3.458 \(\int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=219 \[ \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {(-B+i A) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d} \]

[Out]

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

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Rubi [A]  time = 1.77, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3605, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {(-B+i A) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a - b)^(3/2)*d)) + (2*B*
ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(b^(3/2)*d) - ((I*A + B)*ArcTanh[(Sqrt[I*a + b
]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a + b)^(3/2)*d) + (2*a*(A*b - a*B)*Sqrt[Tan[c + d*x]])/(b
*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]])

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 205

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

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 3605

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

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

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Mathematica [C]  time = 40.22, size = 177751, normalized size = 811.65 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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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(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),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(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 1.99, size = 1560668, normalized size = 7126.34 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^(3/2)/(b*tan(d*x + c) + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(3/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**(3/2),x)

[Out]

Integral((A + B*tan(c + d*x))*tan(c + d*x)**(3/2)/(a + b*tan(c + d*x))**(3/2), x)

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