Optimal. Leaf size=667 \[ \frac{2 i a f (e+f x) \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac{2 i a f (e+f x) \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}+\frac{2 i b f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )}+\frac{2 i b f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )}-\frac{i b f (e+f x) \text{PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac{2 a f^2 \text{PolyLog}\left (3,-i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}+\frac{2 a f^2 \text{PolyLog}\left (3,i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}-\frac{2 b f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \left (a^2-b^2\right )}-\frac{2 b f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^3 \left (a^2-b^2\right )}+\frac{b f^2 \text{PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{2 d^3 \left (a^2-b^2\right )}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )}+\frac{b (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{d \left (a^2-b^2\right )}-\frac{2 i a (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{d \left (a^2-b^2\right )} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 1.14299, antiderivative size = 667, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {4533, 4519, 2190, 2531, 2282, 6589, 6742, 4181, 3719} \[ \frac{2 i a f (e+f x) \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac{2 i a f (e+f x) \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}+\frac{2 i b f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )}+\frac{2 i b f (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )}-\frac{i b f (e+f x) \text{PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac{2 a f^2 \text{PolyLog}\left (3,-i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}+\frac{2 a f^2 \text{PolyLog}\left (3,i e^{i (c+d x)}\right )}{d^3 \left (a^2-b^2\right )}-\frac{2 b f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \left (a^2-b^2\right )}-\frac{2 b f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d^3 \left (a^2-b^2\right )}+\frac{b f^2 \text{PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{2 d^3 \left (a^2-b^2\right )}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )}+\frac{b (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{d \left (a^2-b^2\right )}-\frac{2 i a (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{d \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 4533
Rule 4519
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rule 6742
Rule 4181
Rule 3719
Rubi steps
\begin{align*} \int \frac{(e+f x)^2 \sec (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{\int (e+f x)^2 \sec (c+d x) (a-b \sin (c+d x)) \, dx}{a^2-b^2}-\frac{b^2 \int \frac{(e+f x)^2 \cos (c+d x)}{a+b \sin (c+d x)} \, dx}{a^2-b^2}\\ &=\frac{i b (e+f x)^3}{3 \left (a^2-b^2\right ) f}+\frac{\int \left (a (e+f x)^2 \sec (c+d x)-b (e+f x)^2 \tan (c+d x)\right ) \, dx}{a^2-b^2}-\frac{b^2 \int \frac{e^{i (c+d x)} (e+f x)^2}{a-\sqrt{a^2-b^2}-i b e^{i (c+d x)}} \, dx}{a^2-b^2}-\frac{b^2 \int \frac{e^{i (c+d x)} (e+f x)^2}{a+\sqrt{a^2-b^2}-i b e^{i (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac{i b (e+f x)^3}{3 \left (a^2-b^2\right ) f}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac{a \int (e+f x)^2 \sec (c+d x) \, dx}{a^2-b^2}-\frac{b \int (e+f x)^2 \tan (c+d x) \, dx}{a^2-b^2}+\frac{(2 b f) \int (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right ) d}+\frac{(2 b f) \int (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right ) d}\\ &=-\frac{2 i a (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac{2 i b f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 i b f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac{(2 i b) \int \frac{e^{2 i (c+d x)} (e+f x)^2}{1+e^{2 i (c+d x)}} \, dx}{a^2-b^2}-\frac{(2 a f) \int (e+f x) \log \left (1-i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d}+\frac{(2 a f) \int (e+f x) \log \left (1+i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d}-\frac{\left (2 i b f^2\right ) \int \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right ) d^2}-\frac{\left (2 i b f^2\right ) \int \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right ) d^2}\\ &=-\frac{2 i a (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac{b (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d}+\frac{2 i a f (e+f x) \text{Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac{2 i a f (e+f x) \text{Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 i b f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 i b f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}-\frac{(2 b f) \int (e+f x) \log \left (1+e^{2 i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d}-\frac{\left (2 b f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{i b x}{a-\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac{\left (2 b f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{i b x}{a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac{\left (2 i a f^2\right ) \int \text{Li}_2\left (-i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^2}+\frac{\left (2 i a f^2\right ) \int \text{Li}_2\left (i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^2}\\ &=-\frac{2 i a (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac{b (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d}+\frac{2 i a f (e+f x) \text{Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac{2 i a f (e+f x) \text{Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 i b f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 i b f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}-\frac{i b f (e+f x) \text{Li}_2\left (-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac{2 b f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^3}-\frac{2 b f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^3}-\frac{\left (2 a f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac{\left (2 a f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac{\left (i b f^2\right ) \int \text{Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d^2}\\ &=-\frac{2 i a (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac{b (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d}+\frac{2 i a f (e+f x) \text{Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac{2 i a f (e+f x) \text{Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 i b f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 i b f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}-\frac{i b f (e+f x) \text{Li}_2\left (-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac{2 a f^2 \text{Li}_3\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac{2 a f^2 \text{Li}_3\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac{2 b f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^3}-\frac{2 b f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^3}+\frac{\left (b f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{2 \left (a^2-b^2\right ) d^3}\\ &=-\frac{2 i a (e+f x)^2 \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac{b (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac{b (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d}+\frac{2 i a f (e+f x) \text{Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac{2 i a f (e+f x) \text{Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 i b f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac{2 i b f (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}-\frac{i b f (e+f x) \text{Li}_2\left (-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac{2 a f^2 \text{Li}_3\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac{2 a f^2 \text{Li}_3\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac{2 b f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^3}-\frac{2 b f^2 \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^3}+\frac{b f^2 \text{Li}_3\left (-e^{2 i (c+d x)}\right )}{2 \left (a^2-b^2\right ) d^3}\\ \end{align*}
Mathematica [B] time = 5.54789, size = 1561, normalized size = 2.34 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 0.7, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{2}\sec \left ( dx+c \right ) }{a+b\sin \left ( dx+c \right ) }}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [C] time = 3.32858, size = 5157, normalized size = 7.73 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e + f x\right )^{2} \sec{\left (c + d x \right )}}{a + b \sin{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{2} \sec \left (d x + c\right )}{b \sin \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]