∫ A+B tan(c+dx)_____ tan7 2 (c+dx)∘ ________

3.457 \(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\)

Optimal. Leaf size=256 \[ \frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}+\frac {(-B+i A) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)} \]

[Out]

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

(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(I*a+b)^(1/2)+2/15*(15*A*a^2-8*A*b^2+10*B*a*b)*(a+b*tan(d*x+c

))^(1/2)/a^3/d/tan(d*x+c)^(1/2)-2/5*A*(a+b*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(5/2)+2/15*(4*A*b-5*B*a)*(a+b*tan(

d*x+c))^(1/2)/a^2/d/tan(d*x+c)^(3/2)

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Rubi [A] time = 1.06, antiderivative size = 256, 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} \[ \frac {2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}+\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(-B+i A) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully veriﬁed.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]),x]

[Out]

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

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

[c + d*x]])/(5*a*d*Tan[c + d*x]^(5/2)) + (2*(4*A*b - 5*a*B)*Sqrt[a + b*Tan[c + d*x]])/(15*a^2*d*Tan[c + d*x]^(

3/2)) + (2*(15*a^2*A - 8*A*b^2 + 10*a*b*B)*Sqrt[a + b*Tan[c + d*x]])/(15*a^3*d*Sqrt[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 {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx &=-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \int \frac {\frac {1}{2} (4 A b-5 a B)+\frac {5}{2} a A \tan (c+d x)+2 A b \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{5 a}\\ &=-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\frac {1}{4} \left (-15 a^2 A+8 A b^2-10 a b B\right )-\frac {15}{4} a^2 B \tan (c+d x)+\frac {1}{2} b (4 A b-5 a B) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}-\frac {8 \int \frac {\frac {15 a^3 B}{8}-\frac {15}{8} a^3 A \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^3}\\ &=-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}-\frac {1}{2} (-i A+B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} (i A+B) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\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 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 d}\\ &=-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\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 )}{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 )}{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} 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 )}{\sqrt {i a+b} d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A b-5 a B) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A] time = 5.29, size = 227, normalized size = 0.89 \[ \frac {\frac {2 \sqrt {a+b \tan (c+d x)} \left (\left (15 a^2 A+10 a b B-8 A b^2\right ) \tan ^2(c+d x)-3 a^2 A-a (5 a B-4 A b) \tan (c+d x)\right )}{a^3 \tan ^{\frac {5}{2}}(c+d x)}-\frac {15 \sqrt [4]{-1} (A-i B) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {15 \sqrt [4]{-1} (A+i B) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}}{15 d} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]),x]

[Out]

((-15*(-1)^(1/4)*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sq

rt[-a + I*b] + (15*(-1)^(1/4)*(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*Sqrt[a + b*Tan[c + d*x]]*(-3*a^2*A - a*(-4*A*b + 5*a*B)*Tan[c + d*x] + (15*a^2*A

- 8*A*b^2 + 10*a*b*B)*Tan[c + d*x]^2))/(a^3*Tan[c + d*x]^(5/2)))/(15*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))/(a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab