3.1247 \(\int \frac{(a+b \tan (e+f x))^4}{\sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=248 \[ -\frac{2 b^2 \left (-87 a^2 d^2+40 a b c d+b^2 \left (-\left (8 c^2-15 d^2\right )\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}-\frac{i (a-i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}}+\frac{i (a+i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f \sqrt{c+i d}} \]

[Out]

((-I)*(a - I*b)^4*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (I*(a + I*b)^4*ArcTanh[
Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) - (2*b^2*(40*a*b*c*d - 87*a^2*d^2 - b^2*(8*c^2 - 15
*d^2))*Sqrt[c + d*Tan[e + f*x]])/(15*d^3*f) - (4*b^3*(2*b*c - 7*a*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(1
5*d^2*f) + (2*b^2*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(5*d*f)

________________________________________________________________________________________

Rubi [A]  time = 0.734448, antiderivative size = 248, 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 = {3566, 3637, 3630, 3539, 3537, 63, 208} \[ -\frac{2 b^2 \left (-87 a^2 d^2+40 a b c d+b^2 \left (-\left (8 c^2-15 d^2\right )\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}-\frac{i (a-i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}}+\frac{i (a+i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f \sqrt{c+i d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(a - I*b)^4*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (I*(a + I*b)^4*ArcTanh[
Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) - (2*b^2*(40*a*b*c*d - 87*a^2*d^2 - b^2*(8*c^2 - 15
*d^2))*Sqrt[c + d*Tan[e + f*x]])/(15*d^3*f) - (4*b^3*(2*b*c - 7*a*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(1
5*d^2*f) + (2*b^2*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(5*d*f)

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

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}{\sqrt{c+d \tan (e+f x)}} \, dx &=\frac{2 b^2 (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}+\frac{2 \int \frac{(a+b \tan (e+f x)) \left (\frac{1}{2} \left (-4 b^3 c+5 a^3 d-a b^2 d\right )+\frac{5}{2} b \left (3 a^2-b^2\right ) d \tan (e+f x)-b^2 (2 b c-7 a d) \tan ^2(e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{5 d}\\ &=-\frac{4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}-\frac{4 \int \frac{\frac{1}{4} \left (-8 b^4 c^2+40 a b^3 c d-15 a^4 d^2+3 a^2 b^2 d^2\right )-15 a b \left (a^2-b^2\right ) d^2 \tan (e+f x)+\frac{1}{4} b^2 \left (40 a b c d-87 a^2 d^2-b^2 \left (8 c^2-15 d^2\right )\right ) \tan ^2(e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{15 d^2}\\ &=-\frac{2 b^2 \left (40 a b c d-87 a^2 d^2-b^2 \left (8 c^2-15 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}-\frac{4 \int \frac{-\frac{15}{4} \left (a^4-6 a^2 b^2+b^4\right ) d^2-15 a b \left (a^2-b^2\right ) d^2 \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{15 d^2}\\ &=-\frac{2 b^2 \left (40 a b c d-87 a^2 d^2-b^2 \left (8 c^2-15 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}+\frac{1}{2} (a-i b)^4 \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{1}{2} (a+i b)^4 \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=-\frac{2 b^2 \left (40 a b c d-87 a^2 d^2-b^2 \left (8 c^2-15 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}+\frac{\left (i (a-i b)^4\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+i b)^4\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 b^2 \left (40 a b c d-87 a^2 d^2-b^2 \left (8 c^2-15 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 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 )}{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 )}{d f}\\ &=-\frac{i (a-i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d} f}+\frac{i (a+i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d} f}-\frac{2 b^2 \left (40 a b c d-87 a^2 d^2-b^2 \left (8 c^2-15 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{15 d^3 f}-\frac{4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}{5 d f}\\ \end{align*}

Mathematica [A]  time = 3.26805, size = 235, normalized size = 0.95 \[ \frac{\frac{2 b^2 \left (87 a^2 d^2-40 a b c d+b^2 \left (8 c^2-15 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{d^2}+6 b^2 (a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}+\frac{4 b^3 (7 a d-2 b c) \tan (e+f x) \sqrt{c+d \tan (e+f x)}}{d}-\frac{15 i d (a-i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}+\frac{15 i d (a+i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d}}}{15 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-15*I)*(a - I*b)^4*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sqrt[c - I*d] + ((15*I)*(a + I*b)^4*d
*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/Sqrt[c + I*d] + (2*b^2*(-40*a*b*c*d + 87*a^2*d^2 + b^2*(8*c^
2 - 15*d^2))*Sqrt[c + d*Tan[e + f*x]])/d^2 + (4*b^3*(-2*b*c + 7*a*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/d
+ 6*b^2*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(15*d*f)

________________________________________________________________________________________

Maple [B]  time = 0.098, size = 10033, normalized size = 40.5 \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))^(1/2),x)

[Out]

result too large to display

________________________________________________________________________________________

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))^(1/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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))^(1/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{4}}{\sqrt{c + d \tan{\left (e + f x \right )}}}\, 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))**(1/2),x)

[Out]

Integral((a + b*tan(e + f*x))**4/sqrt(c + d*tan(e + f*x)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (f x + e\right ) + a\right )}^{4}}{\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((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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