3.467 \(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=301 \[ -\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}+\frac {(-B+i A) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \]

[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)^(5/2)/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)^(5/2)/d-2/3*b*(3*A*a^4+17*A*a^2*b^2+8*A*b^4-8*B*a^3*b-2
*B*a*b^3)*tan(d*x+c)^(1/2)/a^3/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)-2*A/a/d/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^
(3/2)-2/3*b*(3*A*a^2+4*A*b^2-B*a*b)*tan(d*x+c)^(1/2)/a^2/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)

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Rubi [A]  time = 1.19, antiderivative size = 301, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3609, 3649, 3616, 3615, 93, 203, 206} \[ -\frac {2 b \left (17 a^2 A b^2+3 a^4 A-8 a^3 b B-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}-\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{3 a^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {(-B+i A) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(5/2)),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)^(5/2)*d) - ((I*A +
B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a + b)^(5/2)*d) - (2*A)/(a*d*Sqrt
[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)) - (2*b*(3*a^2*A + 4*A*b^2 - a*b*B)*Sqrt[Tan[c + d*x]])/(3*a^2*(a^2
+ b^2)*d*(a + b*Tan[c + d*x])^(3/2)) - (2*b*(3*a^4*A + 17*a^2*A*b^2 + 8*A*b^4 - 8*a^3*b*B - 2*a*b^3*B)*Sqrt[Ta
n[c + d*x]])/(3*a^3*(a^2 + b^2)^2*d*Sqrt[a + b*Tan[c + d*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 203

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

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 3609

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*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n
+ 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e +
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ
[a, 0])))

Rule 3615

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

Rule 3616

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

Rule 3649

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*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx &=-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 \int \frac {\frac {1}{2} (4 A b-a B)+\frac {1}{2} a A \tan (c+d x)+2 A b \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}} \, dx}{a}\\ &=-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {4 \int \frac {\frac {1}{4} \left (9 a^2 A b+8 A b^3-3 a^3 B-2 a b^2 B\right )+\frac {3}{4} a^2 (a A+b B) \tan (c+d x)+\frac {1}{2} b \left (3 a^2 A+4 A b^2-a b B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 a^2 \left (a^2+b^2\right )}\\ &=-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {8 \int \frac {\frac {3}{8} a^3 \left (2 a A b-a^2 B+b^2 B\right )+\frac {3}{8} a^3 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a^3 \left (a^2+b^2\right )^2}\\ &=-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(i A-B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}+\frac {(i A+B) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}\\ &=-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(i A-B) \operatorname {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}+\frac {(i A+B) \operatorname {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d}\\ &=-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(i A-B) \operatorname {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^2 d}+\frac {(i A+B) \operatorname {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a-i b)^2 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 )}{\sqrt {i a-b} (a+i b)^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)^{5/2} d}-\frac {2 A}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2 A+4 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4 A+17 a^2 A b^2+8 A b^4-8 a^3 b B-2 a b^3 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 3.76, size = 326, normalized size = 1.08 \[ \frac {-\frac {2 b \left (3 a^2 A-a b B+4 A b^2\right ) \sqrt {\tan (c+d x)}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (3 a^4 A-8 a^3 b B+17 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}+3 \sqrt [4]{-1} a^3 \left (\frac {(a+i b)^2 (A-i B) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}-\frac {(a-i b)^2 (A+i B) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )}{a \left (a^2+b^2\right )^2}-\frac {6 a A}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(5/2)),x]

[Out]

((-6*a*A)/(Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)) - (2*b*(3*a^2*A + 4*A*b^2 - a*b*B)*Sqrt[Tan[c + d*x]
])/((a^2 + b^2)*(a + b*Tan[c + d*x])^(3/2)) + (3*(-1)^(1/4)*a^3*(((a + I*b)^2*(A - I*B)*ArcTan[((-1)^(1/4)*Sqr
t[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] - ((a - I*b)^2*(A + I*B)*ArcTan[((-1
)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b]) - (2*b*(3*a^4*A + 17*a^2*A
*b^2 + 8*A*b^4 - 8*a^3*b*B - 2*a*b^3*B)*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]])/(a*(a^2 + b^2)^2))/(3*a^
2*d)

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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((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),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((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 2.39, size = 2978232, normalized size = 9894.46 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x)

[Out]

result too large to display

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maxima [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((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(5/2)),x)

[Out]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(5/2)), x)

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sympy [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((A+B*tan(d*x+c))/tan(d*x+c)**(3/2)/(a+b*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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