3.2.32 \(\int \frac {\sqrt {c+d \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(a+b \tan (e+f x))^{3/2}} \, dx\) [132]

Optimal. Leaf size=300 \[ -\frac {(i A+B-i C) \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} f}-\frac {(B-i (A-C)) \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} f}+\frac {2 C \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} f}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{b \left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 2.88, antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3726, 3736, 6857, 65, 223, 212, 95, 214} \begin {gather*} -\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}-\frac {\sqrt {c-i d} (i A+B-i C) \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 (a-i b)^{3/2}}-\frac {\sqrt {c+i d} (B-i (A-C)) \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 (a+i b)^{3/2}}+\frac {2 C \sqrt {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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*A + B - I*C)*Sqrt[c - I*d]*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Ta
n[e + f*x]])])/((a - I*b)^(3/2)*f)) - ((B - I*(A - C))*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]])])/((a + I*b)^(3/2)*f) + (2*C*Sqrt[d]*ArcTanh[(Sqrt[d]*Sqrt[a
+ b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/(b^(3/2)*f) - (2*(A*b^2 - a*(b*B - a*C))*Sqrt[c + d*Ta
n[e + f*x]])/(b*(a^2 + b^2)*f*Sqrt[a + b*Tan[e + f*x]])

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 3726

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] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Ta
n[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

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

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 33.57, size = 621058, normalized size = 2070.19 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F(-1)] grade_annotation
time = 180.00, size = 0, normalized size = 0.00 hanged

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)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(3/2),x)

[Out]

int((c+d*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(3/2),x)

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Maxima [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)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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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)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c + d \tan {\left (e + f x \right )}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, 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)+C*tan(f*x+e)**2)/(a+b*tan(f*x+e))**(3/2),x)

[Out]

Integral(sqrt(c + d*tan(e + f*x))*(A + B*tan(e + f*x) + C*tan(e + f*x)**2)/(a + b*tan(e + f*x))**(3/2), 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)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(3/2),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(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c + d*tan(e + f*x))^(1/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/(a + b*tan(e + f*x))^(3/2),x)

[Out]

\text{Hanged}

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