a*b^2*d*(2*c^2 + d^2) + b^3*(8*
c^3 + 5*c*d^2))*Sqrt[c + d*Tan[e + f*x]])/(3*d^3*(c^2 + d^2)*f) - (2*b^2*(3*a*d*(2*b*c - a*d) - b^2*(4*c^2 + d
^2))*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d^2*(c^2 + d^2)*f)

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Rubi [A]  time = 0.910865, antiderivative size = 317, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {3565, 3637, 3630, 3539, 3537, 63, 208} \[ -\frac{2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 f \left (c^2+d^2\right )}-\frac{2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 f \left (c^2+d^2\right )}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{i (a-i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}}+\frac{i (a+i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^4/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

((-I)*(a - I*b)^4*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(3/2)*f) + (I*(a + I*b)^4*ArcTan
h[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((c + I*d)^(3/2)*f) - (2*(b*c - a*d)^2*(a + b*Tan[e + f*x])^2)/(d*(
c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]) - (2*b*(15*a^2*b*c*d^2 - 6*a^3*d^3 - 12*a*b^2*d*(2*c^2 + d^2) + b^3*(8*
c^3 + 5*c*d^2))*Sqrt[c + d*Tan[e + f*x]])/(3*d^3*(c^2 + d^2)*f) - (2*b^2*(3*a*d*(2*b*c - a*d) - b^2*(4*c^2 + d
^2))*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d^2*(c^2 + d^2)*f)

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 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(b*C*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] &&  !LtQ[n, -1]

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(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*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

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))^4}{(c+d \tan (e+f x))^{3/2}} \, dx &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 \int \frac{(a+b \tan (e+f x)) \left (\frac{1}{2} \left (4 b^3 c^2+a^3 c d-9 a b^2 c d+6 a^2 b d^2\right )+\frac{1}{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 (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{d \left (c^2+d^2\right )}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}-\frac{4 \int \frac{\frac{1}{4} \left (-24 a b^3 c^2 d-3 a^4 c d^2+33 a^2 b^2 c d^2-18 a^3 b d^3+2 b^4 c \left (4 c^2+d^2\right )\right )-\frac{3}{4} d^2 \left (4 a^3 b c-4 a b^3 c-a^4 d+6 a^2 b^2 d-b^4 d\right ) \tan (e+f x)+\frac{1}{4} b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \tan ^2(e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d^2 \left (c^2+d^2\right )}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac{2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}-\frac{4 \int \frac{-\frac{3}{4} d^2 \left (a^4 c-6 a^2 b^2 c+b^4 c+4 a^3 b d-4 a b^3 d\right )-\frac{3}{4} d^2 \left (4 a^3 b c-4 a b^3 c-a^4 d+6 a^2 b^2 d-b^4 d\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d^2 \left (c^2+d^2\right )}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac{2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}+\frac{(a-i b)^4 \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)}+\frac{(a+i b)^4 \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac{2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}+\frac{(a+i b)^4 \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (i c-d) f}-\frac{(a-i b)^4 \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (i c+d) f}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac{2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}-\frac{(a-i b)^4 \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) d f}-\frac{(a+i b)^4 \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) d f}\\ &=-\frac{i (a-i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(c-i d)^{3/2} f}+\frac{i (a+i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(c+i d)^{3/2} f}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}-\frac{2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac{2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}\\ \end{align*}

Mathematica [C]  time = 6.32813, size = 415, normalized size = 1.31 \[ \frac{2 b^2 (a+b \tan (e+f x))^2}{3 d f \sqrt{c+d \tan (e+f x)}}+\frac{2 \left (-\frac{2 b^2 (2 b c-5 a d) (a+b \tan (e+f x))}{d f \sqrt{c+d \tan (e+f x)}}+\frac{-\frac{2 \left (29 a^2 b^2 d^2-28 a b^3 c d+8 b^4 c^2-3 b^4 d^2\right )}{d \sqrt{c+d \tan (e+f x)}}+\frac{2 \left (\frac{\left (\frac{3}{2} d^4 \left (-6 a^2 b^2+a^4+b^4\right )-6 a b c d^3 (a-b) (a+b)\right ) \left (\frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c+i d}\right )}{(-d+i c) \sqrt{c+d \tan (e+f x)}}-\frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c-i d}\right )}{(d+i c) \sqrt{c+d \tan (e+f x)}}\right )}{d}+6 a b d^2 (a-b) (a+b) \left (\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}\right )\right )}{d}}{2 d f}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^4/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

(2*b^2*(a + b*Tan[e + f*x])^2)/(3*d*f*Sqrt[c + d*Tan[e + f*x]]) + (2*((-2*b^2*(2*b*c - 5*a*d)*(a + b*Tan[e + f
*x]))/(d*f*Sqrt[c + d*Tan[e + f*x]]) + ((-2*(8*b^4*c^2 - 28*a*b^3*c*d + 29*a^2*b^2*d^2 - 3*b^4*d^2))/(d*Sqrt[c
+ d*Tan[e + f*x]]) + (2*(6*a*(a - b)*b*(a + b)*d^2*(((-I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sq
rt[c - I*d] + (I*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/Sqrt[c + I*d]) + ((-6*a*(a - b)*b*(a + b)*c*
d^3 + (3*(a^4 - 6*a^2*b^2 + b^4)*d^4)/2)*(-(Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)]/((
I*c + d)*Sqrt[c + d*Tan[e + f*x]])) + Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)]/((I*c -
d)*Sqrt[c + d*Tan[e + f*x]])))/d))/d)/(2*d*f)))/(3*d)

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Maple [B]  time = 0.101, size = 20054, normalized size = 63.3 \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))^4/(c+d*tan(f*x+e))^(3/2),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))^4/(c+d*tan(f*x+e))^(3/2),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))^4/(c+d*tan(f*x+e))^(3/2),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 )^{4}}{\left (c + d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))**4/(c+d*tan(f*x+e))**(3/2),x)

[Out]

Integral((a + b*tan(e + f*x))**4/(c + d*tan(e + f*x))**(3/2), 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 )}^{4}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^4/(d*tan(f*x + e) + c)^(3/2), x)