\(\int (a+b \sec (c+d x))^m (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)) \, dx\) [1074]

Optimal result

Integrand size = 48, antiderivative size = 48 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} b (a+b) C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^m \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-m} \tan (c+d x)}{d \sqrt {1+\sec (c+d x)}}+(b B-a C) \text {Int}\left ((a+b \sec (c+d x))^{1+m},x\right ) \]

[Out]

b*(a+b)*C*AppellF1(1/2,-1-m,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*(a+b*sec(d*x+c))^m*2^(1/2)*tan(
d*x+c)/d/(((a+b*sec(d*x+c))/(a+b))^m)/(1+sec(d*x+c))^(1/2)+(B*b-C*a)*Unintegrable((a+b*sec(d*x+c))^(1+m),x)

Rubi [N/A]

Not integrable

Time = 0.36 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx \]

[In]

Int[(a + b*Sec[c + d*x])^m*(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2),x]

[Out]

(Sqrt[2]*b*(a + b)*C*AppellF1[1/2, 1/2, -1 - m, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*(a
+ b*Sec[c + d*x])^m*Tan[c + d*x])/(d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^m) + (b*B - a*C)*De
fer[Int][(a + b*Sec[c + d*x])^(1 + m), x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+b \sec (c+d x))^{1+m} \left (b^2 (b B-a C)+b^3 C \sec (c+d x)\right ) \, dx}{b^2} \\ & = (b C) \int \sec (c+d x) (a+b \sec (c+d x))^{1+m} \, dx+(b B-a C) \int (a+b \sec (c+d x))^{1+m} \, dx \\ & = (b B-a C) \int (a+b \sec (c+d x))^{1+m} \, dx-\frac {(b C \tan (c+d x)) \text {Subst}\left (\int \frac {(a+b x)^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \\ & = (b B-a C) \int (a+b \sec (c+d x))^{1+m} \, dx+\frac {\left ((-a-b) b C (a+b \sec (c+d x))^m \left (-\frac {a+b \sec (c+d x)}{-a-b}\right )^{-m} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \\ & = \frac {\sqrt {2} b (a+b) C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^m \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-m} \tan (c+d x)}{d \sqrt {1+\sec (c+d x)}}+(b B-a C) \int (a+b \sec (c+d x))^{1+m} \, dx \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 38.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx \]

[In]

Integrate[(a + b*Sec[c + d*x])^m*(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2),x]

[Out]

Integrate[(a + b*Sec[c + d*x])^m*(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2), x]

Maple [N/A] (verified)

Not integrable

Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00

\[\int \left (a +b \sec \left (d x +c \right )\right )^{m} \left (B a b -C \,a^{2}+b^{2} B \sec \left (d x +c \right )+b^{2} C \sec \left (d x +c \right )^{2}\right )d x\]

[In]

int((a+b*sec(d*x+c))^m*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x)

[Out]

int((a+b*sec(d*x+c))^m*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x)

Fricas [N/A]

Not integrable

Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C b^{2} \sec \left (d x + c\right )^{2} + B b^{2} \sec \left (d x + c\right ) - C a^{2} + B a b\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+b*sec(d*x+c))^m*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)*(b*sec(d*x + c) + a)^m, x)

Sympy [N/A]

Not integrable

Time = 12.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.98 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=- \int C a^{2} \left (a + b \sec {\left (c + d x \right )}\right )^{m}\, dx - \int \left (- B a b \left (a + b \sec {\left (c + d x \right )}\right )^{m}\right )\, dx - \int \left (- B b^{2} \left (a + b \sec {\left (c + d x \right )}\right )^{m} \sec {\left (c + d x \right )}\right )\, dx - \int \left (- C b^{2} \left (a + b \sec {\left (c + d x \right )}\right )^{m} \sec ^{2}{\left (c + d x \right )}\right )\, dx \]

[In]

integrate((a+b*sec(d*x+c))**m*(B*a*b-C*a**2+b**2*B*sec(d*x+c)+b**2*C*sec(d*x+c)**2),x)

[Out]

-Integral(C*a**2*(a + b*sec(c + d*x))**m, x) - Integral(-B*a*b*(a + b*sec(c + d*x))**m, x) - Integral(-B*b**2*
(a + b*sec(c + d*x))**m*sec(c + d*x), x) - Integral(-C*b**2*(a + b*sec(c + d*x))**m*sec(c + d*x)**2, x)

Maxima [N/A]

Not integrable

Time = 22.50 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C b^{2} \sec \left (d x + c\right )^{2} + B b^{2} \sec \left (d x + c\right ) - C a^{2} + B a b\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+b*sec(d*x+c))^m*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)*(b*sec(d*x + c) + a)^m, x)

Giac [N/A]

Not integrable

Time = 0.81 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C b^{2} \sec \left (d x + c\right )^{2} + B b^{2} \sec \left (d x + c\right ) - C a^{2} + B a b\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+b*sec(d*x+c))^m*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)*(b*sec(d*x + c) + a)^m, x)

Mupad [N/A]

Not integrable

Time = 21.58 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12 \[ \int (a+b \sec (c+d x))^m \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^m\,\left (\frac {B\,b^2}{\cos \left (c+d\,x\right )}-C\,a^2+\frac {C\,b^2}{{\cos \left (c+d\,x\right )}^2}+B\,a\,b\right ) \,d x \]

[In]

int((a + b/cos(c + d*x))^m*((B*b^2)/cos(c + d*x) - C*a^2 + (C*b^2)/cos(c + d*x)^2 + B*a*b),x)

[Out]

int((a + b/cos(c + d*x))^m*((B*b^2)/cos(c + d*x) - C*a^2 + (C*b^2)/cos(c + d*x)^2 + B*a*b), x)