3.13.65 \(\int (a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)} \, dx\) [1265]

Optimal. Leaf size=337 \[ -\frac {i (a-i b)^{5/2} \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {i (a+i b)^{5/2} \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {\sqrt {b} \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}-\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f} \]

[Out]

1/4*(10*a*b*c*d+15*a^2*d^2-b^2*(c^2+8*d^2))*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan(f*x+e))^(1
/2))*b^(1/2)/d^(3/2)/f-I*(a-I*b)^(5/2)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2)/(c+d*tan(f*x
+e))^(1/2))*(c-I*d)^(1/2)/f+I*(a+I*b)^(5/2)*arctanh((c+I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*ta
n(f*x+e))^(1/2))*(c+I*d)^(1/2)/f-1/4*b*(-9*a*d+b*c)*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/d/f+1/2*b^2*
(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2)/d/f

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Rubi [A]
time = 2.84, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3647, 3728, 3736, 6857, 65, 223, 212, 95, 214} \begin {gather*} \frac {\sqrt {b} \left (15 a^2 d^2+10 a b c d-\left (b^2 \left (c^2+8 d^2\right )\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}-\frac {i (a-i b)^{5/2} \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {i (a+i b)^{5/2} \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(a - I*b)^(5/2)*Sqrt[c - I*d]*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d
*Tan[e + f*x]])])/f + (I*(a + I*b)^(5/2)*Sqrt[c + I*d]*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[
a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/f + (Sqrt[b]*(10*a*b*c*d + 15*a^2*d^2 - b^2*(c^2 + 8*d^2))*ArcTanh[(Sqrt[
d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(4*d^(3/2)*f) - (b*(b*c - 9*a*d)*Sqrt[a + b*
Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*d*f) + (b^2*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(2
*d*f)

Rule 65

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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3647

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 3728

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((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 - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3736

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^2)/(1 + ff^2*x^2)), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int (a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)} \, dx &=\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac {\int \frac {\sqrt {c+d \tan (e+f x)} \left (\frac {1}{2} \left (-b^3 c+4 a^3 d-3 a b^2 d\right )+2 b \left (3 a^2-b^2\right ) d \tan (e+f x)-\frac {1}{2} b^2 (b c-9 a d) \tan ^2(e+f x)\right )}{\sqrt {a+b \tan (e+f x)}} \, dx}{2 d}\\ &=-\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac {\int \frac {-\frac {1}{4} b \left (b^3 c^2-8 a^3 c d+14 a b^2 c d+9 a^2 b d^2\right )+2 b 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}{4} b^2 \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 b d}\\ &=-\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac {\text {Subst}\left (\int \frac {-\frac {1}{4} b \left (b^3 c^2-8 a^3 c d+14 a b^2 c d+9 a^2 b d^2\right )+2 b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) x+\frac {1}{4} b^2 \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) x^2}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 b d f}\\ &=-\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac {\text {Subst}\left (\int \left (\frac {b^2 \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right )}{4 \sqrt {a+b x} \sqrt {c+d x}}+\frac {2 \left (b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )+b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) x\right )}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{2 b d f}\\ &=-\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac {\text {Subst}\left (\int \frac {b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )+b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) x}{\sqrt {a+b x} \sqrt {c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{b d f}+\frac {\left (b \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{8 d f}\\ &=-\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac {\text {Subst}\left (\int \left (\frac {-b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )+i b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )}{2 (i-x) \sqrt {a+b x} \sqrt {c+d x}}+\frac {b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )+i b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )}{2 (i+x) \sqrt {a+b x} \sqrt {c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{b d f}+\frac {\left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b \tan (e+f x)}\right )}{4 d f}\\ &=-\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac {\left ((i a+b)^3 (c-i d)\right ) \text {Subst}\left (\int \frac {1}{(i+x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {\left (-b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )+i b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )\right ) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 b d f}+\frac {\left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{4 d f}\\ &=\frac {\sqrt {b} \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}-\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac {\left ((i a+b)^3 (c-i d)\right ) \text {Subst}\left (\int \frac {1}{-a+i b-(-c+i d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {\left (-b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )+i b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )\right ) \text {Subst}\left (\int \frac {1}{a+i b-(c+i d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{b d f}\\ &=-\frac {i (a-i b)^{5/2} \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {i (a+i b)^{5/2} \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {\sqrt {b} \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}-\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\\ \end {align*}

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Mathematica [A]
time = 6.03, size = 565, normalized size = 1.68 \begin {gather*} \frac {\frac {4 b d \left (b \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )+\sqrt {-b^2} \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {-c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+\sqrt {-b^2}} \sqrt {-c+\frac {\sqrt {-b^2} d}{b}}}-\frac {4 b d \left (b \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )-\sqrt {-b^2} \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+\frac {\sqrt {-b^2} d}{b}}}+b^3 (-b c+9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}+2 b^4 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}-\frac {b^{5/2} \sqrt {c-\frac {a d}{b}} \left (-10 a b c d-15 a^2 d^2+b^2 \left (c^2+8 d^2\right )\right ) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{\sqrt {d} \sqrt {c+d \tan (e+f x)}}}{4 b^2 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*b*d*(b*(3*a^2*b*c - b^3*c + a^3*d - 3*a*b^2*d) + Sqrt[-b^2]*(a^3*c - 3*a*b^2*c - 3*a^2*b*d + b^3*d))*ArcTa
nh[(Sqrt[-c + (Sqrt[-b^2]*d)/b]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(
Sqrt[-a + Sqrt[-b^2]]*Sqrt[-c + (Sqrt[-b^2]*d)/b]) - (4*b*d*(b*(3*a^2*b*c - b^3*c + a^3*d - 3*a*b^2*d) - Sqrt[
-b^2]*(a^3*c - 3*a*b^2*c - 3*a^2*b*d + b^3*d))*ArcTanh[(Sqrt[c + (Sqrt[-b^2]*d)/b]*Sqrt[a + b*Tan[e + f*x]])/(
Sqrt[a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[c + (Sqrt[-b^2]*d)/b]) + b^3*(-(b*
c) + 9*a*d)*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]] + 2*b^4*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e +
f*x])^(3/2) - (b^(5/2)*Sqrt[c - (a*d)/b]*(-10*a*b*c*d - 15*a^2*d^2 + b^2*(c^2 + 8*d^2))*ArcSinh[(Sqrt[d]*Sqrt[
a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c - (a*d)/b])]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(Sqrt[d]*Sqrt[c
+ d*Tan[e + f*x]]))/(4*b^2*d*f)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \sqrt {c +d \tan \left (f x +e \right )}\, \left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}}\, dx\]

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))^(5/2),x)

[Out]

int((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^(5/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \sqrt {c + d \tan {\left (e + f x \right )}}\, dx \end {gather*}

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))**(5/2),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))^(5/2)*(c + d*tan(e + f*x))^(1/2),x)

[Out]

int((a + b*tan(e + f*x))^(5/2)*(c + d*tan(e + f*x))^(1/2), x)

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