3.1279 \(\int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+b \tan (e+f x)}} \, dx\)

Optimal. Leaf size=264 \[ \frac {d^{3/2} (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} f}+\frac {d^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}-\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b}}+\frac {i (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a+i b}} \]

[Out]

d^(3/2)*(-a*d+5*b*c)*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan(f*x+e))^(1/2))/b^(3/2)/f-I*(c-I*d
)^(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))/f/(a-I*b)^(1/2)+I*(
c+I*d)^(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))/f/(a+I*b)^(1/2
)+d^2*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/b/f

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Rubi [A]  time = 2.50, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3566, 3655, 6725, 63, 217, 206, 93, 208} \[ \frac {d^{3/2} (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} f}+\frac {d^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}-\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b}}+\frac {i (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a+i b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 93

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 206

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

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]

Rule 217

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 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 3655

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 6725

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

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Mathematica [A]  time = 2.88, size = 432, normalized size = 1.64 \[ \frac {\frac {b \left (\sqrt {-b^2} c \left (c^2-3 d^2\right )-b d \left (d^2-3 c^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}-c} \sqrt {a+b \tan (e+f x)}}{\sqrt {\sqrt {-b^2}-a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {\sqrt {-b^2}-a} \sqrt {\frac {\sqrt {-b^2} d}{b}-c}}+\frac {b \left (\sqrt {-b^2} c \left (c^2-3 d^2\right )+b d \left (d^2-3 c^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {-b^2} d}{b}+c} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {\frac {\sqrt {-b^2} d}{b}+c}}+\frac {\sqrt {b} d^{3/2} (5 b c-a d) \sqrt {c-\frac {a d}{b}} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right )}{\sqrt {c+d \tan (e+f x)}}+b d^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)/(a+b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)/(a+b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{\sqrt {a +b \tan \left (f x +e \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)/(a+b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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