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

Optimal. Leaf size=193 \[ \frac{a^2 (A+i B) \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{5}{2},1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}}+\frac{a^2 (A-i B) \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{5}{2},1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}} \]

[Out]

(a^2*(A + I*B)*AppellF1[1 + m, -5/2, 1, 2 + m, -((b*Tan[c + d*x])/a), (-I)*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*
Sqrt[a + b*Tan[c + d*x]])/(2*d*(1 + m)*Sqrt[1 + (b*Tan[c + d*x])/a]) + (a^2*(A - I*B)*AppellF1[1 + m, -5/2, 1,
2 + m, -((b*Tan[c + d*x])/a), I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*Sqrt[a + b*Tan[c + d*x]])/(2*d*(1 + m)*Sqr
t[1 + (b*Tan[c + d*x])/a])

________________________________________________________________________________________

Rubi [A]  time = 0.451647, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {3603, 3602, 135, 133} \[ \frac{a^2 (A+i B) \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{5}{2},1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}}+\frac{a^2 (A-i B) \tan ^{m+1}(c+d x) \sqrt{a+b \tan (c+d x)} F_1\left (m+1;-\frac{5}{2},1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt{\frac{b \tan (c+d x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(A + I*B)*AppellF1[1 + m, -5/2, 1, 2 + m, -((b*Tan[c + d*x])/a), (-I)*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*
Sqrt[a + b*Tan[c + d*x]])/(2*d*(1 + m)*Sqrt[1 + (b*Tan[c + d*x])/a]) + (a^2*(A - I*B)*AppellF1[1 + m, -5/2, 1,
2 + m, -((b*Tan[c + d*x])/a), I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*Sqrt[a + b*Tan[c + d*x]])/(2*d*(1 + m)*Sqr
t[1 + (b*Tan[c + d*x])/a])

Rule 3603

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] &&  !Integ
erQ[m] &&  !IntegerQ[n] &&  !IntegersQ[2*m, 2*n] && NeQ[A^2 + B^2, 0]

Rule 3602

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] &&  !IntegerQ
[m] &&  !IntegerQ[n] &&  !IntegersQ[2*m, 2*n] && EqQ[A^2 + B^2, 0]

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac{1}{2} (A-i B) \int (1+i \tan (c+d x)) \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} \, dx+\frac{1}{2} (A+i B) \int (1-i \tan (c+d x)) \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\\ &=\frac{(A-i B) \operatorname{Subst}\left (\int \frac{x^m (a+b x)^{5/2}}{1-i x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{(A+i B) \operatorname{Subst}\left (\int \frac{x^m (a+b x)^{5/2}}{1+i x} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{\left (a^2 (A-i B) \sqrt{a+b \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{x^m \left (1+\frac{b x}{a}\right )^{5/2}}{1-i x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt{1+\frac{b \tan (c+d x)}{a}}}+\frac{\left (a^2 (A+i B) \sqrt{a+b \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{x^m \left (1+\frac{b x}{a}\right )^{5/2}}{1+i x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt{1+\frac{b \tan (c+d x)}{a}}}\\ &=\frac{a^2 (A+i B) F_1\left (1+m;-\frac{5}{2},1;2+m;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d (1+m) \sqrt{1+\frac{b \tan (c+d x)}{a}}}+\frac{a^2 (A-i B) F_1\left (1+m;-\frac{5}{2},1;2+m;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt{a+b \tan (c+d x)}}{2 d (1+m) \sqrt{1+\frac{b \tan (c+d x)}{a}}}\\ \end{align*}

Mathematica [F]  time = 28.353, size = 0, normalized size = 0. \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.49, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( A+B\tan \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \tan \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{2} \tan \left (d x + c\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \tan \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )\right )} \sqrt{b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^2*tan(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*tan(d*x + c)^2 + (B*a^2 + 2*A*a*b)*tan(d*x + c))*sq
rt(b*tan(d*x + c) + a)*tan(d*x + c)^m, x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**m*(a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError